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http://www.123eng.com/forum/threads/ge-placement-paper-on-jul-26-2013.157015/ | 1,537,614,640,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158320.19/warc/CC-MAIN-20180922103644-20180922124044-00507.warc.gz | 282,065,004 | 9,849 | # GE Placement-Paper on Jul 26, 2013
Discussion in 'Latest Placement papers' started by krishna, Nov 14, 2013.
1. ### krishnaBanned
Paper pattern
For Electrical – 10 Aptitude and 10 Technical questions.
For Civil – 10 Aptitude and 10 Technical questions.
For Mechanical – 10 Aptitude and 10 Technical questions.
For Electronics – 10 Aptitude questions.
For Computer Science – 10 Aptitude and 10 Technical questions in C.
Duration of test 30 minutes. Each correct answers carries 1mk .
Note: Negative marking of ¼ mark exists for each wrong answer.
Getting through the written test is quite easy. Then we had Technical interview followed by HR interview.
This paper is for Electrical students. Hope aptitude is almost same for all branches, only technical paper is different.
Aptitude paper
1. The time taken by boat to row upstream 2km and downstream 6km is same. Calculate the speed of boat in still water.
a) 4kmph b) 8kmph c) 2kmph
Ans a
2. A, B, C can finish a piece of work in 12 days. A alone can finish the same work in 20 days. B alone can finish the same work in 18 days. In how days C alone can complete the same work.( numbers are not accurate but question is of this type).
3. The fungus in a vessel doubles for every 5minutes.at 3.00 P.M the fungus in the vessel is 120. at what time the fungus rate is 960.
Ans 3.15 P.M
4. The size of the room is 120x80 and the size of each tile is 2.4x1.8. Find minimum number of tiles required to cover the floor (number are not accurate)
5. A triangle inscribed in a circle is given and asked to find out the angle. (A bit tough question among all the questions)
6. AB and CD are two chords of a circle. AB=2*CD. The perpendicular distance ofchords AB and CD from the center of the circle are a, b respectively. Express AB in terms of a, b.
7. A pipe can fill tank in 3 hrs. A hole a bottom of tank can empty the tank in 6 hrs. After ½ hr the hole at the bottom of tank is opened. Find the time taken to fill the tank completely. (Numbers are not accurate. Problem of this model is given)
8. Problem on figures is given. Some sequence of figures (around 5) are given and asked to find the next figure that comes in sequence.
Technical paper
1. In dc motor the motion of conductor is given by
a) Flemming’s right hand rule
b) Fleming’s left hand rule
c) Cork screw rule
Ans b.
2. The resistance of conductor is R. If the conductor length gets doubled and cross section is reduced by half a times then the resistance of conductor is
a).R remains same
b) R gets reduced by 2
c) R increases by 2
d) R increases by 4
Ans d.
3. Two plates of cross section 2mm2 are separated by a distance of 1cm. the capacitance between these plates is
( I don’t remember exact figures, question of this to compute the value of capacitance C = €A/d)
4. No of poles, speed of the alternator, etc are given and asked to find the voltage induced in the rotor ckt of the alternator.
5. In parallel circuit under resonance
a) current is maximum
b) impedance is minimum
c) voltage magnification takes place
d) current magnification takes place
6. 3 resistances of R/2 are connected in star , the magnitude of resistances when connected in delta is
Ans 3R/2
7. Problem on dual networks
Sorry I could not recall remaining questions. For aptitude R.S.Agarwal aptitude
R.S.Agarwal verbal and non verbal reasoning are more than enough. For technical just
prepare basic concepts in networks and machines.
Technical interview
This is the basic elimination round. My interview lasted for about 45 to 50min.
Before attending the interview just have a very good brush up in all the electrical subjects especially power systems protection, machines, power system operation and control, power electronics and drives.
Answer all questions with great confidence . Be cool. GE people are very friendly. Dont give up any question quite easily Try till last they will help you at most of times. Try to answer all the questions. Once you get through this round they will ask you to stay back for next round.
HR interview.
This is very very cool round among all the rounds.95% its not elimination round. Here they test your communication skills and your problem handling capability.
For me it was just 30-40min.For one of my friend it lasted for about 1and ½ hr.
2. ### krishnaBanned
GE PAPER ONH 5TH NOVEMBER AT KHARAGPUR
(1) Written test -----
(a) Aptitude test common for all branch.
(b) Five type of paper in Stress analysis, CFD, Chemical Engg, Electrical & Instrumentation, C- language test.
I attended CFD ----- there were 10 questions in aptitude and 10 in Fluid and heat- - transfer.
Some questions are as follows based on my memory –
(1) What is the probability of getting a two digit number which is multiple of 2 and 3 both?
(2) One insect is trying to climb on a pole and in each second it goes for some height but it slips down for some distance …..in how much time it will get on the top…? (I am missing some numerical values but….i hope u have got my point)
(3) One helicopter start to fly from Delhi at 2 pm with speed 600 kph for the distance 2400km but after start of its journey its on of engine got failed and then pilot reduces its speed 400 kph ….now he become 80min late….Find that at what time engine failed?
(4) Calculate the max area of rectangle that is inscribed in a 14 cm radius circle?
(5) One question form BAR CHART?
(6) One question from work and wages?
(7) Max number that is devisable by three consecutive even numbers?
(8) ---
(9) ---
(10) --- Thermal & Fluid
(1) What is Kirchoff’s law?
(2) What is the adiabatic flame temperature?
(3) What is the analogy of Grasoff’s number in forced convection (Ans –Pr)?
(4) What is the Neumann’s BC & Drichlet’s BC?
(5) What is the work that is obtained when the pressure and volume changes from P1to P2 and V1 to V2?
(6) What is the max to average velocity ratio in turbulent tube flow?
(7) Relation in viscous sub layer ….( Ans U+=y+)?
(8) Raleigh Number’s significance?
(9) ---
(10) ---
After I got selected in written ….I appeared for Tech Interview and then HR interview
Tech interview is basically elimination process and questions were from the Project work .
HR interview is 95% selective process… | 1,565 | 6,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-39 | latest | en | 0.907809 |
https://www.hackmath.net/en/calculator/fraction?input=9%2F12%C3%B76 | 1,628,100,619,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154897.82/warc/CC-MAIN-20210804174229-20210804204229-00599.warc.gz | 758,190,620 | 10,797 | # Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 9/12÷6 = 1/8 = 0.125
Spelled result in words is one eighth.
### How do you solve fractions step by step?
1. Divide: 9/12 : 6 = 9/12 · 1/6 = 9 · 1/12 · 6 = 9/72 = 9 · 1 /9 · 8 = 1/8
Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 6/1 is 1/6) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. In the next intermediate step, , cancel by a common factor of 9 gives 1/8.
In words - nine twelfths divided by six = one eighth.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Equation with x
Solve the following equation: 2x- (8x + 1) - (x + 2) / 5 = 9
• Fruits
Amy bought a basket of fruits 1/5 of them were apples,1/4 were oranges, and the rest were 33 bananas. How many fruits did she buy in all?
• Class 8.A
Three quarters of class 8.A went skiing. Of those who remained at home one third was ill and the remaining six were on math olympic. How many students have class 8.A?
• Expressions
Let k represent an unknown number, express the following expressions: 1. The sum of the number n and two 2. The quotient of the numbers n and nine 3. Twice the number n 4. The difference between nine and the number n 5. Nine less than the number n
• Three 43
Three brothers inherited a cash amount of 62,000, and they divided it among themselves in the ratio of 5:4:1. How much more is the largest share than the smallest share?
• Rainfall
A rectangular garden of 25m in length and width 20m in width fall 4mm of water. Express by a fraction in basic form what part of the 60-hectolitre tank we would fill with this water.
• University bubble
You'll notice that the college up slowly every other high school. In Slovakia/Czech republic a lot of people studying political science, mass media communication, social work, many sorts of management MBA. Calculate how many times more earns clever 25-yea
• LCD 2
The least common denominator of 2/5, 1/2, and 3/4
• Election mathematics
In elections, 12 political parties received this shares of voters: party A 56.2 %party B 8.5 %party C 8.2 %party D 6.2 %party E 6.1 %party F 5.5 %party G 3.2 %party H 2.1 %party I 2 %party J 1 %party K 1 % Calculate what the shares acquired in the parliam
• Equation - inverse
Solve for x: 7: x = 14: 1000
• Cube, cuboid, and sphere
Volumes of a cube and a cuboid are in ratio 3: 2. Volumes of sphere and cuboid are in ratio 1: 3. At what rate are the volumes of cube, cuboid, and sphere?
• MO Z9–I–2 - 2017
In the VODY trapezoid, VO is a longer base and the diagonal intersection K divides the VD line in a 3:2 ratio. The area of the KOV triangle is 13.5 cm2. Find the area of the entire trapezoid.
• Sea water
Seawater contains about 4.3% salt. How many dm3 of distilled water we must pour into 5 dm3 of sea water to get water with 1.8% salt? | 1,670 | 5,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2021-31 | longest | en | 0.845573 |
https://studylib.net/doc/7357996/deductive-logic-is-a-form-of-reasoning-that-assures-us | 1,627,818,179,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154175.76/warc/CC-MAIN-20210801092716-20210801122716-00083.warc.gz | 550,611,758 | 12,339 | # Deductive logic is a form of reasoning that assures us
```Valid and Invalid Logic
A valid argument can be considered as an argument that provides sufficient grounds for
accepting a conclusion as true if the stated premises of the argument are also true.
Valid arguments fall into two categories: deductive and inductive. In the first part of the
course we will examine deductive logic. In the latter part, inductive logic.
Deductive logic is a form of reasoning that assures us that if we start with true premises,
apply the rules of valid deductive logic, then our conclusions will also be true. In other
words, we will not come to false conclusions derived from the truth if we use deductive
logic. There are logical forms that when used will allow us to derive only truths if we
start with truths. These are called valid logical forms. If a false conclusion can derive
from true premises, then the logical form is said to be invalid.
Valid Logical Forms
Rule of Detachment -- RD
The first valid logical form we will use is called the Rule of Detachment.
The RD (Rule of Detachment) basically reaffirms the implication.
Recall the implication states that if proposition P is true then it follows that proposition Q
must also be true.
Implication: If P is true then Q is true.
Example:
Premise: All dogs have hair. (Implication)
Premise: Astro is a dog.
Conclusion: Astro has hair.
If both premises are true, we are guaranteed that the conclusion is true.
In symbolic form:
P Q
P
Q
Example:
Joe is an athlete.
Athletes have athlete’s foot fungis.
Joe has athlete’s foot fungis.
In symbolic form:
P
P Q
Q
Note: The order of the premises is not important.
A list of premises followed by a conclusion is called a logical argument.
Note: There are many logical arguments. RD is a valid logical argument form.
Many other logical argument forms are not valid.
``` | 428 | 1,867 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-31 | latest | en | 0.909602 |
https://www.essayfountain.com/m3a1/ | 1,627,218,708,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151672.96/warc/CC-MAIN-20210725111913-20210725141913-00183.warc.gz | 795,819,323 | 10,956 | # m3a1
One way to use the second proportion of each z score as a p value is when comparing databases to see if a data point is more likely in one than the other.
To see how this works, generate a second normal distribution and compare the p values of the same data points as above (33, 35, 43, 22, 17) when µ = 25 and σ = 5 in Figure 2 to the p values they take on when µ = 25 and σ = 8 (same mu, different sigma). Generate the new set of z scores for the data points and compare the corresponding probabilities. Use a blank bell curve to plot µ, calculate the new values for the data at each whole number z score and the location of each data point (33, 35, 43, 22, 17) so you can see how location varies depending on sigma (σ).
For example, when µ = 25 and σ = 5 and a data point of 30, z = 1.00 and p = .160. However, when µ = 25 and σ = 8, the same data point of 30 has z = 0.63 and p = .260. Because z = (30-25)/8 = 0.63, the associated proportions/probabilities from z-score table are .2357 and .2643, respectively. Rounding off the second proportion to three decimal places, p = .260.
Now generate a third distribution, but this time reduce the scatter of data: µ = 25 and σ = 2. Using the same data points from above (33, 35, 43, 22, 17), generate z scores for and compare the corresponding probabilities.
For example, when µ = 25 and σ = 2, the data point of 30 has z = 2.50 and p = .010. Because z = (30-25)/2 = 2.50, the associated proportions/probabilities from z-score table are .4938 and .0062, respectively. Rounding off the second proportion to three decimal places, p = .010.
Use the data in the table below for the following 4 problems. For each, use the stated µ and σ to calculate the z score, get the p value (3 decimal places), and write out a formal APA statement of conclusion for each child. When you are finished with that, then answer the following question for each pair of parameters: Which child or children, if any, appeared to come from a significantly different population than the one used in the null hypothesis? What happens to the “significance” of each child’s data as the data are progressively more dispersed?
NOTE: Problems #1 and #2 are PRACTICE PROBLEMS with the answers available via the link beside them.
Problems #3 and #4 are GRADED PROBLEMS
Please submit all 4 problems using the Module 3 Assignment 1 Template found in the Doc Sharing area.
Problem 1. µ = 100 seconds and σ = 10 (practice problem – link to answer)
Problem 2. µ = 100 seconds and σ = 20 (practice problem – link to answer)
Problem 3. µ = 100 seconds and σ = 30 (graded problem)
Problem 4. µ = 100 seconds and σ = 40 (graded problem)
Given the research scenario, data points, set of population parameters and alpha set at p = .05, is the student able to generate the correct:
Don't use plagiarized sources. Get Your Custom Essay on
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Pair of hypotheses for each data point
A z statistic and p value for each data point
Decision about the null hypothesis for each data point
APA-formatted statement of results for each data point
Please use Module 3_Assignment_1_Template located in Shared Documents. Save the template so that you can make changes.
Please report z scores to two decimals and p values to three decimals. If the p value is less than .001, report it as p < .001.
Submit your assignment by the due date assigned to the Submissions Area, listed as, LastName_FirstInitial_M3A1.
Assignment 1 Grading CriteriaMaximum PointsPair of hypotheses for each data point20A z statistic and p value for each data point20Decision about the null hypothesis for each data point20APA-formatted statement of results for each data point20Total:80
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The price is based on these factors: | 1,314 | 5,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2021-31 | longest | en | 0.894672 |
https://physics.stackexchange.com/questions/10870/calculating-position-in-space-assuming-general-relativity | 1,606,414,975,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188899.42/warc/CC-MAIN-20201126171830-20201126201830-00130.warc.gz | 429,904,359 | 33,624 | # Calculating position in space assuming general relativity
Suppose two pointed masses are given in space. Suppose further that one of the masses has a given velocity at (local) time 0. Is there a way to compute its position in a future time?
Neglecting general relativity, I will simply compute an integral, but with general relativity, we see that the metric of the space changes with time, so I need to compute an integral with respect to a measure that changes along time.
Can this be done? If so, how?
Thank you!
$$\frac{\mathrm{d}^2x^\lambda}{\mathrm{d}t^2} + \Gamma^{\lambda}_{\mu\nu}\frac{\mathrm{d}x^\mu}{\mathrm{d}t}\frac{\mathrm{d}x^\nu}{\mathrm{d}t} = 0$$
This is essentially the general relativistic equivalent of Newton's second law: it's a differential equation that governs how a test particle's position changes in time. The connection coefficients $\Gamma^{\lambda}_{\mu\nu}$ (also called Christoffel symbols) can be calculated from the metric. So if the metric is known, even if it's time-dependent, you can calculate the connection coefficients in your desired reference frame, plug them in, and find a solution to the geodesic equation that tells you how the particle will move. There may not be an analytic solution, but in all but the most extreme cases you can either solve the equation numerically, or make some approximation that might make it analytically solvable.
$$G^{\mu\nu} = 8\pi T^{\mu\nu}$$ | 360 | 1,430 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-50 | latest | en | 0.904154 |
http://mathhelpforum.com/pre-calculus/91392-circle-geomtery-help-print.html | 1,495,917,838,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609054.55/warc/CC-MAIN-20170527191102-20170527211102-00157.warc.gz | 288,474,430 | 3,104 | # circle geomtery help
• Jun 1st 2009, 04:44 AM
Tweety
circle geomtery help
The circle C has centre (5, 2) and passes through the point (7, 3).
(a) Find the length of the diameter of C.
$2x \sqrt{4+1} = 2\sqrt{5}$
(b) Find an equation for C.
$(x-5)^2 + (y-2)^2 = 5$
(c) Show that the line y = 2x − 3 is a tangent to C and find the coordinates
of the point of contact.
Need help with question 'c'.
I know that the angle between a tangent and radius is 90 , but how would I show that the product of their gradients is -1, if I dont know the point of contact?
Also how do I find the point were it touches the circle?
• Jun 1st 2009, 05:14 AM
mr fantastic
Quote:
Originally Posted by Tweety
The circle C has centre (5, 2) and passes through the point (7, 3).
(a) Find the length of the diameter of C.
$2x \sqrt{4+1} = 2\sqrt{5}$
(b) Find an equation for C.
$(x-5)^2 + (y-2)^2 = 5$
(c) Show that the line y = 2x − 3 is a tangent to C and find the coordinates
of the point of contact.
Need help with question 'c'.
I know that the angle between a tangent and radius is 90 , but how would I show that the product of their gradients is -1, if I dont know the point of contact?
Also how do I find the point were it touches the circle?
Here's one approach out of many possible approaches: Show that the two equations
$y = 2x - 3$
$(x-5)^2 + (y-2)^2 = 5$
have only one solution.
• Jun 1st 2009, 05:23 AM
Tweety
Quote:
Originally Posted by mr fantastic
Here's one approach out of many possible approaches: Show that the two equations
$y = 2x - 3$
$(x-5)^2 + (y-2)^2 = 5$
have only one solution.
So if I solved this simutaneously, wouldn't that mean that these to equations intersect at that one point, instead of touching?
• Jun 1st 2009, 05:24 AM
mr fantastic
Quote:
Originally Posted by Tweety
So if I solved this simutaneously, wouldn't that mean that these to equations intersect at that one point, instead of touching?
If a line intersects a circle at one point then it is tangent to the circle at that point. | 610 | 2,026 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-22 | longest | en | 0.940744 |
https://math.answers.com/Q/How_do_i_convert_13_liters_to_millimeters | 1,701,240,547,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100056.38/warc/CC-MAIN-20231129041834-20231129071834-00590.warc.gz | 466,116,799 | 46,141 | 0
# How do i convert 13 liters to millimeters?
Updated: 9/25/2023
Wiki User
7y ago
Best Answer
You do not.
A millimetre is a measure of distance in 1-dimensional space, whereas a litre is a measure of volume in 3-dimensional space. The two measure different things and, according to the basic rules of dimensional analysis, conversion from one to the other is not valid.
Wiki User
7y ago
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Related questions
### What is 0.67 liters equal to millimeters?
liters does not convert to millimeters
### How do you convert millimeters in to liters?
You don't. Millimeters is a length, liters is a volume.
### How many liters in 1000000 millimeters?
Millimeters are distance, liters are volume. You can't convert from one to the other.
### What is the result of converting 340 millimeters into liters?
You can not convert millimeters (a measure of length) into liters (a measure of volume).
### How man millimeters are in 3 liters?
You can't convert that.
### How do you convert millimeters into liters?
You can't convert millimeters into liters - the first is a measure of length, and the second is a measure of volume.
### How many millimeters to equal one liter?
Meters and liters are different units of measure that do not convert, so the question is unanswerable.
### Why does meters times centimeters times millimeters equal Liters?
If you have three measurements of length (meters, centimeters, millimeters) and you multiply them together, you will end up with a measure of volume (liters). Of course, if you were to actually do this, you would want to convert all of the lengths into a common unit (convert centimeters and millimeters to meters or convert meters and centimeters to millimeters).
### How many liters are in one hundred millimeters?
0.10 LitersAnswer 2No. Sorry!Liters measury capacity, millimeters length.You cannot convert inches into gallons
### Write 340 millimeters in liters?
You can't. Millimeters is a measurement of length or distance. Liters is a measurement of volume or fluid volume. You can't convert from one to the other.
### How much liters is 750millmeters times 2?
You can't convert millimeters (a unit of length) to liters (a unit of volume).
### 3600 millimetres how much in letres?
Millimeters is a measure of length; liters is a measure of volume. You can't convert between the two.Millimeters is a measure of length; liters is a measure of volume. You can't convert between the two.Millimeters is a measure of length; liters is a measure of volume. You can't convert between the two.Millimeters is a measure of length; liters is a measure of volume. You can't convert between the two. | 654 | 2,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2023-50 | latest | en | 0.902567 |
http://openstudy.com/updates/5102fe0ce4b0ad57a562bb98 | 1,448,845,866,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398460519.28/warc/CC-MAIN-20151124205420-00183-ip-10-71-132-137.ec2.internal.warc.gz | 176,008,406 | 9,736 | ## deer126 2 years ago What is the solution to the proportion 3/m = 12/15 A. m = 2.4 B. m = 3.75 C. m = 4.5 D. m = 4.75
3*15=12*m
divide both sides by 12
3. deer126
this isnt for homeowrk its a question somone asked me and could someone please give me an answer? | 97 | 266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2015-48 | longest | en | 0.943402 |
https://avatest.org/2023/02/20/topology-daixie-math625-homotopies-of-maps-and-spaces/ | 1,726,586,903,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00838.warc.gz | 96,345,116 | 20,903 | Posted on Categories:Topology, 拓扑学, 数学代写
# 数学代写|拓扑学代写TOPOLOGY代考|MATH625 Homotopies of Maps and Spaces
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## 数学代写|拓扑学代写TOPOLOGY代考|Homotopies of Maps and Spaces
In the last chapter, we discussed homotopies of maps between $[0,1]$ and a topological space $X$. We can generalize this to maps between two arbitrary topological spaces $X$ and $Y$. We say that two maps $f, g: X \rightarrow Y$ are homotopic if we can continuously deform one into the other. We can express this notion more formally, in a similar manner to how we defined homotopies of maps between $[0,1]$ and $X$ :
Definition $9.1$ Suppose $X$ and $Y$ are two topological spaces, and $f, g: X \rightarrow Y$ are two continuous maps. Then a homotopy between $f$ and $g$ is a continuous map $H:[0,1] \times X \rightarrow Y$ satisfying the following properties:
$H(0, x)=f(x)$ for all $x \in X$,
$H(1, x)=g(x)$ for all $x \in X$.
If there is a homotopy between $f$ and $g$, then we say that $f$ and $g$ are homotopic. We write $f \sim g$ when $f$ and $g$ are homotopic.
Example Let $X$ be the interval $[0,1]$, and let $Y$ be the single point 0 . Then $X$ and $Y$ are homotopy equivalent. To see this, we need to define maps $f: X \rightarrow Y$ and $g: Y \rightarrow X$. We define $f(x)=0$ for all $x \in X$, and $g(0)=0$ (for the only point 0 in $Y$ ). Then $(g \circ f)(x)=0$ for all $x \in X$. To see that this is homotopic to the identity map $h(x)=x$, we need to construct a homotopy $H:[0,1] \times X \rightarrow$ $X$ between them. Our homotopy will be defined by $H(s, x)=s x$. Then we have $H(0, x)=0=(g \circ f)(x)$, and $H(1, x)=x=h(x)$. So this is a homotopy between $(g \circ f)(x)$ and the identity function on $X$.
Now we have to show that $f \circ g$ is homotopic to the identity function on $Y$. But this is easier, because both functions are the same function that sends the only point in $Y$ to itself. The homotopy $J$ between them is defined by $J(s, x)=0$.
## 数学代写|拓扑学代写TOPOLOGY代考|Computing the Fundamental Group of a Circle
So far, it is not yet clear whether the fundamental group is an interesting invariantthat is, does it ever distinguish spaces? Are there any spaces at all with nontrivial fundamental group? In case the name didn’t give it away, here’s a spoiler: yes! We will show that the circle has nontrivial fundamental group.
Before we do this, let us see intuitively why we ought to believe that the circle has nontrivial fundamental group. Suppose our circle is the set $\mathbb{S}^1=\left{(x, y): x^2+y^2=\right.$ 1} $\subset \mathbb{R}^2$. Let us pick as our basepoint the point $p=(1,0)$. Let us consider the loop $\alpha$ on the circle; $\alpha$ is a map $\alpha:[0,1] \rightarrow \mathbb{S}^1$ so that $\alpha(0)=\alpha(1)=p$, and we will choose it to be the loop $\alpha(t)=(\cos 2 \pi t, \sin 2 \pi t)$, so it is a loop of constant speed that goes around the circle once in the counterclockwise direction.
This loop appears not to be homotopic to the trivial loop: it seems that this loop goes around once, and the trivial loop goes around 0 times. But how can we prove that, by doing some clever homotopy, we can’t shrink it down to a point?
There are several ways of proving this, and the different techniques highlight different properties of fundamental groups. In this section, we’ll see a way to do it using a first example of covering spaces, while in the next chapter we’ll see a different proof. We won’t talk more about covering spaces in general in this book, but the procedure we employ here to compute fundamental groups is very general and can be used to compute the fundamental group of any reasonably nice space.
The outline of the proof is the following: We want to start with a loop on the circle, lift it up to some other space, and see what the lifted version of the loop looks like.
## 数学代写|拓扑学代写TOPOLOGY代考|Homotopies of Maps and Spaces
$H(0, x)=f(x)$ 对全部 $x \in X ，$
$H(1, x)=g(x)$ 对全部 $x \in X$.
$H:[0,1] \times X \rightarrow X$ 它们之间。我们的同伦定义为 $H(s, x)=s x$. 然后我们有 $H(0, x)=0=(g \circ f)(x)$ 和 $H(1, x)=x=h(x)$. 所以这是之间的同伦 $(g \circ f)(x)$ 和身份函数 $X$.
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 | 1,815 | 4,924 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-38 | latest | en | 0.704075 |
https://buycheapcollegeessays.com/2021/10/12/first-review-the-purpose-of-500mb-charts-and-how-to/ | 1,643,273,285,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00151.warc.gz | 207,041,820 | 13,988 | # First, review the purpose of 500mb charts and how to
First, review the purpose of 500mb charts and how to read wind barbs.
1 (+1): Analyze the 500mb chart above. If a low-pressure system is located in Texas (find the yellow “X”), what direction will the low-pressure system most likely move: East, North, South, or West?
2 (+1): Over California, examine the movement of air using the 500mb chart.
What direction will the low-pressure system (marked by the yellow “X” in the image) most likely move: Northeast, Southeast, Southwest, or Northwest?
3 (+1): Now let us connect Module 5’s work on air masses to this week’s work. Identify the air mass (marked by the yellow “X” in the image) that will be moving over California. Hint: go back to the last module for this and connect the topics
4 (+1): Using the characteristics of the air mass you indicated in the question above, make a general weather prediction for California (rainy or clear, hot or cold will be good enough, no more detail is needed).
### Part 2 Surface Map Analysis
This is a surface weather map:
See all the symbols? They tell a story about the weather at each location. First, we have to decide what they mean.
Examine the following table (all temperatures in Celsius)
Station 1
Station 2
Station 3
4 (+1): The temperature at station 1 is __________
5 (+1): The dew point at station 2 is __________
6 (+1): The air pressure at station 1 is __________ Hint, it is not 986, go watch the video lesson
7 (+1): The air pressure at station 3 is __________ Hint, it is not 002, go watch the video lesson
8 (+1): What is the wind direction at station 3? Hint: remember, we name winds based on where they come from
9 (+1): Think back to relative humidity and the relationship between temperature and dew point. Which station has the highest relative humidity?
10 (+1): The wind speed at station 1 is __________
Examine the following image
11 (+1): Identify which location is closest to the center of the low-pressure system AND indicate the pressure reading. Hint: the location with the lowest pressure will be closest to the center of the low-pressure system
12 (+1): Identify which location is closest to the cold front and describe your reasoning in one or two sentences.
13 (+1): Identify which location is closest to the warm front and describe your reasoning in one or two sentences.
14 (+1): You are a forecaster and handed the map above. Which of the four locations would most likely experience strong thunderstorms? Identify the location and describe your reasoning in one or two sentences. Hint, think back to cold and warm fronts.
### Part 3 Hurricane Forecasting
In hurricane season, Floridians are very familiar with these types of images often called “spaghetti models.”
15 (+1): “Spaghetti models” represent a type of forecasting called __________________.
16 (+1): Describe how these types of forecasts are made and describe their importance in weather forecasting, especially hurricanes.
### Part 4: Optical phenomena
17 (+1): Most tornadoes in the United States move towards the east due to the movement of mid-latitude cyclones and prevailing wind patterns. Assume it is nearly 4:30 pm (Note, the time is important, think about where the sun would be in the sky) and you walk outside and see the following:
Using your knowledge about how rainbows form and the movement of tornadoes (usually move towards the east), identify if the tornado is heading away from you, towards you, or can you not determine the direction of the tornado based on this image and describe your reasoning in one or two sentences?
18 (+1): Provide a general description of rainbow formation.
19 (+1): Examine the image below:
This image shows the optical phenomenon of (a/an) _________________.
20 (+1): Review how the optical phenomena shown in #19 forms. Identify which season this image was most likely taken (choose: Spring, Summer, Fall or Winter) and describe your reasoning in one or two sentences. | 888 | 3,996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2022-05 | latest | en | 0.912453 |
https://www.meritnation.com/ask-answer/question/find-two-consecutive-positive-integers-sum-of-whose-squares/quadratic-equations/3217141 | 1,511,581,300,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809392.94/warc/CC-MAIN-20171125032456-20171125052456-00667.warc.gz | 825,043,186 | 18,709 | Select Board & Class
Subject: Math , asked on 1/11/12
# find two consecutive positive integers sum of whose squares is 365
Certified by MeritNation Expert
Let the consecutive positive integers be x and x + 1.
Either x + 14 = 0 or x − 13 = 0, i.e., x = 14 or x = 13
Since the integers are positive, x can only be 13.
x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers will be 13 and 14.
This conversation is already closed by Expert
• 16
Let The Numbers Be x And x+1
Squares x2 And (x+1)2
Sum = 365
Hence
x2+(x+1)2 = 365
x 2 +x2+1+2x = 365
2x2+2x+1 = 365
2x2+2x-364 = 0
x2+x-182 = 0
-b+Square Root (b2-4ac) / 2a And -b-Square Root (b2-4ac) / 2a
-1+Square Root (1+728) / 2 And -1-Square Root (1+728) / 2
-1+27/2 And -1-27/2
13 And -14
As Positive Integers So 13
Other Integer = 13+1 = 14
• 10
Hope it helps.....
• 2
182 and 183
• -3
Syllabus | 352 | 887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-47 | latest | en | 0.765441 |
https://www.brainfit.world/answers-february-2021/ | 1,716,760,375,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058973.42/warc/CC-MAIN-20240526200821-20240526230821-00100.warc.gz | 587,210,901 | 42,708 | Select Page
February 2021
## Keep up those challenges every day!
Here are the answers to the February 2021 Brain Practice
### The Ball Challenge
Tennis Ball = 2
Explanation
Equation 1: 10 + 10 + 10 = 30 (Rugby Ball = 10)
Equation 2: (5 x 5) + 10 = 35 (White Baseball = 5)
Equation 3: (4 x 4) + 10 = 26 (Green Baseball = 4)
Let us assume that the value of ‘Tennis Ball’ be ‘T’;
Equation 4:
5 + (4 x T) + 10 = 23
=> 4T + 15 = 23
=> 4T = 23 – 15
=> 4T = 8
=> T = 8/4
=> T = 2
Therefore; the value of ‘Tennis Ball’ = 2.
1. STRAIN
2. RESTRAIN
3. RESTRICT
4. DISTRICT
5. DISTRUST
6. DISTRESS
7. DISTRACT
8. REACT
9. REACH
10. PREACH
## Can you solve these?
• A girl who was just learning to drive went down a one-way street in the wrong direction, but didn’t break the law. How is that possible?
She was walking!
• How can you share 5 apples with 5 people and still have one left in the basket?
One person carries the apple in the basket.
•
• There are several suggested answers to this one. Which is correct, do you think?
– Grammar rules, the first sentence is in the present tense. So when it says he has 6 eggs, he has 6 eggs. The other sentences are in the past tense. (This is the officially correct answer ….)
However:
– The exact answer is 2 because if two eggs were broken from 6 eggs there will be 4 more and then she cooked and ate two extra eggs. 6-4=2 so there will be 2 more left
– There are four left… he broke two, cooked the same two, and ate the same two.
You decide for yourself! | 480 | 1,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-22 | latest | en | 0.956574 |
https://samacheer-kalvi.com/samacheer-kalvi-11th-business-maths-guide-chapter-5-ex-5-9/ | 1,716,154,032,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057922.86/warc/CC-MAIN-20240519193629-20240519223629-00014.warc.gz | 454,759,768 | 7,917 | Students can download 11th Business Maths Chapter 5 Differential Calculus Ex 5.9 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.
Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 5 Differential Calculus Ex 5.9
Samacheer Kalvi 11th Business Maths Differential Calculus Ex 5.9 Text Book Back Questions and Answers
Question 1.
Find y2 for the following functions:
(i) y = e3x+2
(ii) y = log x + ax
(iii) x = a cosθ, y = a sinθ
Solution:
(i) y = e3x+2
(ii) y = log x + ax
(iii) x = a cosθ, y = a sinθ
$$\frac{d x}{d \theta}$$ = a(-sinθ) = -a sinθ …….. (i)
$$\frac{d y}{d \theta}$$ = a(cosθ)
Question 2.
If y = 500e7x + 600e-7x, then show that y2 – 49y = 0.
Solution:
y = 500e7x + 600e-7x
(or) y2 – 49y = 0
Question 3.
If y = 2 + log x, then show that xy2 + y1 = 0.
Solution:
y = 2 + log x
Question 4.
If = a cos mx + b sin mx, then show that y2 + m2y = 0.
Solution:
y = a cos mx + b sin mx
y1 = a $$\frac{d}{d x}$$ (cos mx) + b $$\frac{d}{d x}$$ (sin mx)
[∵ $$\frac{d}{d x}$$ (sin mx) = cos mx $$\frac{d}{d x}$$ (mx) = (cos mx) . m]
= a(-sin mx) . m + b(cos mx) . m
= -am sin mx + bm cos mx
y2 = -am(cos mx) . m + bm(-sin mx) . m
= -am2 cos mx – bm2 sin mx
= -m2 [a cos mx + b sin mx]
= -m2y
∴ y2 + m2y = 0
Question 5.
If y = $$\left(x+\sqrt{1+x^{2}}\right)^{m}$$, then show that (1 + x2) y2 + xy1 – m2y = 0
Solution:
$$y_{1}=\frac{m y}{\sqrt{1+x^{2}}}$$
Squaring both sides we get,
$$y_{1}^{2}=\frac{m^{2} y^{2}}{\left(1+x^{2}\right)}$$
(1 + x2) ($$y_{1}^{2}$$) = m2y2
Differentiating with respect to x, we get
(1 + x2) . 2(y1) (y2) + (y1)2 (2x) = 2m2yy1
Dividing both sides by 2y1 we get,
(1 + x2) y2 + xy1 = m2y
⇒ (1 + x2) y2 + xy1 – m2y = 0
Question 6.
If y = sin(log x), then show that x2y2 + xy1 + y = 0.
Solution:
y = sin(log x)
y1 = cos(log x) $$\frac{d}{d x}$$ (log x)
y1 = cos(log x) . $$\frac{1}{x}$$
∴ xy1 = cos(log x)
Differentiating both sides with respect to x, we get
xy2 + y1(1) = -sin(log x) . $$\frac{1}{x}$$
⇒ x[xy2 + y1] = -sin(log x)
⇒ x2y2 + xy1 = -y
⇒ x2y2 + xy1 + y = 0 | 937 | 2,197 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-22 | latest | en | 0.574911 |
http://mathhelpforum.com/pre-calculus/19153-simple-absolute-values.html | 1,480,775,705,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540932.10/warc/CC-MAIN-20161202170900-00464-ip-10-31-129-80.ec2.internal.warc.gz | 177,785,901 | 10,330 | # Thread: simple absolute values
1. ## simple absolute values
how do you rewrite |3x+4|-|8-16| without absolute values? thanks!
2. You cannot remove absolute values unless you know the specific value. Thus, $|3x+4|$ cannot be simplified.
3. Originally Posted by asnxbbyx113
how do you rewrite |3x+4|-|8-16| without absolute values? thanks!
first, let's find the turning point (i think that's what it's called). where does 3x + 4 become negative? for $x \in (- \infty, -4/3)$, of course. ok. now we have to split the function in 2.
$|3x + 4| - |8 - 16| = |3x + 4|-8$
now because of the absolute values, we need to account for when the function is negative and when it is positive. we already know where it is negative.
so, if $f(x) = |3x + 4| - 8$
then, $f(x) = \left \{ \begin {array}{cc} 3x + 4 - 8, & \mbox { if } x \ge - \frac 43 \\ -(3x + 4) - 8, & \mbox { if } x < - \frac 43 \end{array}\right.$
$\Rightarrow f(x) = \left \{ \begin {array}{cc} 3x -4, & \mbox { if } x \ge - \frac 43 \\ -3x -12, & \mbox { if } x < - \frac 43 \end{array}\right.$ | 374 | 1,058 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2016-50 | longest | en | 0.602406 |
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Contemporary's GED Mathematics
# GED Score Conversion Chart
To see if you are ready for the actual GED Mathematics Test, use the conversion chart below to determine your GED score. Count the number of questions you answered correctly on the GED Practice Test and find that number in the top row below. Read down to find your GED standard score in the bottom row.
Correct Answers 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 Standard Score 800 710 640 580 540 520 500 480 470 460 450 440 430 410 400 390 380 370 360 340 320 290 250 220 200
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Use the following chart to determine the areas you need to review before taking the test. Make note of any items you answered incorrectly and find them in the chart. Pay particular attention to areas where you missed half or more of the questions. For those questions that you missed, review the indicated skill pages in Contemporary's GED Mathematics.
Content Area Item Number NUMBER SENSE AND OPERATIONS Fractions(see pages 103–136) 4, 18 Ratio and Proportion(see pages 137–148) 10 Percent(see pages 149–182) 13 Word Problems(see pages 51–74) 3, 9 MEASUREMENT AND GEOMETRY Perimeter, Circumference, Area, and Volume(see pages 235–258) 11, 12 Similarity, Angles, Triangles, and Pythagorean Relationship(see pages 259–275) 1, 8, 14, 25 DATA ANALYSIS, STATISTICS, PROBABILITY Graphs and Tables(see pages 197–211) 6, 20, 21 Statistics(see pages 217–219) 7, 19 Probability(see pages 212–216) 2, 15 ALGEBRA, FUNCTIONS, AND PATTERNS Writing and Solving Algebra Equations(see pages 281–322) 5, 16, 17, 23, 24 Coordinate Plane, Graphing Equations, Slope and Intercept(see pages 323–335) 22 | 596 | 2,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-10 | latest | en | 0.692605 |
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posted by .
Sketch the graph of a function that is continuous at x=5 but not differentiable at x=5.
• Calculus -
horizontal till x=5, then fromx=5 upward sloping line.
## Similar Questions
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Sketch a graph of a function f(x) that is differentiable and that satisfies the following conditions. c) f'(-3) = 0 and f'(1) = 0 Please show me, step by step, how to sketch the problem!
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sketch graph of a function f that is differentiable and that satisfies the following
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sketch graph of a function f that is differentiable and that satisfies the following conditions:(1)f'(x)>0, when x<-5 (2)f'(x)<0, when -5<x<1 and when x>1 (3)f'(-5) =0and f'(1)=0 (4)f(-5)=6and f(1)=2.
4. ### grade 12 AP calculus
Sketch a possible graph of a continuous function f that has domain [0,3], where f(0)=-3 and the graph of y=f'(x) is shown. Thank you so much for the help. It is making the concepts far more clear.
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Consider the function f(x)= x −1≤ x< 0 f(x)=tan(x) 0≤ x ≤ π/4 a. Draw a neat sketch of this function. b. Is f(x) continuous at x = 0?
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More Similar Questions | 523 | 1,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-34 | latest | en | 0.869348 |
https://hchapman.org/classes/f2019-301/cw/counting_2/ | 1,702,027,731,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100739.50/warc/CC-MAIN-20231208081124-20231208111124-00082.warc.gz | 356,777,873 | 3,863 | # Harrison Chapman
## Math 301: Counting problems and the pigeonhole principle
Class date: Friday, September 13, 2019
1. How many different strings can you make by rearranging the letters of MATHEMATICS? For example, one is THEMMAASTIC.
There are 11 letters in MATHEMATICS. So, if all letters were distinct, there would be $$11!$$ words we could make. But, there are two M’s, T’s, and A’s that are each indistinguishable. Each of those three letters can appear in any of $$2!$$ ways each. So, the answer is; $\frac{11!}{(2!)^3}.$
2. On an $$n \times n$$ grid, how many paths connect the lower-left corner to the upper-right corner, if only unit-length up steps and unit-length right steps are allowed? Here are examples for sizes $$n = 1,2,3$$:
Since we can only walk up or right, notice that every path must consist of $$2n$$ steps. Furthermore, we can only make it to the top right corner if we take the same number of up-steps as right-steps (that is, $$n$$ of them). So, the answer is $\binom {2n}n.$
This is a neat problem, since it’s not obvious from the get-go that we can solve it with binomial coefficients!
3. If 11 numbers are chosen from between 1 and 100, show that two of them have difference less than 10.
If we make buckets be the mutually exclusive sets $$\{x \in \mathbb R : 10(k-1) < x \le 10k\}$$ for $$k \in \mathbb N, 1 \le k \le 10$$, we have partitioned the numbers $$(0,100]$$ into 10 mutually exclusive buckets. Notice that if we take any two numbers $$x,y$$ where $$x > y$$ in the same bucket, their difference is $$x-y < 10k - 10(k-1) = 10.$$
If we choose 11 numbers from 1–100, we can then put each number in its appropriate bucket. There are 11 numbers and 10 buckets, so the pigeonhole principle says that at least one bucket has two numbers; that is, at least two numbers have difference less than 10.
One last thing—notice the way we’ve proved this works whether we’re picking integers or real numbers. | 538 | 1,945 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-50 | latest | en | 0.903015 |
http://openstudy.com/updates/557d8c39e4b07028ea60144a | 1,477,623,688,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721555.36/warc/CC-MAIN-20161020183841-00467-ip-10-171-6-4.ec2.internal.warc.gz | 187,246,890 | 92,483 | ## anonymous one year ago will medal question below
1. anonymous
|dw:1434291285071:dw|
2. anonymous
@butterflydreamer
3. butterflydreamer
is the last number a 7 or a 1?
4. anonymous
7
5. CGGURUMANJUNATH
|dw:1434290279530:dw|
6. CGGURUMANJUNATH
|dw:1434290331067:dw|
7. CGGURUMANJUNATH
|dw:1434290365941:dw|
8. butterflydreamer
okay :) You have to use the quadratic formula for this: |dw:1434291441022:dw| From your equation, can you tell me what's a, b and c? Then plug it into the formula :)!
9. butterflydreamer
oops i was too slow LOL
10. anonymous
a=2 b=-3 c=-7
11. CGGURUMANJUNATH
|dw:1434290411013:dw|
12. CGGURUMANJUNATH
|dw:1434290471648:dw|
13. anonymous
in the formula why is b squared 3 and not -3 @butterflydreamer
14. butterflydreamer
b squared? We know b = -3, so b^2 = -3 * -3 = 9 (multiplying two negatives make a positive)
15. anonymous
sorry i meant b
16. anonymous
|dw:1434291775453:dw|
17. anonymous
i thought that would be -3 ?
18. butterflydreamer
ohhhh! okay :) Looking at our formula we have: |dw:1434291773407:dw|
19. anonymous
im talking about the b squared in the formula @butterflydreamer
20. anonymous
it should be -3 cos b is -3
21. butterflydreamer
are you talking about in @CGGURUMANJUNATH 's working? |dw:1434291913118:dw|
22. anonymous
yes
23. CGGURUMANJUNATH
|dw:1434290833150:dw|
24. butterflydreamer
oh yeah, it should be (-3)^2 :) But -3 squared and 3 squared both give you 9... just a typo probably
25. CGGURUMANJUNATH
|dw:1434290844945:dw|
26. CGGURUMANJUNATH
|dw:1434290878100:dw|
27. CGGURUMANJUNATH
|dw:1434290905645:dw|
28. anonymous
ok
29. anonymous
|dw:1434292314596:dw|
30. anonymous
@butterflydreamer
31. butterflydreamer
remember that the entire this is divided by 4 so we actually have: |dw:1434292371847:dw|
32. anonymous
x=-1.3 to 1 dp or x=2.8
33. CGGURUMANJUNATH
|dw:1434291344388:dw||dw:1434291383387:dw|
34. butterflydreamer
that's corrreect :)!
35. anonymous
thanks both @butterflydreamer and @CGGURUMANJUNATH
36. butterflydreamer
you're welcome :))!
37. CGGURUMANJUNATH
|dw:1434291441704:dw| | 751 | 2,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2016-44 | longest | en | 0.616878 |
https://www.coursehero.com/file/p261d/To-avoid-coefficient-growth-altogether-we-again-use-the-idea-of-Algorithm-M/ | 1,540,299,770,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516135.92/warc/CC-MAIN-20181023111223-20181023132723-00117.warc.gz | 905,714,307 | 43,414 | Assessment 2 knuth book
# To avoid coefficient growth altogether we again use
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To avoid coefficient growth altogether, we again use the idea of Algorithm M. Instead of solving the given problem directly, we solve one or more related problems in Zp[x~, • • • , x~_~], and then reconstruct the desired result. Let Fi' and F ' 2 be viewed as polynomials in g[x~, • • • X~-l], where 9 denotes the polynomial domain Zp[xv]. Although F~' and F( are primitive as elements of Zp[xl, • .. , x~], they need not be primitive as elements of 9[x~, • • • , x~_~]. Let c~, c2, c, F1, F2 , G, H1, H2 , f~ , f~ , g, h~ , h~ , O, G, lP~ , F~ , H~ , and/t~ be defined as in Algorithm M. Now, however, all of the lower ease symbols denote polynomials in a = Z~[x~], while the upper ease symbols denote polynomials in ~l[xl, • • • , X~_l]. For any fixed b ~ Zp, let g~ denote the field of polynomials in ~l modulo the ir- reducible polynomial (x~ -- b). Since for polynomials f ~ ~, the quantity f(x~) mod (xv -- b) is equal tof(b) ~ Z~, we see that ~ is precisely Z~. Algorithm P is essentially identical to Algorithm M, except that v is replaced by v - 1, Z is replaced by 9, the prime p ~ Z is replaced by the irreducible polynomial (x~ - b) ~ g, and Z, is replaced by :¢~ = Zp. In this situation, (31) is replaced by F(~) F1 mod (x~ b~), 1 = (49) while the polynomials 0 (i), H~), and Zp[xl , ... , x~-l]. Furthermore, (34 ~(1) and when e = d, we have for i = 1, /~2 mod( x, - bi), /~(i) 2 satisfy (32) and (33) in gb~[Xl, "" • , x~-i] = ) becomes 0 mod (x~ - b~), (50) ~(i) __- =- • " , n, in place of (35). q = II (xv -- bl), i=l 0 mod (xv - bi), /71 mod (x~ - bd, (51) /I~ mod (x~ - bl), Also, (36) becomes (52)
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494 w.S. BROWN while G*, Hi*, and H2* [see (37)] are the unique polynomials (in ~[Xa , "" , xv-,]) with eoeffieients (in 9) of degree (in Xv) less than n, such that G* ~ G (i) rood (xv - bi), H,* ------ //~) rood (x, - bi), (53) H2* ~ " (i) 1/12 mod (x, - b~), fori= 1,...,n. Now as soon as e = d, weseefrom (51)and (53) that (38 holds. When we also achieve n > , = max (0v (0), o~ (/it,), 0~ (/72) ), (54) where 0v denotes the degree in x,, it follows that (40) holds. To obtain the final results, we then use (41) and (42) as in Algorithm ~[. Although the preceding discussion is sufficient in principle to define Algorithm P, the interested reader may find it instructive to compare the following detailed description with the earlier presentation (Section 4.3) of Algorithm M. (1) If v 1, then F, and F' = ' 2 are elements of 9 invoke Algorithm U to comput G' = gcd(F,', F(), and return. Otherwise use Algorithm U to compute 0 = eont(Fl' ) c2 = cont(F2'), c = gcd(o, c2). (2) Set F~ = Fl'/C,, F~ = F2'/c2. (3) Set fl = lc(F,), f~ = lc(F2), 0 = gcd(f,, f2). (4) Set n = 0, e = min (~(F,), O(F~)). (5) Set ~, = 0~(0) + O~(F,), ~2 = 0~(0) + O~(F2), ~ = max(p,, P2). It follows that h = 0~ (f,) = 0~ (G) + 0v (/4,), ~2 = 0~ (f~) = 0~ (G) + 0~. (/t2), and v > ,. (6) Let b be a new element of Z, such that (x~ - b) ~ fir2. If Zp is exhausted, then p is too small and the algorithm fails. (7) Set 0 = 0 mod(x~ -- b), IP~ = OF, mod(x, - b), F2 = 0F2 mod(x~ -- b). (8) Invoke Algorithm P reeursively to compute G = g" ged (F~, F2), lq, = IPl/G, and tq2 = F2/G, all in 9b[xl, "" , X~_l] = Zp[xi, ... , xv ,]. These relations imply thatle(G) = g, andl)(G) > d. (9) If it(G) = 0, set G = 1, Hi = Fx, He = F2, and skip to Step (15). If i} (G) > e, go back to Step (6). If i} (G) < e, set n = 0, e = ~ (G).
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# Equations - Introductory Statistics Formulas N i =1...
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Introductory Statistics – Formulas Population Mean : 1 = N i i x x x N N μ = = ; Sample Mean : 1 n i i x x x n n = = = Range : R = Maximum Value Minimum Value Quartile locations ( Minitab ): 1 2 3 ( 1) 2( 1) 3( 1) @ ; @ ; @ 4 4 4 n n n Q Q Q + + + . Quartile values: 1 . ( ) j i i i Q x dd x x + = + , where i is the integer portion of the quartile location and dd is the decimal portion of quartile location (i.e., if Q 3 @ 47.75, then i = 47 and . dd = 0.75). Interquartile range : IQR = Q 3 Q 1 Potential Outlier Limits : Lower Limit = Q 1 - 1.5 IQR ; Upper Limit = Q 3 + 1.5 IQR Population Variance : 2 2 ( ) x x x N μ σ = ; Pop. Standard Deviation : 2 ( ) x x x N μ σ = Population Standard Deviation (Short-cut formula): 2 2 x x x N σ μ = Sample Variance : 2 2 ( ) 1 x x x s n = ; Sample Standard Deviation : 2 ( ) 1 x x x s n = Sample Standard Deviation (short-cut formula) : ( ) 2 2 / 1 x x x n s n = Standardized Pop. Data Value : x x x z μ σ = ; Standardized Sample Data Value : x x x z s = T’s Theorem (Chebychev’s Rule) :
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# More about the step size choice
## Get insight into the step size choice for the numerical programs of Ti-84 plus
### 1: How to select the appropriate step size for solving first order differential equations solved with the Ti-84 plus
Gaining insight into the numerical processes is import for electrical engineers who apply numerical programs such as Simulink and Multisim during their work.
In the numerical programs written for the Ti-84 plus CE calculator, such as the first order differential equation solver DV1ETRP3.8XP, a step size must also be chosen before the program starts calculations. A good choice of the step size is important because if the step size is too small, then calculation time takes a long time and memory overflow can occur. If the step size is too large, then the results are no longer reliable. In the figure below, the calculated results with a good step size and with a wrong step size are shown. If a wrong step size is chosen, then calculated results are incorrect and the plot may appear inaccurate. However, it is not always so simple to determine that the results are incorrect. Using practical examples, we will clarify what is a good choice for the step size and not through complicated theories. Also, a method for controlling the reliability and accuracy of the calculated results will be explained
To understand the explanation, knowledge of solving first-order differential equations analytically is necessary. Insight into the effects of step size on the accuracy of first order differential equation solver results is also useful for other numerical solvers written for the Ti calculator, such as second order differential equation solver or numerical integrals. Let us start with researching the step size for linear first order differential equation solvers. The general equation of a linear first order differential equation Is :
To obtain good numerical results, the step size must be at least ten times smaller than the time constant. For a constant or slowly changing f(x), the requirements related to the time constant are sufficient.
Suppose a=0.1, b=0.7 and f(x)=2(constant). The step size choice is 0.1/(10*.7)=0.01429 sec. Comparing the results of the numerical calculations with the analytical solution shows that the program gives good results with this rather large step size. A fast changing f(x) however, sets stricter requirements on the step size. We will demonstrate this with the following example.
Suppose that the input signal f(x)= 2 sin(2π60x) instead of a constant f(x)=2, and we use the same time step as before = 0.01429 sec., then the program produces wrong results. As the frequency of the input is 60 Hz, which is a period of time of 0.0166 sec and almost equals the step size, it is clear that no reliable results can be obtained.
A smaller step size is necessary. From experience 20 samples of a sinusoidal input signal and trapezoidal solver are sufficient to get reliable results which means that the step size is decreased to T/20= 1/(f*20) =1/(60*20)= 0.0008333 sec.
This is 17.15 times smaller than the time used at constant f(x) as input. Here below, the results calculated with a large step size (blue) and with a small step size (red) are shown. The results totally differ. The red results are OK.
For other signals than sinusoidal signals, we advise taking a step size equal to Xmax/999.
For example, we take f(x)= (x^2)/1E-6 in the area 0<x<4E-3. Xmax=4E-3.
Below, you can see the difference in results between a step size of Xmax/100 and Xmax/999
The program DV1ETRP3 can save up to 999 points, so a step size of 4E-3/999= 4.004E-6 gives 999 calculated points.
A step size 10 times smaller shows that the calculated and simulated results agree good
Summarizing the rules for the step size for a first order differential equation :
1: determine the time constant a/b. The step size should be a/(10b) for slow-changing function f(x).
2: For rapidly changing f(x) a step size of Xmax/100 is usually sufficient but can be chosen smaller. The maximum of saved points is 999. Then the step size is Xmax/999. A smaller step size is possible, but be aware that only 999 points will be saved.
2: How to select the appropriate step size for second order differential equations solved with the Ti-84
Too large a step size gives inaccurate results with second order differential solver as well. Similar rules for the step size for the second order differential equation solver can be derived.
1: Calculate dx = (√(A/C)) /20 and dx =(B/C)/10 and take the smallest step size of both.
2 For rapidly changing f(x), take a step size of dx= Xmax/999. To get confidence in the results, you can repeat the calculations with a step size of dx= Xmax/2000. If the results between both are small (less than 4 %) than you can trust the results.
The maximum of saved points is 999. Then the step size = Xmax/999. A smaller step size is possible, but be aware that only 999 points are saved.
bottom of page | 1,191 | 5,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-26 | latest | en | 0.91141 |
https://agda.readthedocs.io/en/v2.5.3/language/mixfix-operators.html | 1,679,618,329,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945218.30/warc/CC-MAIN-20230323225049-20230324015049-00062.warc.gz | 114,942,986 | 6,059 | # Mixfix Operators¶
A name containing one or more name parts and one or more `_` can be used as an operator where the arguments go in place of the `_`. For instance, an application of the name `if_then_else_` to arguments `x`, `y`, and `z` can be written either as a normal application `if_then_else_ x y z` or as an operator application `if x then y else z`.
Examples:
```_and_ : Bool → Bool → Bool
true and x = x
false and _ = false
if_then_else_ : {A : Set} → Bool → A → A → A
if true then x else y = x
if false then x else y = y
_⇒_ : Bool → Bool → Bool
true ⇒ b = b
false ⇒ _ = true
```
## Precedence¶
Consider the expression `true and false ⇒ false`. Depending on which of `_and_` and `_⇒_` has more precedence, it can either be read as `(false and true) ⇒ false = true`, or as `false and (true ⇒ false) = true`.
Each operator is associated to a precedence, which is an integer (can be negative!). The default precedence for an operator is 20.
If we give `_and_` more precedence than `_⇒_`, then we will get the first result:
```infix 30 _and_
-- infix 20 _⇒_ (default)
p-and : {x y z : Bool} → x and y ⇒ z ≡ (x and y) ⇒ z
p-and = refl
e-and : false and true ⇒ false ≡ true
e-and = refl
```
But, if we declare a new operator `_and’_` and give it less precedence than `_⇒_`, then we will get the second result:
```_and’_ : Bool → Bool → Bool
_and’_ = _and_
infix 15 _and’_
-- infix 20 _⇒_ (default)
p-⇒ : {x y z : Bool} → x and’ y ⇒ z ≡ x and’ (y ⇒ z)
p-⇒ = refl
e-⇒ : false and’ true ⇒ false ≡ false
e-⇒ = refl
```
## Associativity¶
Consider the expression `true ⇒ false ⇒ false`. Depending on whether `_⇒_` is associates to the left or to the right, it can be read as `(false ⇒ true) ⇒ false = false`, or `false ⇒ (true ⇒ false) = true`, respectively.
If we declare an operator `_⇒_` as `infixr`, it will associate to the right:
```infixr 20 _⇒_
p-right : {x y z : Bool} → x ⇒ y ⇒ z ≡ x ⇒ (y ⇒ z)
p-right = refl
e-right : false ⇒ true ⇒ false ≡ true
e-right = refl
```
If we declare an operator `_⇒’_` as `infixl`, it will associate to the left:
```infixl 20 _⇒’_
_⇒’_ : Bool → Bool → Bool
_⇒’_ = _⇒_
p-left : {x y z : Bool} → x ⇒’ y ⇒’ z ≡ (x ⇒’ y) ⇒’ z
p-left = refl
e-left : false ⇒’ true ⇒’ false ≡ false
e-left = refl
```
## Ambiguity and Scope¶
If you have not yet declared the fixity of an operator, Agda will complain if you try to use ambiguously:
```e-ambiguous : Bool
e-ambiguous = true ⇒ true ⇒ true
```
```Could not parse the application true ⇒ true ⇒ true
Operators used in the grammar:
⇒ (infix operator, level 20)
```
Fixity declarations may appear anywhere in a module that other declarations may appear. They then apply to the entire scope in which they appear (i.e. before and after, but not outside). | 892 | 2,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-14 | latest | en | 0.470538 |
https://www.w3resource.com/python-exercises/list-advanced/python-list-advanced-exercise-9.php | 1,726,513,483,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00563.warc.gz | 981,007,965 | 27,964 | Python List - Maximum sum sub-sequence in a list
# Python List Advanced Exercise - Maximum sum sub-sequence in a list
## Python List Advanced: Exercise-9 with Solution
Write a Python a function to find the maximum sum sub-sequence in a list. Return the maximum value.
Sample Solution:
Python Code:
``````# Define a function to find the maximum sum subsequence in an array
def max_sum_subsequence(arr):
n = len(arr)
# Initialize a dynamic programming (DP) array 'dp' with all zeros
dp = [0 for i in range(n)]
# Set the first element of 'dp' to the first element of 'arr'
dp[0] = arr[0]
# Iterate through the array from the second element
for i in range(1, n):
# Calculate the maximum sum at the current position by comparing the element itself and the sum from the previous position
dp[i] = max(arr[i], dp[i-1] + arr[i])
# Return the maximum value in the 'dp' array, which represents the maximum sum subsequence
return max(dp)
# Create a list of numbers
nums = [1, 2, 3]
# Print the original list of numbers
print("Original list:")
print(nums)
# Call the max_sum_subsequence function with the list and store the result in 'result'
result = max_sum_subsequence(nums)
# Print the result, which is the maximum sum subsequence of the list
print("Maximum sum sub-sequence of the said list:")
print(result)
# Create another list of numbers
nums = [1, 2, 4, 3, 5, 4, 6, 9, 2, -10]
# Print the original list of numbers
print("\nOriginal list:")
print(nums)
# Call the max_sum_subsequence function with the second list and store the result in 'result'
result = max_sum_subsequence(nums)
# Print the result, which is the maximum sum subsequence of the list
print("Maximum sum sub-sequence of the said list:")
print(result)
# Create another list of numbers with negative values
nums = [1, 2, -4, 3, 5, 4, 6, 9, 2, -10]
# Print the original list of numbers
print("\nOriginal list:")
print(nums)
# Call the max_sum_subsequence function with the third list and store the result in 'result'
result = max_sum_subsequence(nums)
# Print the result, which is the maximum sum subsequence of the list
print("Maximum sum sub-sequence of the said list:")
print(result)
# Create another list of numbers with a negative value in the middle
nums = [1, 2, 4, 3, 5, -24, 6, 9, -2]
# Print the original list of numbers
print("\nOriginal list:")
print(nums)
# Call the max_sum_subsequence function with the fourth list and store the result in 'result'
result = max_sum_subsequence(nums)
# Print the result, which is the maximum sum subsequence of the list
print("Maximum sum sub-sequence of the said list:")
print(result)
```
```
Sample Output:
```Original lists:
[1, 2, 3]
Maximum sum sub-sequence of the said list:
6
Original lists:
[1, 2, 4, 3, 5, 4, 6, 9, 2, -10]
Maximum sum sub-sequence of the said list:
36
Original lists:
[1, 2, -4, 3, 5, 4, 6, 9, 2, -10]
Maximum sum sub-sequence of the said list:
29
Original lists:
[1, 2, 4, 3, 5, -24, 6, 9, -2]
Maximum sum sub-sequence of the said list:
15
```
Flowchart:
What is the time complexity and space complexity of the following Python code?
``````def max_sum_subsequence(arr):
n = len(arr)
dp = [0 for i in range(n)]
dp[0] = arr[0]
for i in range(1, n):
dp[i] = max(arr[i], dp[i-1] + arr[i])
return max(dp)
``````
Time complexity - The time complexity of the said code is O(n), where n is the length of the input array "arr". The code iterates through the input array to fill the dp array, which takes O(n) time, and then finds the maximum value in the "dp" array, which also takes O(n) time. Therefore, the overall time complexity is O(n).
Space complexity - The space complexity of the said code is also O(n). This is because the code creates a new array "dp" of length n, which takes O(n) space. Therefore, the overall space complexity is O(n).
Python Code Editor:
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
| 1,082 | 3,936 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-38 | latest | en | 0.733378 |
https://math.andyou.com/477 | 1,709,245,705,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474853.43/warc/CC-MAIN-20240229202522-20240229232522-00346.warc.gz | 377,969,660 | 5,927 | ### 10.3 Professional Sports
<
>
• Math Help
Here are some observations about the win probability graph on page 477 (also shown below).
• Remember that this graph is just for the team that has the ball.
• Here is what the lines represent.
• +7: The team with the ball is winning by 7 points. • +3: The team with the ball is winning by 3 points. • +1: The team with the ball is winning by 1 point. • -1: The team with the ball is losing by 1 point. • -3: The team with the ball is losing by 3 points. • -7: The team with the ball is losing by 7 point.
• To find the win probability for the team without the ball, subtract the win probability of the team with the ball in the graph from 100%. Because the probability of winning for a team with the ball that is 3 points ahead with 10 minutes remaining is about 85%, the probability of winning for a team without the ball that is 3 points behind with 10 minutes remaining is about 100% - 85% = 15%.
• Notice that the graph drops off quickly for a team with the ball that is losing by 7 points with about 20 minutes remaining. There are similar drop offs for a team with the ball that is losing by 1 or 3 points with about 15 minutes to go.
• Notice that for any point in the game, the team with the ball and a lead has a greater than 50% chance of winning.
• Notice that from between 20 minutes remaining and 10 minutes remaining, the probability of a team with the ball winning while losing by 1 point is greater than the probability of a team with the ball winning while winning by 1 point. This seems contradictory to common sense because you would expect the probability while winning to be greater than the probability while losing. One factor that may influence this is that some teams play too conservatively with a small lead and play more aggressive when they are losing.
• Consumer Suggestion
To watch video clips of exciting NFL football games, visit NFL.com. Here you can search by NFL team, season, and other criteria to find game clips of your favorite team.
• Checkpoint Solution
You could use the graph to make decisions such as whether to go for a 4th down and whether to kick a field goal or go for a touchdown depending on the stage of the game. The graph could also help you determine whether to play aggressively or conservatively depending on where you stand and how late it is in the game. For instance, if you lead by 1 point, you do not want to play too conservatively with 10-15 minutes remaining because the win probability graph shows a slight dip during that time. On the other hand, if you are down by 1 near the beginning of the 4th quarter, you should adopt an aggressive strategy and try to take the lead. If you are down by 7, you want to take the lead before the 20-minute mark (close to the end of the 3rd quarter) because there is a dramatic drop-off in win probability. If you are down by 1 or 3, the same applies for a few minutes into the 4th quarter.
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``` ____ _____ ______ ___ ____ | 769 | 3,377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-10 | latest | en | 0.971768 |
https://rxcal.org/practice/practice/1720181114055128-45-rxcal-140.php?id=140&domain=19&version=1 | 1,675,046,798,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499790.41/warc/CC-MAIN-20230130003215-20230130033215-00286.warc.gz | 497,921,122 | 7,480 | ### Type 3 - Find the quantity of pure drug
9. Altering Product Strength 9.2) Alligation 9.2.1) Alligation
9.2.1.1) Alligation Normal 1
Total tried: Correct: Wrong:
How many grams of a pure drug powder should you add to 360 g of 45 % of an ointment of the same drug so that the blended strength becomes 76%?
Click on the button below to see the answer and explanations
lb equals 465 g kg
Tic-tac-toe, Short form:
[[100 % (? \quad g),\quad\quad\quad\quad, 31 pt],[\quad\quad\quad\quad,76 % , \quad\quad\quad\quad],[45 % (360 \quad g),\quad\quad\quad\quad,24 pt]]
Therefore, 31 parts of pure drug should be mixed with 24 parts of the mixture which has a concentration of 45 % and the weight of 360 grams.
(360 \quad g)/(24 \quad parts)=x/(31 \quad parts) \quad therefore x = 465 \quad g. Ans. | 255 | 811 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-06 | longest | en | 0.704523 |
https://www.jiskha.com/questions/1655196/how-to-find-the-integral-of-3x-pi-cos3xdx-from-pi-over-6-pi-over-3 | 1,537,457,915,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156513.14/warc/CC-MAIN-20180920140359-20180920160759-00245.warc.gz | 779,556,887 | 4,918 | # calculus 2
how to find the integral of (3x+pi)cos3xdx from [pi over 6, pi over 3}
1. integrate by parts
int u dv = u v - int v du
dv = cos 3 x dx
so
v = (1/3) sin 3x
u = 3 x + pi
du = 3 dx
so
u v - (3/3)integral of sin 3x dx
(3x+pi)(sin 3x)- int sin3x dx
at pi/3 minus at pi/6
posted by Damon
2. i really appreciate the work it helps me a lot thank you
posted by bern
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6. ### Calculus
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A. Find the integral of the following function. Integral of (x√(x+1)) dx. B. Set up and evaluate the integral of (2√x) for the area of the surface generated by revolving the curve about the x-axis from 4 to 9. For part
8. ### calc
how do you start this problem: integral of xe^(-2x) There are two ways: 1) Integration by parts. 2) Differentiation w.r.t. a suitably chosen parameter. Lets do 1) first. This is the "standard method", but it is often more tedious
9. ### Calc BC
1. Find the indefinite integral. Indefinite integral tan^3(pix/7)sec^2(pix/7)dx 2. Find the indefinite integral by making the substitution x=3tan(theta). Indefinite integral x*sqrt(9+x^2)dx 3. Find the indefinite integral.
10. ### calc asap!
can you help me get started on this integral by parts? 4 S sqrt(t) ln(t) dt 1 please help! thanks! Integral t^(1/2)Ln(t)dt = 2/3 t^(3/2)Ln(t)- 2/3 Integral t^(1/2) dt = 2/3 t^(3/2)Ln(t) - 4/9 t^(3/2) Simpler method: Integral
More Similar Questions | 850 | 2,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-39 | latest | en | 0.811445 |
https://discuss.leetcode.com/topic/60891/another-easy-to-understand-recursive-solution-with-simple-explanation | 1,513,374,090,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948579567.73/warc/CC-MAIN-20171215211734-20171215233734-00776.warc.gz | 542,807,498 | 8,690 | # Another easy-to-understand recursive solution with simple explanation !
• `````` target = 21
[
[1, 4, 7, 11, 15],
[2, 5, 8, 12, 19],
[3, 6, 9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]
1. Search in the most right column
15
19
22-->start, top
24
30
2. Search in the row [3, 6, 9, 16, 22]
|
startR, right
3. Search in the sub matrix
[
[3, 6, 9, 16],
[10, 13, 14, 17],
[18, 21, 23, 26]
]
``````
""
``````public boolean searchMatrix(int[][] matrix, int target) {
int right = matrix[0].length-1;
int top = 0;
return findRecursive(matrix, target, right, top);
}
private boolean findRecursive(int[][] matrix, int target, int right, int top){
if(top==matrix.length) return false;
if(right<0) return false;
int start = top;
int end = matrix.length-1;
//Search in the most right column
while(start<=end){
int mid = start + (end-start)/2;
if(target == matrix[mid][right]) return true;
else if(target < matrix[mid][right]) end = mid-1;
else start = mid+1;
}
top = start;
if(top==matrix.length) return false;
int startR = 0;
int endR = right;
//Search in one row
while(startR<=endR){
int midR = startR + (end - start)/2;
if(target == matrix[top][midR]) return true;
else if(target < matrix[top][midR]) endR = midR - 1;
else startR = midR+1;
}
return findRecursive(matrix, target, endR, top+1);
}
``````
""
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 499 | 1,439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2017-51 | latest | en | 0.595308 |
https://ghc.haskell.org/trac/ghc/wiki/TypeFunctionsSynTC/PlanMSRevised2 | 1,508,686,081,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825308.77/warc/CC-MAIN-20171022150946-20171022170946-00534.warc.gz | 708,809,579 | 7,300 | ## WHY WE NEED TO NORMALIZE EQUATIONS
Consider
```H (G Int) = Int (1)
G Int = F (G Int) (2)
```
where F, G, H are type function constructors. I will omit evidence (construction steps) for the moment. The above equations imply
```H (F .... F (G Int)) = Int (3)
```
Indeed, we can repeatidly apply the second equation on the first equation to obtain the third equation. But this also shows that we cannot naively apply the following reduction step
``` T1[F t] = t'
--> T1[s] = t'
```
where F t = s. We write T1[] to denote a type with a hole. NOTE: The above reduction step is proposed in Plan MC and Plan MS.
Otherwise, we find that
``` H (G Int) = Int
--> H (F (G Int)) = Int
--> ...
```
So, the question is whether there is a way out to avoid non-termination.
As argued by Plan MS revised there is a way out if we "normalize" equations. "Normalization" of equations works by "flattening" equations such that all equations are of the form
```F n = n'
```
where n and n' only refer to types NOT containing type function constructors. For example, the equations
```H (G Int) = Int (1)
G Int = F (G Int) (2)
```
are normalized to the form
```H a = Int
G Int = a
G Int = b
F c = b
G Int = c
```
We can consider variables a, b and c as existentially quantified. Notice that logically speaking
```H (G Int) = Int /\ G Int = F (G Int)
iff
exists a, b, c.
H a = Int /\
G Int = a /\
G Int = b /\
F c = b /\
G Int = c /\
```
On the equations in normal form we apply the reduction step
``` F n1 = n2
F n1 = n3
-->
F n1 = n2
F n1 = n3
n2 = n3
```
Points to note:
• this corresponds exactly to the FD-CHR reduction step
• n2 = n3 is a "pure" unification constraint. Recall that by (normalization) assumption, n2 and n3 do NOT refer to type function constructors
• we can either discard F n1 = n3 (or F n1 = n2), or we remember having applied the reduction step on F n1 = n2 and F n1 = n3 to avoid infinite application of the same reduction step
Let's apply this reduction step to the above normal form. We find that
```H a = Int
G Int = a *
G Int = b * we highlight equations involved in the reduction
F c = b
G Int = c
-->
H a = Int
G Int = a *
b = a
F c = a -- cause b = a we replace F c = b by F c = a
G Int = c *
-->
H a = Int
G Int = a
b = a
c = a
F a = a -- cause c = a we find F a = a
```
At this point no further reductions are possible. Notice the "cycle" in F a = a. In essence,
```H a = Int
G Int = a
b = a
c = a
F a = a
```
represents the canonical (ie fully reduced) normal form of all equations which are derivable
```H (G Int) = Int
G Int = F (G Int)
```
## HOW DOES NORMALIZATION FIT TOGETHER WITH EVIDENCE CONSTRUCTION
Let's assume that
```g1 : H (G Int) = Int
g2 : G Int = F (G Int)
```
are given equations where g1 and g2 are the respective evidence values and we would like to compute g3 (wanted) such that
```g3 : H (F (G Int)) = Int
```
For the construction of wanted evidence from given evidence it seems counter-intuitive to normalize (given) equations. Based on the normalization procedure we turn
```g1 : H (G Int) = Int
g2 : G Int = F (G Int)
```
into the normal form
```g1' : H a = Int
g2' : G Int = a
g3' : G Int = b
g4' : F c = b
g5' : G Int = c
```
It's clear how to build g1 and g2 from g1',g2',g3',g4' and g5'. But the other direction is not obvious.
## EVIDENCE CONSTRUCTION IN THE PRESENCE OF NORMAL FORMS
So much for how we put equations into normal form. Now we look at the real algorithm. Let's consider a slightly different example, which we will use from now on. We are given the equations
```g1 : F (G Int) = Int
g2 : G Int = F (G Int)
```
It should be clear that the above yields the canonical normal form SLPJ: what is a "canonical normal form", and why do we want it? Our goal, after all, is to compute evidence for "wanted" from "given"
```g1' : F Int = Int
g2' : G Int = Int
```
where
```g1' = sym (F g2') trans g1 = sym (F (g2 trans g1)) trans g1
g2' = g2 trans g1
```
SLPJ: when you say "it should be clear", you mean "the solution we seek is this", right?
We write 'trans' for the transitive relation on coercions, 'sym' is symmetry etc.
The challenge is how to compute
```g1' : F Int = Int
g2' : G Int = Int
```
for some g1' and g2' from the normal form?
We apply the following strategy:
(Ia) Put the given equations into normal form.
(Ib) Exhaustively apply reduction steps on these nonrmal-form
equations.
(II) Then, we normalize the proof terms of the equations
in the final store (to obtain proof terms which only refer to the proof term variables from the original equation set).
We first consider (Ib).
There are two kind of reduction steps on equations
1) FD-CHR reduction step:
```g: F n1 = n2
g': F n1 = n3
-->
g: F n1 = n3
(sym g) trans g' : n2 = n3
```
We discard g': F n1 = n3 because we can build g' using `g: F n1 = n3` and `(sym g) trans g' : n2 = n3`.
2) Substitution step (replacing variables by equals):
```g : a = n1
g' : F a = n2
-->
g : a = n2
(sym (F g)) trans g' : F n1 = n2
```
We omit the other cases, e.g. `n1 = a`, `F n2 = a` etc, and reductions such as `[n1] = [n2] --> n1 = n2`.
NOTE: both reduction steps maintain the normal form property
Here's the normal form for our example, attached with evidence.
```g1' : F a = Int
d1 : G Int = a
g2' : G Int = b
d2 : F c = b
d3 : G Int = c
```
We use the convention that d refers to some "anonymous" evidence.
Notice that we can express g1' and g2' in terms of d1,d2,d3 and g1,g2.
```g1' = sym (F d1) trans g1
g2' = g2 trans (F d3) trans d2
```
SLPJ: for me, g1' is like d1, a piece of anonymous evidence. The evidence I have is g1. I can construct g1 from g1' and d1:
``` g1 = (F d1) trans g1'
```
Now that also implies that I can construct g1' from g and d1, as you say; and that I can construct d1 from g1 and g1'! But the starting point is surely g1. I found this confusing.
We exhaustively apply reduction steps on the above normal form.
```g1' : F a = Int
d1 : G Int = a *
g2' : G Int = b
d2 : F c = b
d3 : G Int = c *
--> FD-CHR step
g1' : F a = Int
d1 : G Int = a
g2' : G Int = b
d2 : F c = b *
(sym d3) trans d1 : c = a *
--> Substitution step
g1' : F a = Int
d1 : G Int = a *
g2' : G Int = b *
(sym (F ((sym d3) trans d1))) trans d2 : F a = b
(sym d3) trans d1 : c = a
--> FD-CHR step
g1' : F a = Int
g2' : G Int = b
(sym (F ((sym d3) trans d1))) trans d2 : F a = b *
(sym d3) trans d1 : c = a
(sym g2') trans d1 : b = a *
--> Substitution step
g1' : F a = Int *
g2' : G Int = b
d4 : F a = a *
(sym d3) trans d1 : c = a
(sym g2') trans d1 : b = a
where d4 = (sym (F ((sym d3) trans d1))) trans d2 trans (sym g2') trans d1
--> FD-CHR step
g1' : F a = Int *
g2' : G Int = b
(sym d3) trans d1 : c = a
(sym g2') trans d1 : b = a
(sym d4) trans g1' : a = Int *
--> Substitution step
(sym (F d5)) trans g1' : F Int = Int
g2' : G Int = b *
(sym d3) trans d1 : c = a
(sym g2') trans d1 : b = a *
(sym d4) trans g1' : a = Int *
where d5 = (sym d4) trans g1'
--> Substitution step
(sym (F d5)) trans g1' : F Int = Int
g2' trans ((sym g2') trans d1) trans d5 : G Int = Int
(sym d3) trans d1 : c = a
(sym g2') trans d1 : b = a
(sym d4) trans g1' : a = Int
```
No further reductions are applicable at this point. Note that
```d4 = (sym (F ((sym d3) trans d1))) trans d2 trans (sym g2') trans d1
d5 = (sym d4) trans g1'
g1' = sym (F d1) trans g1
g2' = g2 trans (F d3) trans d2
```
Next, we consider (II).
We will use the equation and some laws to "normalize" the proof terms attached to equations `F Int = Int` and `G Int = Int` such that the proof terms only mention g1 and g2.
We make use of the following laws:
```sym (F (g1 trans g2)) = (F (sym g2)) trans (F (sym g1))
sym (F g) = F (sym g)
F (g1 trans g2) = (F g1) trans (F g2)
sym (sym g) = g
```
The above laws imply the following property
Property: For each `g : s = t` we can achieve the normal form
``` d1 trans .... trans dn : s = t
```
where each di is either of the form `sym (F1 ... Fn n) or (F1 ... Fn n)`, where n is either a variable or a primitive type such as Int etc.
SLPJ give intuition and/or proof for this property. Why is the n the last parameter?
Let's normalize the proof terms attached to the equations `F Int = Int` and `G Int = Int`.
We have that
```(sym (F d5)) trans g1' : F Int = Int
g2' trans ((sym g2') trans d1) trans d5 : G Int = Int
where
d4 = (sym (F ((sym d3) trans d1))) trans d2 trans (sym g2') trans d1
d5 = (sym d4) trans g1'
g1' = sym (F d1) trans g1
g2' = g2 trans (F d3) trans d2
```
During the normalization of proof terms process we also apply the "cancellation" law that
```g1 trans (sym g2) trans g2 = g1
```
We start off with `(sym (F d5)) trans g1' : F Int = Int`
```(sym (F d5)) trans g1'
--> d5 = (sym d4) trans g1'
(sym (F ((sym d4) trans g1'))) trans g1'
--> push sym inside
(F (sym g1')) trans (F d4) trans g1'
--> g1' = sym (F d1) trans g1
and normalize
(sym (F g1)) trans (F (F d1)) trans (F d4) trans g1'
--> d4 = (sym (F ((sym d3) trans d1))) trans d2 trans
(sym g2') trans d1
= (sym (F d1)) trans (F d3) trans d2 trans
(sym g2') trans d1
and normalize
(sym (F g1)) trans (F (F d1)) trans
(sym (F (F d1)) trans (F (F d3)) trans (F d2) trans
(sym (F g2')) trans (F d1) trans g1'
--> g1' = sym (F d1) trans g1
g2' = g2 trans (F d3) trans d2
and normalize
(sym (F g1)) trans (F (F d1)) trans
(sym (F (F d1)) trans (F (F d3)) trans (F d2) trans
(sym (F d2)) trans (sym (F (F d3))) trans (sym (F g2)) trans
(F d1) trans sym (F d1) trans g1
-- ********* Here a miracle happens
-- Magically, a (F d2) meets (sym (F d2)) etc
--> "cancellation" law
(sym (F g1)) trans (sym (F g2)) trans g1
```
A similar calculation for `g2' trans ((sym g2') trans d1) trans d5 : G Int = Int` yields
```g2' trans ((sym g2') trans d1) trans d5
-->*
g2 trans g1
```
Hence, we have verified that
```(sym (F g1)) trans (sym (F g2)) trans g1 : F Int = Int
and
g2 trans g1 : G Int = Int
```
This is the result we were hoping for!
Note that `(sym (F g1)) trans (sym (F g2)) trans g1` is equivalent to the more "compact" form `sym (F (g2 trans g1)) trans g1`.
## SUMMARY
The method outlined above performs reductions on equations in normal form. These equations are logically equivalent to the original constraint set, for our running example, we have that
```F (G Int) = Int /\ G Int = F (G Int)
iff
exists a, b, c.
F a = Int /\
G Int = a /\
G Int = b /\
F c = b /\
G Int = c
```
When reducing the equations in normal form, we obtain proof terms in terms of
```g1' : F a = Int
d1 : G Int = a
g2' : G Int = b
d2 : F c = b
d3 : G Int = c
rather than
g1 : F (G Int) = Int
g2 : G Int = F (G Int)
```
The trick is that we yet need to normalize proof terms. See the above calculations. Thus, we find that
```(sym (F g1)) trans (sym (F g2)) trans g1 : F Int = Int
and
g2 trans g1 : G Int = Int
```
In general, we will need to include axioms and wanted equations. But there is really no problem with such extensions. Of course, the details have yet to be worked out. | 3,779 | 11,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-43 | latest | en | 0.915365 |
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# Triangle Sum Theorem
## Interior angles add to 180 degrees
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Triangle Sum Theorem
by CK-12 //basic
Apply the Triangle Sum Theorem.
4
## Triangle Sum Theorem
Learn about the sum of the angles of triangles.
0
## Understanding the Angle Measures of Triangles
Learn to understand the angle measures of triangles.
0
0
## Triangle Sum Theorem and Exterior Angle Theorem
by Jaclyn Tant //basic
Recognize the sum of the interior angles of a triangle as 180 degrees
0
## Geometry and Equations (A-7)
by LISD Secondary Math //basic
0
## Triangle Sum Theorem
Apply the Triangle Sum Theorem.
0
## Triangle Sum Theorem: Introduction
Recognize the sum of the interior angles of a triangle as 180 degrees
0
## Triangle Sum Theorem Continued
This concept teaches students the Triangle Sum Theorem, and how to use it.
0
## Triangle Sum Theorem.
by Scott Lamie //basic
Apply the Triangle Sum Theorem.
0
• PLIX
## Identifying Sets of Pythagorean Triples, Triangle Sum Theorem: Bike Ramp
Triangle Sum Theorem Interactive
0
• PLIX
## Triangle Sum Theorem
Triangle Sum Theorem Interactive
0
• PLIX
## Triangle Sum Theorem: Creating Right Triangles
Triangle Sum Theorem Interactive
0
• PLIX
## Angle Measures in Given Triangles: Lighthouse
Triangle Sum Theorem Interactive
0
• Video
## Triangle Sum Theorem Principles - Basic
by CK-12 //basic
This video provides the student with a walkthrough on the Triangle Sum Theorem.
0
• Video
## Triangle Sum Theorem Examples - Basic
by CK-12 //basic
This video provides the student with a walkthrough of one or more examples from the concept "Triangle Sum Theorem".
0
• Video
## Triangle Sum Theorem Principles
This video gives more detail about the mathematical principles presented in Triangle Sum Theorem.
0
• Video
## Triangle Sum Theorem Examples
This video shows how to work step-by-step through one or more of the examples in Triangle Sum Theorem.
0
• Video
## Proof - Sum of Measures of Angles in a Triangle is 180
by CK-12 //basic
Proves that the sum of a triangle is 180.
0
• Video
## Animation of the Sum of the Interior Angles of a Triangle
by CK-12 //basic
Illustrates that the sum of interior angles of a triangle is 180 degrees.
0
• Video
## Proving the Triangle Sum Theorem
by CK-12 //basic
Take a look at proving the triangle sum theorem.
0
## Triangle Sum Theorem Quiz
Ten question quiz. Students must solve for missing angles in triangles using the triangle sum theorem and basic arithmetic and their algebra skills.
0
• Practice
0%
## Triangle Sum Theorem Practice
1
• Interactive Exercise
## Triangle Sum Theorem
0
• Interactive Exercise
## Triangle Sum Theorem Visual
0
• Interactive Exercise
## Triangle Sum Theorem Visual 2
0
• Interactive Exercise
## Find The Measure of the Third Angle
0
• Interactive Exercise
## Find The Measure of the Third Angle 2
0
• Interactive Exercise
## Find The Measure of the Third Angle of an Isoceles Triangle
0
• Interactive Exercise
## Use Algebra and The Triangle Sum Theorem
Solve for x in the triangle.
0
• Interactive Exercise
## Use Algebra and The Triangle Sum Theorem 2
Solve for x in the triangle.
0
• Interactive Exercise
## Use Algebra and The Triangle Sum Theorem 3
Use the triangle sum theorem to find the measure of an unknown angle.
0
• Interactive Exercise
## Use Algebra and The Triangle Sum Theorem 4
Use the triangle sum theorem to find the measure of an unknown angle.
0
• Interactive Exercise
## Given Two Congruent Triangles, Solve for the Unknown Interior Angle
Checks your understanding of the Triangle Sum Theorem.
0
• Interactive Exercise
## Given Two Congruent Triangles, Solve for the Unknown Interior Angle (part 2)
Checks your understanding of the Triangle Sum Theorem.
0
• Critical Thinking
## Triangle Sum Theorem Discussion Questions
A list of student-submitted discussion questions for Triangle Sum Theorem.
0
## Triangle Sum Theorem Pre Read
To activate prior knowledge, to generate questions about a given topic, and to organize knowledge using a KWL Chart.
0
## Triangle Sum Theorem Post Read
To activate prior knowledge, to generate questions about a given topic, and to organize knowledge using a KWL Chart.
0
## Triangle Sum Theorem Stop and Jot
Learn new vocabulary words and help remember them by coming up with your own sentences with the new words using a Stop and Jot table.
0
## Recognize the Sum of the Interior Angles of a Triangle as 180 Degrees Post Read
Develop understanding of concepts by studying them in a relational manner. Analyze and refine the concept by summarizing the main idea, creating visual aids, and generating questions and comments using a Four Square Concept Matrix.
0
## Understanding the Angle Measures of Triangles
Summarize the main idea of a reading, create visual aids, and come up with new questions using a Four Square Concept Matrix.
0
• Real World Application
## Triangle Sum Theorem
Do the angles in a triangle always add up to 180 degrees? How are the lengths of the sides of a triangle controlled by the sizes of the angles in Spherical Geometry?
0
• Study Guide
## Triangle Relationships Study Guide
This study guide reviews terminology for triangles, different types of triangles, and interior and exterior angles theorems.
1
## Angle Sum Theorem for a Triangle
by CK-12 //basic
A simple interactive that allows you to play with a triangle and see how the angles change when you move on of the vertices.
0
## Angle Activities
by Kyle Bardman //basic
1
## Angle Activities
by Isha //basic
0
• Flashcards
## Understanding the Angle Measures of Triangles
by Ellaine Chou //basic
0
• Flashcards
## Triangle Sum Theorem
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# Cant get a grip on integration. watch
1. So I started an integration module and I feel quite lost. It is the basics of integration and is basically showing how to define an integral in terms of area under a graph (definite) etc. I feel like I can do the questions set which just involve dissecting a given interval and taking the endpoint of all the intervals and adding them together and then take the limit as you get infinitly many strips. But I don't understand how we can take the right end point say and then use that, would the left end point give the same area if used to calculate it that way? Presumably it involves showing that both the lower end point and the upper end point have the same limit in the sum???
I have seen many proofs using sup and inf and delta epsilon proofs (again I think this may be proof of the limits) and I don't really understand them to be honest. Especially one proof that shows that the inf of the upper sum must be less or equal to the sup of the lower sum. Is this trying to show the limit is the same?
Also how can you define an indefinite integral as I only know how to define a definite integral using area, (is there any other way rather than saying it is just the opposite of differentiation).
I really like calculus and want to understand it better but it is confusing me very much.
I understand that this post is not all the clearest but maybe if someone can help me in a small way or link me some good resources that I can check out.
It doesnt help also that my lecturer is chinese and I find it very difficult to understand him as english is not both of our first languages haaaaaa.
Thank you guys!!!!
2. (Original post by poorform)
So I started an integration module and I feel quite lost. It is the basics of integration and is basically showing how to define an integral in terms of area under a graph (definite) etc. I feel like I can do the questions set which just involve dissecting a given interval and taking the endpoint of all the intervals and adding them together and then take the limit as you get infinitly many strips. But I don't understand how we can take the right end point say and then use that, would the left end point give the same area if used to calculate it that way? Presumably it involves showing that both the lower end point and the upper end point have the same limit in the sum???
I have seen many proofs using sup and inf and delta epsilon proofs (again I think this may be proof of the limits) and I don't really understand them to be honest. Especially one proof that shows that the inf of the upper sum must be less or equal to the sup of the lower sum. Is this trying to show the limit is the same?
Also how can you define an indefinite integral as I only know how to define a definite integral using area, (is there any other way rather than saying it is just the opposite of differentiation).
I really like calculus and want to understand it better but it is confusing me very much.
I understand that this post is not all the clearest but maybe if someone can help me in a small way or link me some good resources that I can check out.
It doesnt help also that my lecturer is chinese and I find it very difficult to understand him as english is not both of our first languages haaaaaa.
Thank you guys!!!!
My second course of analysis was just on calculus but unlike you I hate it ...
I plodded along through the course pulling my hair at times, and during exam leave I memorized (no word of a lie) around 50 theorems and lemmas.
I did get a comfortable first on that course, so plod along ...
(what University are you in?)
3. (Original post by poorform)
So I started an integration module and I feel quite lost. It is the basics of integration and is basically showing how to define an integral in terms of area under a graph (definite) etc. I feel like I can do the questions set which just involve dissecting a given interval and taking the endpoint of all the intervals and adding them together and then take the limit as you get infinitly many strips. But I don't understand how we can take the right end point say and then use that, would the left end point give the same area if used to calculate it that way? Presumably it involves showing that both the lower end point and the upper end point have the same limit in the sum???
I have seen many proofs using sup and inf and delta epsilon proofs (again I think this may be proof of the limits) and I don't really understand them to be honest. Especially one proof that shows that the inf of the upper sum must be less or equal to the sup of the lower sum. Is this trying to show the limit is the same?
This all depends on the exact details of how you've defined the integral. That exact definition will have the details required to make sure that you do get the same answer if you take the left end point, or the right end point (or the middle of each interval, for that matter).
It's very hard to give more info without knowing exactly how you've been taught an integral is defined.
Also how can you define an indefinite integral as I only know how to define a definite integral using area, (is there any other way rather than saying it is just the opposite of differentiation).
In analysis terms, an integral is essentially "area under a curve". But if you consider
this will behave very similarly to how you expect the "A-level style" indefinite integral to. (Note that the lower limit c is arbitrary - so this is where you get the arbit constant from on the LHS).
4. (Original post by DFranklin)
This all depends on the exact details of how you've defined the integral. That exact definition will have the details required to make sure that you do get the same answer if you take the left end point, or the right end point (or the middle of each interval, for that matter).
It's very hard to give more info without knowing exactly how you've been taught an integral is defined.
In analysis terms, an integral is essentially "area under a curve". But if you consider
this will behave very similarly to how you expect the "A-level style" indefinite integral to. (Note that the lower limit c is arbitrary - so this is where you get the arbit constant from on the LHS).
Ok thanks I am using the Riemann integral I believe I can't find the exact notes from my prof but I have been using this pdf as the definitions and theorems are the same with just slightly different notation.
https://www.math.ucdavis.edu/~hunter/m125b/ch1.pdf
Seems like good content to me, I spent about an hour looking at it yesterday and trying to get to grips with some of it and I'm definitely getting more comfortable I will see if anyone has anthing to add now I have given more info and try having a go at some exercises and prrofs myself later.
It seems like a lot of it relies on definitions of limits and epsilon delta proofs I believe.
5. I completely forgot to mention it was the Riemann integral apologies.
6. (Original post by poorform)
Ok thanks I am using the Riemann integral I believe I can't find the exact notes from my prof but I have been using this pdf as the definitions and theorems are the same with just slightly different notation.
https://www.math.ucdavis.edu/~hunter/m125b/ch1.pdf
OK, so this is a fairly standard definition involving taking the sup / inf of f on each interval.
But what I now don't understand is how this relates to:
(Original post by poorform)
I feel like I can do the questions set which just involve dissecting a given interval and taking the endpoint of all the subintervals and adding them together and then take the limit as you get infinitly many strips. But I don't understand how we can take the right end point say and then use that, would the left end point give the same area if used to calculate it that way?
Because this doesn't seem to mention sup / inf at all. I strongly suspect you are just evaluating f at a point in each subinterval. In which case if f isn't actually Riemann integrable then it is possible that evaluating at a different point in each subinterval could give a different result.
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https://chemteam.info/GasLaw/Gas-Charles-Problems1-10.html | 1,725,828,471,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651017.43/warc/CC-MAIN-20240908181815-20240908211815-00784.warc.gz | 153,968,591 | 2,506 | ### Charles' LawProblems #1 - 10
Notes:
I used:
V1 / T1 = V2 / T2
to set up the solution for the first few.
Sometimes, you will see the symbolic equation in cross-multiplied form:
V1T2 = V2T1
I set up some solutions toward the end using various permutations of the cross-multiplied form.
In all the problems below, the pressure and the amount of gas are held constant.
Problem #1: Calculate the decrease in temperature (in Celsius) when 2.00 L at 21.0 °C is compressed to 1.00 L.
Solution:
(2.00 L) / 294.0 K) = (1.00 L) / (x)
cross multiply to get:
2x = 293
x = 147.0 K
Converting 147.0 K to Celsius, we find -126.0 °C, for a total decrease of 147.0 °C, from 21.0 °C to -126.0 °C.
Problem #2: 600.0 mL of air is at 20.0 °C. What is the volume at 60.0 °C?
Solution:
(600.0 mL) / (293.0) = (x) / (333.0 K)
x = 682 mL
Problem #3: A gas occupies 900.0 mL at a temperature of 27.0 °C. What is the volume at 132.0 °C?
Solution:
(900.0 mL) / (300.0 K) = (x) / (405.0 K)
x = 1215 mL
Problem #4: What change in volume results if 60.0 mL of gas is cooled from 33.0 °C to 5.00 °C?
Solution:
(60.0 mL) / (306.0 K) = (x) / (278.00 K)
Cross multiply to get:
306x = 16680
x = 54.5 mL <--- that's the ending volume, which is NOT the answer
The volume decreases by 5.5 mL.
Problem #5: Given 300.0 mL of a gas at 17.0 °C. What is its volume at 10.0 °C?
Solution:
In cross-multiplied form, it is this:
V1T2 = V2T1
V2 = (V1T2) / T1 <--- divided both sides by T1
x = [(300.0 mL) (283.0 K)] / 290.0 K
Problem #6: A gas occupies 1.00 L at standard temperature. What is the volume at 333.0 °C?
Solution:
In cross-multiplied form, it is this:
V1T2 = V2T1
V2 = (V1) [T2 / T1] <--- notice how I grouped the temperatures together
x = (1.00 L) [(606.0 K) / (273.0 K)]
x = 2.22 L
Problem #7: At 27.00 °C a gas has a volume of 6.00 L. What will the volume be at 150.0 °C?
Solution:
Two different set-ups:
(6.00 L) / (300.0 K) = (x) / (423.0 K)
or
(6.00 L) (423.0 K) = (x) (300.0 K)
x = 8.46 L
Problem #8: At 225.0 °C a gas has a volume of 400.0 mL. What is the volume of this gas at 127.0 °C?
Solution:
From #6:
V2 = (V1) [T2 / T1]
x = (400.0 mL) [(400.0 K) / (498.0 K)
x = 321 mL
Here's the "traditional" way:
(400.0 mL) / (498.0 K) = (x) / (400.0 K)
Problem #9: At 210.0 °C a gas has a volume of 8.00 L. What is the volume of this gas at -23.0 °C?
Solution:
(8.00 L) / (483.0 K) = (x) / (250.0 K)
Note how you can have a negative Celsius temperature, but not a negative Kelvin temperature.
Problem #10: When the volume of a gas is changed from ___ mL to 852 mL, the temperature will change from 315 °C to 452 °C. What is the starting volume?
Solution:
Write Charles Law and substitute values in:
V1 / T1 = V2 / T2
x / 588 K = 852 mL / 725 K
(x) (725 K) = (852 mL) (588 K)
x = 691 mL
Note the large °C values, trying to get you to forget to add 273. Remember, only Kelvin temperatures are allowed in the calculations.
Bonus Problem: An open "empty" 2 L plastic pop container, which has an actual inside volume of 2.05 L, is removed from a refrigerator at 5 °C and allowed to warm up to 21 °C. What volume of air measured at 21 °C, will leave the container as it warms?
Solution:
2.05 L / 278 K = V2 / 294 K
Calculate V2. The volume that "escapes" is V2 minus 2.05 L
Usually, a Charles' Law problem asks for what the volume is at the end (the V2 in this question) or at the start, before some temperature change. This question asks you for the difference between V1 and V2. It's not hard to solve, it's just that it doesn't get asked very often in a Charles' Law setting. | 1,264 | 3,616 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-38 | latest | en | 0.828027 |
http://mathhelpforum.com/pre-calculus/103388-not-sure-what-i-should-do-print.html | 1,527,254,257,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867092.48/warc/CC-MAIN-20180525121739-20180525141739-00555.warc.gz | 180,693,828 | 3,162 | # Not sure what I should do...
• Sep 20th 2009, 06:14 PM
Morgan82
Not sure what I should do...
A racehorse is running a 10-furlong race. (1 furlong = 220 yards)
As the horse passes each furlong marker (F), its time is recorded.
F| 0 1 2 3 4 5 6 7 8 9 10
T| 0 20 33 46 59 73 86 100 112 124 135
a) What is the approx speed of the horse as it passes the 3rd marker?
(Do I need to come up with an equation?...and find the derivative, to get the speed at that point?)
• Sep 20th 2009, 06:29 PM
tutor.mathitutor
You can calculate the average speed by dividing the distance by time
= (3x220/33)yards/sec = (220/11)yards/sec
= 20 yards/sec
• Sep 20th 2009, 11:11 PM
pflo
Quote:
Originally Posted by Morgan82
A racehorse is running a 10-furlong race. (1 furlong = 220 yards)
As the horse passes each furlong marker (F), its time is recorded.
F| 0 1 2 3 4 5 6 7 8 9 10
T| 0 20 33 46 59 73 86 100 112 124 135
a) What is the approx speed of the horse as it passes the 3rd marker?
(Do I need to come up with an equation?...and find the derivative, to get the speed at that point?)
In order to get the actual rate of travel at the third marker you would be best off getting a derivative of an equation, but it would be pretty hard to come up with an equation. The essential word here is you only need the 'approx' speed.
The 'average rate of change' can be calculated from the data. One way to do this is to say the horse has gone 660 yards in 46 seconds, which would be 14.35 yards per second (you can convert units to mph if you want). This would represent the average speed over the first 3 furlongs. A better approximation might be to look at the average speed between the 2nd and 4th furlong markers. It took the horse 26 seconds to cover that 440 yards, which would indicate a speed of only 16.92 yards per second.
It looks like the horse is losing speed over this first part of the race! | 580 | 1,889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-22 | latest | en | 0.959225 |
https://www.physicsforums.com/threads/projectile-motion-question-with-angles-and-velocity.222124/ | 1,575,605,630,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540484477.5/warc/CC-MAIN-20191206023204-20191206051204-00397.warc.gz | 836,570,796 | 15,269 | # Projectile motion question with angles and velocity
1. Homework Statement
Joe Carter hit a fly ball during the spring training. It just cleared a 10m high vertical fence on the way down at 45 degrees as shown and struck the level ground 8m beyond the fence. Calculate the speed of the ball when it left his bat at ground level (start and end points are the same, ignore air resistance). Answer in m/s.
2. Homework Equations
∆x = vit
5 kinimatics equations for y direction
3. The Attempt at a Solution
Since air resistance is negligible, and the ball starts and ends at the same level, vi = vf
a = 9.8 m/s down
We can break equation into 2 parts - part 1 before fence and part 2 after
For Part 2
∆dix = 8m
∆diy = 10m
a = 9.8 m/s (down)
vf = vi of part 1
vf = ?
It seems we're missing time and vf, two variables of the equation for part 2, and several variables for part 1. If you guys can help me figure this out that'd be sweet.
Related Introductory Physics Homework Help News on Phys.org
Wherez the figure?
Use coordinates with the spot where the ball cleared the fence as origin. The time coordinate will also be 0 when this happens. Call the speed of the ball at this point v.
This is the initial speed.
Since we know the angle we have v_x = cos(45)v and v_y = -sin(45)v.
note that cos(45) = sin(45)
Now write down expression for x and y as a function of t.
x(t) = ....
y(t) = ....
The ball will hit the ground at x=8, y=-10. call the time this happens T.
if you then set x(T) = 8 and y(T) =-10 you get 2 equations for v and T.
The v you find from that is at the top of the fence, so you'll need to compute v_x(T) and v_y(T) to get the speed at the point the ball hits the ground. | 466 | 1,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-51 | latest | en | 0.907957 |
http://math.stackexchange.com/questions/245080/weird-integration | 1,467,348,370,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783400031.51/warc/CC-MAIN-20160624155000-00156-ip-10-164-35-72.ec2.internal.warc.gz | 202,471,471 | 18,527 | # Weird integration.
I have an example of an equation where I don't know how the right hand side is derived from left hand side. It looks very weird as the differential changes. Notice that $m$ and $c$ are constants.
$$\frac{m}{2} \int \frac{\textrm{d}v^2}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{-mc^2}{2} \int \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \mathrm d\left(1-\frac{v^2}{c^2}\right)$$
Which rules apply here? Could anyone explain this or provide me with links to any good websites? Than you.
-
Since you say that the differential changes, there seems to be a typo -- the "differential" is actually missing on the left-hand side. – joriki Nov 26 '12 at 17:29
It looks like the substitution $u=1-\frac{v^2}{c^2}$, but if so it is not done correctly, the $v^2$ on top has mysteriously disappeared. How does it finish, so one can check whether it is a typo or a mistake? – André Nicolas Nov 26 '12 at 17:30
Something looks wrong here: $$d\left(1-\frac{v^2}{c^2}\right)=-\frac{2v}{c^2}$$and the $\,c^2\,$ in the denominator and the minus sign are "balaanced" by the constant in the right side, yet the $\,2v\,$ doesn't appear anymore... – DonAntonio Nov 26 '12 at 17:33
@joriki I did add the differential on the LHS. It was a typo yes. – 71GA Nov 26 '12 at 17:40
This is a part of a derivation i took from Wikipedia article "Relativistic kinetic energy" here – 71GA Nov 26 '12 at 17:46
What you want is to get a diferential that contains whats inside the square root in the denominator, so you create a derivative that contains both $dv^2$ and $d\left(1-\frac{v^2}{c^2}\right)$. It is easy to do so, and we see that this one can be used: $$\frac{d\left(1-\frac{v^2}{c^2}\right)}{d(v^2)}=-\frac{1}{c^2}$$ Isolating $d(v^2)$ we get: $$d(v^2)=-c^2 d\left(1-\frac{v^2}{c^2}\right)$$
-
In first equation, how did you know that LHS equals $-\frac{1}{c^2}$? – 71GA Nov 26 '12 at 20:10
Oh never mind i just nave to derive this as it was (please confirm) $$\frac{\textrm d \left(1-\frac{x}{c^2}\right)}{\textrm d x}$$ and this equals $$0 - \frac{1}{c^2} = - \frac{1}{c^2}$$ – 71GA Nov 26 '12 at 21:24
Yes, its just a simple derivative. – Ivan Lerner Nov 27 '12 at 3:21
$$\mathrm dv^2=c^2\mathrm d\frac{v^2}{c^2}=-c^2\mathrm d\left(-\frac{v^2}{c^2}\right)=-c^2\mathrm d\left(1-\frac{v^2}{c^2}\right)\;.$$
-
Thank you. This is what i needed. And how i needed it. – 71GA Nov 26 '12 at 21:14
There is ony one disturbing thing to me... Why does it hold that $$\textrm d \left(-\frac{v^2}{c^2}\right) = \textrm{d} \left(1 - \frac{v^2}{c^2} \right )$$ – 71GA Nov 26 '12 at 21:19
Added: The answer belows tries to address the OP as it was, assuming it was $\,dv\,$ and not $\,dv^2\,$ as it happened to be later.
Imo, it must be
$$\frac{m}{2}\int\frac{v^2}{\sqrt{1-\frac{v^2}{c^2}}}dv=\frac{m}{2}\int\frac{v}{\sqrt{1-\frac{v}{c^2}}}\left(-\frac{2v}{c^2}\right)\left(-\frac{c^2}{2}\right)dv=$$
$$=-\frac{mc^2}{4}\int\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}d\left(1-\frac{v^2}{c^2}\right)$$
-
The differential on the LHS is $\mathrm d v^2$ and not $\mathrm d v$ you used. I am sorry for the confusion, but it was a typo i fixed now. – 71GA Nov 26 '12 at 17:42 | 1,182 | 3,139 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2016-26 | latest | en | 0.872714 |
https://www.eduzip.com/ask/question/in-the-figure-pqr-is-a-straight-line-find-x-then-complete-the-fol-520919 | 1,642,914,264,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304134.13/warc/CC-MAIN-20220123045449-20220123075449-00550.warc.gz | 800,642,364 | 8,825 | Mathematics
# In the figure, $PQR$ is a straight line. Find $x$. Then complete the following :(i) $\angle AQB=$(ii) $\angle BQP=$(iii) $\angle AQR=$
##### SOLUTION
In the given fig., $\angle{PQA}, \angle{AQB}$ and $\angle{BQR}$ forms a linear pair.
Therefore,
$\angle{PQA} + \angle{AQB} + \angle{BQR} = 180°$
$\left( x + 20° \right) + \left( 2x + 10° \right) + \left( x - 10° \right) = 180°$
$4x + 20° = 180°$
$\Rightarrow x = \cfrac{180° - 20°}{4} = 40°$
Therefore,
$\angle{AQB} = 2x + 10° = 2 \times 40° + 10° = 90°$
$\angle{BQP} = \left( x + 20° \right) + \left( 2x + 10° \right) = 3x + 30° = 3 \times 40° + 30° = 150°$
$\angle{AQR} = \left( 2x + 10° \right) + \left( x - 10° \right) = 3x = 3 \times 40° = 120°$
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1 Verified Answer | Published on 09th 09, 2020 | 641 | 1,887 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2022-05 | latest | en | 0.727237 |
https://puzzling.stackexchange.com/questions/77737/from-2019-to-digits | 1,718,866,881,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861883.41/warc/CC-MAIN-20240620043158-20240620073158-00498.warc.gz | 414,924,458 | 38,479 | # From 2019 to digits
Is it possible to obtain the digits from 0 to 9 starting from 2019 and using its digits in the same order, together with the usual operations +, *, -, /, concatenation of digits, and the less usual operators ^, !, sqrt(), int()? For example, 1 = 20-19. Unary minus is allowed too.
I manage to use only basic operations and elevation to a power for all digits except 4 and 5, but maybe somebody will do better!
$$0 = 2 \cdot 0 \cdot 1 \cdot 9$$
$$1 = 20 - 19$$
$$2 = 2^0 + 1^9$$
$$3 = 2+0+1^9$$
$$4 = \lfloor \sqrt{20} \rfloor + \lfloor 1/9 \rfloor$$
$$5 = 2 + 0 \cdot 1 + \sqrt{9}$$
$$6 = -(2+0+1) + 9$$
$$7 = -(2+0 \cdot 1) + 9$$
$$8 = -(2 \cdot 0 + 1) + 9$$
$$9 = -(2 \cdot 0 \cdot 1) + 9$$
• @deepthought Fixed. Commented Dec 23, 2018 at 21:29
• Nice! 4 may also be $2\cdot 0 + 1 + \sqrt 9$
– mau
Commented Dec 23, 2018 at 21:48
• A few more for 4 and 5: $\lfloor\sqrt{\sqrt{20\times19}}\rfloor=4$, $\lceil\sqrt{\sqrt{20\times19}}\rceil=5$ , $\lceil\sqrt{2+0+1+9}\rceil=4$ , $\lfloor\sqrt{2+0+19}\rfloor=4$ , $\lceil\sqrt{2+0+19}\rceil=5$
– JMP
Commented Dec 23, 2018 at 22:02
• and $\lfloor\sqrt{201}\rfloor-9=5$
– JMP
Commented Dec 23, 2018 at 22:05
• Fun fact: $-\lfloor -x \rfloor = \lceil x \rceil.$ Commented Dec 23, 2018 at 22:55 | 545 | 1,274 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-26 | latest | en | 0.714157 |
http://urbanability.org/central_limit_theorem.html | 1,653,051,665,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662532032.9/warc/CC-MAIN-20220520124557-20220520154557-00123.warc.gz | 55,688,584 | 32,627 | • General
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# Central Limit Theorem (CLT)
## What Is the Central Limit Theorem (CLT)?
In probability theory, the central limit theorem (CLT) states that the distribution of a sample variable approximates a normal distribution (i.e., a “bell curve”) as the sample size becomes larger, assuming that all samples are identical in size, and regardless of the population's actual distribution shape.
Put another way, CLT is a statistical premise that, given a sufficiently large sample size from a population with a finite level of variance, the mean of all sampled variables from the same population will be approximately equal to the mean of the whole population. Furthermore, these samples approximate a normal distribution, with their variances being approximately equal to the variance of the population as the sample size gets larger, according to the law of large numbers.
Although this concept was first developed by Abraham de Moivre in 1733, it was not formalized until 1930, when noted Hungarian mathematician George Polya dubbed it the Central Limit Theorem.
### Key Takeaways
• The central limit theorem (CLT) states that the distribution of sample means approximates a normal distribution as the sample size gets larger, regardless of the population's distribution.
• Sample sizes equal to or greater than 30 are often considered sufficient for the CLT to hold.
• A key aspect of CLT is that the average of the sample means and standard deviations will equal the population mean and standard deviation.
• A sufficiently large sample size can predict the characteristics of a population more accurately.
1:22
## Understanding the Central Limit Theorem (CLT)
According to the central limit theorem, the mean of a sample of data will be closer to the mean of the overall population in question, as the sample size increases, notwithstanding the actual distribution of the data. In other words, the data is accurate whether the distribution is normal or aberrant.
As a general rule, sample sizes of around 30-50 are deemed sufficient for the CLT to hold, meaning that the distribution of the sample means is fairly normally distributed. Therefore, the more samples one takes, the more the graphed results take the shape of a normal distribution. Note, however, that the central limit theory will still be approximated in many cases for much smaller sample sizes, such as n=8 or n=5.
The central limit theorem is often used in conjunction with the law of large numbers, which states that the average of the sample means and standard deviations will come closer to equaling the population mean and standard deviation as the sample size grows, which is extremely useful in accurately predicting the characteristics of populations.
## The Central Limit Theorem in Finance
The CLT is useful when examining the returns of an individual stock or broader indices, because the analysis is simple, due to the relative ease of generating the necessary financial data. Consequently, investors of all types rely on the CLT to analyze stock returns, construct portfolios, and manage risk.
Say, for example, an investor wishes to analyze the overall return for a stock index that comprises 1,000 equities. In this scenario, that investor may simply study a random sample of stocks to cultivate estimated returns of the total index. To be safe, use at least 30-50 randomly selected stocks across various sectors, should be sampled for the central limit theorem to hold. Furthermore, previously selected stocks must be swapped out with different names to help eliminate bias.
### Article Sources
Investopedia requires writers to use primary sources to support their work. These include white papers, government data, original reporting, and interviews with industry experts. We also reference original research from other reputable publishers where appropriate. You can learn more about the standards we follow in producing accurate, unbiased content in our editorial policy.
1. George Polya. "Uber den Zentralen Grenzwertsatz der Wahrscheinlichkeit-Srechnung und das Momentenproblem." Mathematische Zeitschrift, Volume 8, 1920, Pages 171-181, 1920.
2. Encyclopaedia Britannica. "Abraham de Moivre." Accessed Aug. 4, 2021.
3. Sheldom M. Ross. "Introductory Statistics," Section 7.4. Academic Press, 2017. | 907 | 4,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-21 | latest | en | 0.925708 |
https://www.kopykitab.com/blog/rs-aggarwal-class-8-maths-chapter-10-ex-10-1-solutions/ | 1,642,480,207,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300722.91/warc/CC-MAIN-20220118032342-20220118062342-00322.warc.gz | 920,975,832 | 26,978 | # RS Aggarwal Class 8 Maths Chapter 10 Ex 10.1 Solutions 2022 | Download Free PDF
RS Aggarwal Class 8 Maths Chapter 10 Ex 10.1 Solutions: In this exercise, the students will study how to calculate the cost price & the selling price. The solutions cover all solutions to every question in a simple & easy method. The word problems are prepared to step by step with detailed explanations that enable the students to easily learn the topics & formulae. Practicing these solutions is the best and more reliable way to obtain higher marks in the exams.
RS Aggarwal Class 8 Maths Chapter 10 Ex 10.1 Solutions considered as a helpful study refresher that provides a quick revision of each concept along with important formulas included in this exercise. These solutions are very helpful for the students to study the concepts in a simple & structured manner. This also enhances their analytical abilities in order to acquire excellent marks in the exam.
## Download RS Aggarwal Class 8 Maths Chapter 10 Ex 10.1 Solutions
RS Aggarwal Class 8 Maths Chapter 10 Ex 10.1 Solutions
## Important Definition for RS Aggarwal Class 8 Maths Chapter 10 Ex 10.1 Solutions
The solutions provide in-depth knowledge about all concepts included in this exercise in a simple manner. The students can refer to this solutions exercise which they can easily download and access at any time and anywhere. The students can develop problem-solving abilities and can able to solve any type of questions easily with the help of these solutions.
RS Aggarwal Class 8 Maths Chapter 10 Ex 10.1 Solutions assist the students to know the best approach to solve all the problems of this chapter without any difficulty. These solutions enable the students to quickly revise all the concepts during Maths final exams.
• Formulas to calculate the cost price
(i) If the Selling Price of the product & gain on its selling is given, the Cost Price of the product will be Selling Price minus gain i.e Cost Price = Selling Price – Gain
(ii) If the Selling Price of an article & loss on its selling is given, the Cost Price of the article will be Selling Price plus loss i.e Cost Price = Selling Price + Loss
• Formula for Selling Price
(i) Selling Price = {(100 + Gain %) / 100} x Cost Price
(ii) Selling Price = {(100 – Loss %) / 100} x Cost Price
(iii) Selling Price = Cost Price + Profit
(iv) Selling Price = Cost Price – loss
## Benefits of RS Aggarwal Class 8 Maths Chapter 10 Ex 10.1 Solutions
• These solutions are intended to assist the students in their Maths final exam preparation.
• The solutions assist the students to develop the knowledge of the Maths subject in a simple and interactive way.
• These solutions clear all doubts of the students related to the calculation of the cost price & the selling price.
• The students can refer to the exercise solutions which are appropriate & reliable that enables them to understand this exercise in a better way.
• The students can easily download & access the solutions PDF at any time and anywhere.
Know more at the official website. | 657 | 3,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2022-05 | latest | en | 0.921017 |
https://www.abcassignmenthelp.com/comb-sort-algorithm-assignment-help | 1,685,571,407,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647459.8/warc/CC-MAIN-20230531214247-20230601004247-00077.warc.gz | 705,522,912 | 29,100 | # Comb Sort Algorithm Assignment Help
Are you stressed about the allocated assignment on the subject of Comb Sort Algorithm Assignment Help? Are you thinking about will you be able to deliver the assignment on time? Algorithm Assignment is a huge subject and it exclusively requires time and knowledge to prepare an assignment.
ABC Assignment Help is a reliable Comb Sort Algorithm Assignment Help service provider helping a number of scholars who are facing difficulties in preparing an assignment on Algorithm.
## What is Comb Sort Algorithm?
1) Comb sort improves on bubble sort by using gap of size more than 1.
2) The gap starts with a large value and shrinks by a factor of 1.3 in every iteration until it reaches the value 1.
EXAMPLE: To start with we have the following array: 4 6 2 3 7 1 8 0 9 5
Iteration 1: Gap : 8
Input 4 6 2 3 7 1 8 0 9 5
Index 0 with index8- we find the elements in order, hence no swap needed.
4 6 2 3 7 1 8 0 9 5
Index 1 with index 9 – we find the elements not in order, hence swap needed.
Iteration 2:- Gap : 6
Input 4 5 2 3 7 1 8 0 9 6
Index 0 with index 6 – we find the elements in order, hence no swap needed.
4 5 2 3 7 1 8 0 9 6
Index 1 with index 7 – we find the elements not in order, hence swap needed.
4 0 2 3 7 1 8 0 9 6
Index 2 with index 8 – we find the elements in order, hence no swap needed.
4 0 2 3 7 1 8 5 9 6
Index 3 with index 9 – we find the elements in order, hence no swap needed.
RESULT 4 0 2 3 7 1 8 5 9 6
Iteration 3:- Gap : 4
Input 4 0 2 3 7 1 8 5 9 6
Index 0 with index 4 – we find the elements in order, hence no swap needed.
4 0 2 3 7 1 8 5 9 6
Index 1 with index 5 – we find the elements in order, hence no swap needed.
4 0 2 3 7 1 8 5 9 6
Index 2 with index 6 – we find the elements in order, hence no swap needed.
4 0 2 3 7 1 8 5 9 6
Index 3 with index 7 – we find the elements in order, hence no swap needed.
4 0 2 3 7 1 8 5 9 6
Index 4 with index 8 – we find the elements in order, hence no swap needed.
4 0 2 3 7 1 8 5 9 6
Index 5 with index 9 – we find the elements in order, hence no swap needed.
RESULT 4 0 2 3 7 1 8 5 9 6
Iteration 4:- Gap : 3
Input 4 0 2 3 7 1 8 5 9 6
Index 0 with index 3- we find the elements not in order, hence swap needed.
3 0 2 4 7 1 8 5 9 6
Index 1 with index 4 – we find the elements in order, hence no swap needed.
3 0 2 4 7 1 8 5 9 6
Index 2 with index 5 – we find the elements not in order, hence swap needed.
3 0 1 4 7 2 8 5 9 6
Index 3 with index 6 – we find the elements in order, hence no swap needed.
3 0 1 4 7 2 8 5 9 6
Index 4 with index 7- we find the elements not in order, hence swap needed.
3 0 1 4 5 2 8 7 9 6
Index 5 with index 8 – we find the elements in order, hence no swap needed.
3 0 1 4 5 2 8 7 9 6
Index 6 with index 9 – we find the elements not in order, hence swap needed.
RESULT 3 0 1 4 5 2 6 7 9 8 | 1,070 | 2,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-23 | latest | en | 0.797638 |
https://www.physicsforums.com/threads/calculate-balloon-volume-from-helium-mass-199-kg.587638/ | 1,726,251,086,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00449.warc.gz | 872,179,619 | 16,747 | # Calculate Balloon Volume from Helium Mass: 199 kg
• 40856
In summary, the total mass of helium in a balloon is 199 kg. There is no simple formula for determining the volume of the balloon without knowing the radius. However, the book states that each cubic foot of helium can lift 0.069 lbs, so this information may be useful. One possible approach is to convert the mass of helium to pounds and divide it by 0.069, which would give a volume of 35.621 cubic feet.
40856
The total mass of helium in a balloon is 199 kg. What is the volume of the balloon? Enter only the numeric portion of the answer, not the units.
That is the problem. I am having trouble locating a simple forumla for this. All I can seem to find involve having the radius of the balloon which I do not have. In the book it says each cubic foot of helium can lift .069 lb, so this may play a factor but I am unsure of how exactly. I have thought about converting 199KG to lbs and dividing by .069. Is this thinking correct?
Helium = 0.179 so maybe I can take 199kg times .179 and that 35.621 kg is this correct?
## 1. How do I calculate the volume of a balloon from its helium mass?
To calculate the volume of a balloon from its helium mass, you will need to use the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the universal gas constant, and T is temperature. You will also need to know the molar mass of helium, which is 4.0026 g/mol.
## 2. What is the ideal gas law and how does it relate to calculating balloon volume from helium mass?
The ideal gas law is a mathematical equation that describes the behavior of an ideal gas. It states that the product of pressure and volume is proportional to the number of moles of gas, the universal gas constant, and the temperature. This law can be applied to calculate the volume of a balloon from its helium mass, as long as the conditions of the gas are ideal.
## 3. Can I use the ideal gas law to calculate the volume of any gas-filled balloon?
No, you can only use the ideal gas law to calculate the volume of a gas-filled balloon if the conditions of the gas are ideal. This means that the gas must be at a constant temperature and pressure, and the particles must have no volume or interactions with each other.
## 4. What other factors besides helium mass affect the volume of a balloon?
Besides helium mass, the volume of a balloon can also be affected by the type of material the balloon is made of, the temperature of the gas inside the balloon, and the external pressure on the balloon. These factors can impact the elasticity and density of the balloon, which can affect its volume.
## 5. Is there a limit to the volume a balloon can reach based on its helium mass?
Yes, there is a limit to the volume a balloon can reach based on its helium mass. This is because the ideal gas law only applies to an ideal gas, and as the volume of the balloon increases, the gas inside becomes less ideal and the law becomes less accurate. Eventually, the balloon will reach a point where it can no longer expand and its volume will remain constant.
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7K | 812 | 3,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-38 | latest | en | 0.936943 |
https://www.edge.org/print/conversation/nassim_nicholas_taleb-the-hard-problem | 1,653,782,758,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663021405.92/warc/CC-MAIN-20220528220030-20220529010030-00455.warc.gz | 836,166,387 | 8,969 | # THE HARD PROBLEM[1]
Slide file [2]
[NASSIM TALEB:] My specialty and fear and obsession is with small probabilities. We don't quite understand small probabilities. Let's discuss. [slide [3]]
You often see in the papers things saying events we just saw should happen every ten thousand years, hundred thousand years, ten billion years. Some faculty here in this university had an event and said that a 10-sigma event should happen every, I don't know how many billion years. Do you ever regard how worrisome it is, when someone makes a statement like that, "it should happen every ten thousand years," particularly when the person is not even two thousand years old?
So the fundamental problem of small probabilities is that rare events don't show in samples, because they are rare. So when someone makes a statement that this in the financial markets should happen every ten thousand years, visibly they are not making a statement based on empirical evidence, or computation of the odds, but based on what? On some model, some theory.
So, the lower the probability, the more theory you need to be able to compute it. Typically it's something called extrapolation, based on regular events and you extend something to what you call the tails.
So my telescope problem is as follows. [slide [4]]
Consider the following. The smaller the probability, the less you observe it in a sample, therefore your error rate in computing it is going to be very high. Actually, your relative error rate can be infinite, because you're computing a very, very small probability, so your error rate is monstrous and probably very, very small. The problem we have now is that we don't care about probabilities. Probabilities are something you do in the classroom. What you care about is what you are going to use the probability for, what I have here called lambda of the impact of the event, the effect of the event, the consequence of the event.
So you're interested in the pair, probability times impact. So notice something, that the vicious aspect is that the pair, probability times effect by lambda, as the probability becomes smaller, the impact is bigger. A thousand year flood is going to be definitely vastly more consequential than the five year flood, or a hundred year flood.
So you care about that, and that pi times lambda becomes a lot more stochastic when pi gets small. In other words, very consequential events, we know absolutely nothing about and that matter a lot, particularly in some domains in which the probabilities don't fall very quickly.
This is what got me into this obsessive disorder [slide [5]] — when I was a trader during the stock market crash of 1987, as I was a derivatives trader. The beauty of derivatives, or the ugliness of derivatives, is that they compound every single error you can have because it's a square, based on a higher order function of random variables.
And you see here the variation of a derivatives portfolio over twenty years. One day represents 97 percent of the variations. And that day is not forecastable, OK, and that's the problem of small probabilities. There are domains in which small probabilities aren't everything – very, very small probabilities, and we know very little about them.
What are these domains? [slide [6]] There's a regular classification. I won't get into the details of the way these things are made, but there is type-1, I call something that reaches central limit in real time, and type-2 that doesn't reach it in real time. The classical distinction is between the thing they call Gaussian or Gaussian family or power law families — it doesn't work very well. It's much more effective to have what we call in real time, a real time central limit. In other words, the rare event in type-2 will dominate the properties regardless of what theory you're using.
Now what worries me. I'm here at the epicenter of the education, of Economics at Harvard, and there is something at the foundation of Economics that is completely bogus. Completely. And let me tell you what it is. Its not the problem that people are rational or irrational or people keep bickering about more problems. It's the fact that, at the center of the techniques used, in econometrics and statistics used, at the center there is this assumption that doesn't seem to hold theoretically — I mean philosophically a priori, defaulting to the type-1 is completely arbitrary and unjustified, the second one empirically. And people tell me they have techniques to adjust for fat tails. (Fat tails mean the contribution of the rare events). Not only can we not predict rare events, we cannot even figure out what role they play in the data.
This graph shows you the following. [slide [7]] I took every economic variable I could find data for that covers 40 plus years, from stock, exchange rates, stock markets — everything that had a lot of data — everything I could find — 20 some million pieces of data — and for most of these variables, one day in 40 years represent 90 percent of a measure called Kurtosis . Kurtosis tells you how much something is not Gaussian (conditional on the distribution being type-1). In other words, you don't even know how non-Gaussian we are. That is a severe problem.
For example, take the stock market crash in 1987: one day in 40 years represents 80-some percent of the Kurtosis, of that measure.
So we are in trouble, very severe trouble. [slide [8]] Even if you used another method called power law, which is beautiful on paper, much more rigorous scientifically, much more adapted to our world, we still have another problem: that its beautiful on paper, but when you come to calibrate, small differences in calibration lead to massively different probabilities for rare events — very, very small, minute differences in calibration. And the errors we have in measuring what these tails are, are monstrous.
Measurability. [slide [9]] Number one. That what we call the Norm L2, nobody should use when you have open ended payoff in economics, nobody should use at all something called variance, Least-Square methods, standard deviations, Sharpe ratio , portfolio theory, even the word correlation. These don't work in the domains of type-2. They don't work. That explains why the system collapsed because banks were using…. I spent 15 years fighting risk management methods in banks, they were all based on standard deviation portfolio theory because they say Nobels were proposing these things.
Second point. Social science has made some distinctions, which I don't think are very rigorous, between what they call Knightian measurable risk and what they call not measurable uncertainty. Several problems. [slide [10]]
Number one, when you talk about uncertainty, you don't say measuring risk, you don't measure a risk, like I'm measuring this table. I'm estimating something that happened would happen in the future on one hand, and on the other one, I am measuring a certain physical entity. I'm measuring with an error of type-1 and type-2. The other problem with the Knightian stuff is that all risks in the tails, all small probabilities, are not measurable, regardless of whether you know what's going on or you don't know what's going on.
This brings me to the consequences. [slide [11]] Visibly forecasting. I don't know if you guys are aware, but you can check all forecasts in economics and they don't work. Why don't they work? Because of the rare events. Unless of course you don't have exposure to rare events, which, with these random variables, is not the case.
I have to specify some sub problems and then go to my main problems [slide [12]].
The hard problem is the classical problem of induction, inverse problems, you don't observe the generator; you don't know what processes you're dealing with. [slide [13]] There are a series of smaller problems that we can probably solve for, something called the Ludic Fallacy, the fact that probabilities that are not as observable in real life as they are in games. There are a lot of problems associated with my general thing, but my main point is that small probabilities are not measurable.
Now, what do you do? My solution is in the following way. [slide [14]]
There are two kinds of decisions you make in life, decisions that depend on small probabilities, and decisions that do not depend on small probabilities. For example, if I'm doing an experiment that is true-false, I don't really care about that pi-lambda effect, in other words, if its very false or false, it doesn't make a difference. So for example if I'm doing tests on individuals in medicine, I'm not really affected by tale events, but if I'm studying epidemics, then the random variable how many people are effected becomes open-ended with consequences so therefore it depends on fat tails. So I have kinds of decisions. One simple, true-false, and one more complicated, like the ones we have in economics, financial decision-making, a lot of things, I call them M1, M1+.
And then we have, as we saw, two types of worlds [slide [12]], I have a world of thin tails and a world of fat tails. Two worlds. And I discovered the following, and this I learned from dealing with statisticians, is that if you tell them "your techniques don't work," they get angry, but if I tell them, "this is where your techniques work," they buy you a diet coke, and try to be nice, say hello in the elevator. So, I tried to make a map of where there is, what I call, a Black Swan problem. [slide [15]] It is when you have complex decisions, decisions that are effected by how important the move is, and we can't really compute small probabilities.
So, the four quadrants. That fourth quadrant is the limits of where we can do things, and I think what we need to do here — that's my hard problem — is try to define how to be a robust, what robustness means to Black Swans, and how to be robust to the Black Swans, in this quadrant. [slide [15]]
We cannot escape it unfortunately in finance, ever since we left the stone-age, our random variables became more and more complex. We cannot escape it. We can become more robust. There are techniques to become more robust. The notion of robustness became my recent obsession. How to be a robust? It's not clear; it's the hard problem — and that's my problem. | 2,230 | 10,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-21 | latest | en | 0.970057 |
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Tutorial problems Introduction to MATLAB Problem1 A simple sine plot: plot y = sin x, 0 <= x <= 2*pi, taking 100 linearly spaced points in the given interval. Label the axes and put a title Solution 1: MATLAB Program x=linspace(0,2*pi,100); figure(1) plot(x,sin(x)) xlabel('x'), ylabel('sin(x)') title('sine plot') grid on Problem 2 An exponentially decaying sine plot: plot y = e -0.4x sin(x), 0 <= x <= 4*pi, taking 10, 50, and 100 points in the interval. Solution 1: MATLAB Program x =linspace(0,4*pi,10); % 10 points y = exp(-0.4*x).*sin(x); plot(x,y) hold on x =linspace(0,4*pi,50); % 50 points y = exp(-0.4*x).*sin(x); plot(x,y) hold on x =linspace(0,4*pi,100); % 100 points y = exp(-0.4*x).*sin(x); plot(x,y) xlabel('X') ylabel('Exponential decay function') Problem 3: Draw a circle with given radius, using Script Programming in MATLAB Solution 3 MATLAB Program % circle a script file to draw a unit circle r = input(‘Enter the radius of the circle’) theta = linspace(0,2*pi,100); x = r * cos(theta) ; y = r * sin(theta) ; plot(x,y) axis(‘equal’) title(‘Circle with unit radius’)
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### Transcript of 6 MatLab Tutorial Problems
Tutorial problems Introduction to MATLAB
Problem1 A simple sine plot: plot y = sin x, 0 <= x <= 2*pi, taking 100 linearly spaced points in the given interval. Label the axes and put a title Solution 1: MATLAB Program x=linspace(0,2*pi,100); figure(1) plot(x,sin(x)) xlabel('x'), ylabel('sin(x)') title('sine plot') grid on Problem 2 An exponentially decaying sine plot: plot y = e-0.4x sin(x), 0 <= x <= 4*pi, taking 10, 50, and 100 points in the interval. Solution 1: MATLAB Program x =linspace(0,4*pi,10); % 10 points y = exp(-0.4*x).*sin(x); plot(x,y) hold on x =linspace(0,4*pi,50); % 50 points y = exp(-0.4*x).*sin(x); plot(x,y) hold on x =linspace(0,4*pi,100); % 100 points y = exp(-0.4*x).*sin(x); plot(x,y) xlabel('X') ylabel('Exponential decay function') Problem 3: Draw a circle with given radius, using Script Programming in MATLAB Solution 3 MATLAB Program % circle a script file to draw a unit circle r = input(‘Enter the radius of the circle’) theta = linspace(0,2*pi,100); x = r * cos(theta) ; y = r * sin(theta) ; plot(x,y) axis(‘equal’) title(‘Circle with unit radius’)
Problem 4 Write a function that outputs a conversion table for celsius and fahrenheit temperatures. Input of the function should be two numbers Ti and Tf specifying the lower and upper range of the table in celsius.
Hint: 3259
+= CF
Solution 4 MATLAB Program function temptable = ctof(tinitial, tfinal); % CTOF : function to convert temperature from C to F % Call syntax : % temptable = ctof(tinitial, tfinal); % ------------- C = [tinitial:tfinal]; F = (9/5)*C + 32; temptable = [C; F]'; Problem 5 Wrtie a function factorial to compute the factorial n! for any integer n. The input should be the number n and the output should be n!. You might use a for loop or a while loop to do the calculations. (Note: you can also use the built in function prod to calculate a factorials) Solution 5: MATLAB Program function factn = factorial(n); % Factorial : function to compute a factorial n! % call syntax : % factn = factorial(n); % ---------------------- factn = 1; for k = n:-1:1 factn = factn*k; end Problem 6: Wrtie a function file crossprod to compute the cross product of two vectors u, and v, given u = (u1, u2, u3), v = (v1, v2, v3), and u X v = (u2v3 – u3v2, u3v1 – u1v3, u1v2 – u2v1). Check your function by taking cross products of pair of unit vectors: (i, j), (j, k) etc . [i = (1 0 0), j = (0 1 0), k = (0 0 1). Do not use the built in function here. Solution 6: MATLAB Program function w =crossprod(u,v); % Cross product : function to compute w u X v for vectors u and v % Call syntax : % w = crossprod(u,v); % ----------------------- if length(u) > 3 | length(v) > 3 % check if u OR v has > 3 elements error('Ask Euler. This cross product is beyond me') end w = [u(2) * v(3) - u(3) * v(2); % first element of w u(3) * v(1) - u(1) * v(3); % second element of w u(1) * v(2) - u(2) * v(1)]; % third element of w
Problem 7: Write a function to compute the sum of a geometric series 1 + r + r2 + r3 + …. + rn for a given r and n. Input to the function must be r and n and output must be the sum of the series. Solution 7: MATLAB Program function s =gseriessum(r,n); % GSSERIESSUM : function to compute the sum of a geometric series % The series is 1+r+r^2+r^3+ ------ +r^n % Call syntax : % s = gseriessum(r,n); % ----------------------- nvector = 0 : n; series = r.^nvector; s = sum(series); Problem 8: The interest you get at the end of n years, at a flat annual rate of r%, depends on how the interest is compunded. If the interest is added to your account k times a year, and the principal amount your invested is X0, then at the end of n years you would have X = X0 (1 + r / k )kn amount money in your account. Write a function to compute the interest (X – X0) on your account for a given X, n, r, and k. Solution 8: MATLAB Program function [capital, interest] = compound(capital, years, rate, timescomp); % compound : function to compute the compounded capital and the interest % Call syntax : % [capital, interest] = compound(capital, years, rate,timescomp) % ----------------------- X0 = capital; n = years; r = rate; k = timescomp; if r > 1 disp('check your interest rate. For 8 % enter 08, not 8.') end capital = X0*(1 + r/k) - (k*n); interest = capital - X0; Problem 9: Enter following three matrices,
⎥⎦
⎤⎢⎣
⎡ −=⎥
⎤⎢⎣
⎡=⎥
⎤⎢⎣
⎡=
3555
4321
9362
CBA
1. Compute D = A + B + C 2. Compute E = A*B*C 3. Compute F = C*B*A and compare E and F
Solution 9: MATLAB Program % Enter matrices A, B, C A = [2 6; 3 9]; B = [1 2; 3 4]; C = [-5 5; 5 3]; D = A + B + C E = A*B*C F = C*B*A
Problem 10 Create a big matrix with submatrices : The following matrix G is created by putting matrices A, B, and C given. Do the following operations on matrix G created above
⎥⎥⎥⎥⎥⎥⎥⎥
⎢⎢⎢⎢⎢⎢⎢⎢
=
350000550000004300002100000093000062
G
1. Delete the last row and last column of the matrix 2. Extract the first 4 X 4 submatrix from G 3. Replace G(5,5) with 4
Solution 10: MATLAB Program Problem 11 The following data is obtained from an experiment aimed at measuring the spring constant of a given spring. Different masses m are hung from the spring and the corresponding delfections δ of the spring from its unstreached configurtation are measured. Find value of stiffness coefficient of spring M (g) 5.0 10.0 20.0 50.0 100.0 δ (mm) 15.5 33.07 53.39 140.24 301.03
Solution 11: MATLAB Program % Spring constant fron experimental data m = [5 10 20 50 100]; g = 9.81; d = [15.5 33.07 53.39 140.24 301.03]; F = m /1000*g; a = polyfit(d,F,1); d_fit = 0:10:300; F_fit = polyval(a, d_fit); plot(d,F,'o',d_fit,F_fit) xlabel('Displacement (mm)') ylabel('Force (N)') k = a(1); text(100,0.32,['\leftarrow Spring Constant K = ', num2str(k)]); Problem 12 The following table shows the time versus pressure variation readings from a vacuum pump. Fit the curve, P(t) = P0 e –t/T t 0 0.5 1.0 5.0 10.0 20.0 P 760 625 528 85 14 0.16
Solution 12:
Taking log of both sides, we have TtPP −= )0ln()ln( or 01
^ataP +=
Where )0ln(0,/11),ln(^
PaandTaPP =−==
MATLAB Program % EXPFIT : Exponenetial curve fit % For the following data for (t,p) fit % an exponential curve % P = P0 * exp(-t/tau) % Problem is solved by taking log and % then using a linear fit (1st order % polynomial % Orginal Data t = [0 0.5 1 5 10 20]; p = [760 625 528 85 14 0.16]; % Prepare new data for linear fit tbar = t; pbar = log(p); % Fit a 1st order polynomial % thorugh (tbar, pbar) a = polyfit(tbar,pbar,1); % Evaluate constant P0 and tau P0 = exp(a(2)); tau = -1/a(1); %plot new curve and the data on linear scale tnew=linspace(0,20,20); pnew=P0*exp(-tnew/tau); figure(1), plot(t,p,'o',tnew,pnew), grid xlabel('Time (Sec)'), ylabel('Pressure (torr)') % Plot the new curve and the data on semilog scale lpnew=exp(polyval(a,tnew)); figure(2), semilogy(t,p,'o',tnew,lpnew), grid xlabel('Time (Sec)'), ylabel('Pressure (torr)') Problem 13 Find solution of the following set of linear algebraic equations as,
23321433
132
=++=++=++
zyxzyx
zyx
Write the equations in matrix form and solve for x = [x y z]T using left division Solution 13: MATLAB Program % Tutorial Problem 13 % Solution of algebraic equations A = [1 2 3; 3 3 4; 2 3 3]; B = [1 1 2]'; X = inv(A)*B; disp('Roots of equations are'),X Problem 14: Figure shows the loading on a semicircular member, The radius of semicircular member is 25 inches. Write a MATLAB program to plot the internal forces, namely the axial forces, shearing forces and bending moment as functions of α for -90 < α < 90
200 lb
V M C
300 lb
200 lb
25 inch
Solution 14: For -90 < α < 90, C = 200 sin α + 150 cos α ; V = 200 cos α - 150 sin α ; M = -200 r sin α + 100 r ( 1- cos α) Matlab Program % Variation o axial force, shearing force and bending moment alpha = -90:90; alpha = pi*alpha/180; r = 25; C = 200.*sin(alpha) + 150.*cos(alpha); V = -200.*cos(alpha) + 150.*sin(alpha); M = -200.*r.*sin(alpha)+ 100.*r.*(1-cos(alpha)); figure(1) subplot(2,1,1), plot(alpha*180/pi,C,alpha*180/pi,V), grid xlabel('Alpha in (deg)'), ylabel('Axial and Shearing Force (lb)') axis([-90,90,-400,400]), legend('Axial Force','Shearing Force') subplot(2,1,2), plot(alpha*180/pi,M) xlabel('Alpha in (deg)'), ylabel('Bending Moment (lb-inch)'), grid axis([-90,90,-5000,10000]) Problem 15: The motion of a particle is defined by equation, X = 35 t2 – 110 t Y = 115 t2 – 42 t3 Where, X and Y displacmenet of the particle, t is time in sec For interval 0 < t < 25 s, write a MATLAB program to plot
1. the path of the particle in XY plane 2. the components of velocity Vx and Vy and magnitude of the velocity V 3. the components of the acceleration Ax and Ay and magnitude of the acceleration
Solution 15: MATLAB Program % Tutorial 15 t = 0:0.5:25 X = 35*t.^2-110*t; Y = 115*t.^2-42*t.^3; Vx = 70*t-110; Vy = 230*t-126*t.^2; V = sqrt(Vx.^2+Vy.^2); Ax = 70; Ay = 230-252*t; A = sqrt(Ax.^2+Ay.^2); figure(1) subplot(2,2,1), plot(X,Y), grid xlabel('X (mm)'),ylabel('Y (mm)') subplot(2,2,2), plot(t,X,t,Y), grid xlabel('Time (sec)'),ylabel('X & Y (mm)') legend('X (mm)','Y (mm)') subplot(2,2,3), plot(t,Vx,t,Vy,t,V), grid xlabel('Time (sec)'),ylabel('Vx, Vy, V (mm/s)') legend('Vx (mm/s)','Vy (mm/s)', 'V (mm/s)') subplot(2,2,4), plot(t,Ax,t,Ay,t,A), grid xlabel('Time (sec)'),ylabel('Ax, Ay, A (mm/s)') legend('Ax (mm/s)','Ay (mm/s)', 'A (mm/s)')
Problem 16: The acceleration due to gravity due to earth at any height h above the surface of the earth is given by,
22 /
)(sm
hRGMg+
−=
Where, G the gravitational constant ( 6.67 X 10-11 Nm2/kg2) M mass of the earth (5.99 X 1024 kg) R mean radius of the earth (6372 km) h height above the surface of the earth (m) Write a MATLAB program to calculate and plot the acceleration due to gravity for 0 < h < 45000 km in 500 km increments Solution 16: MATLAB Program % Tutorial 16 % Shows variation of g with height upto 45000 km G = 6.67E-11; M = 5.99E24; R = 6372; h = 0:0.5:500000; g = G.*M./(R + h).^2; g = g/1E6; figure(1) plot(h/1000,g), grid xlabel('Height above a earth surface (km)') ylabel('Acceleration due to gravity (m/s^2)') axis([0, 50, 0, 10]) Problem 17:
The satellite orbit in polar coordinates is given by, )cos(1 ϑε−
=Pr
r distance of satellite from the center of the earth θ angle of the satellite from center of the earth P a parameter specifying the size of the orbit ε a parameter specifying eccentricity of the orbit Write a MATLAB program to plot the orbit of a satellite for ε = 0, 0.25, 0.5, and 0.75 Solution 17: MATLAB Program % Tutorial 17 P = 1100; eps = 0.0; theta = 0:2*pi/200:2*pi; r = P./(1-eps.*cos(theta)); figure(1) subplot(2,2,1), polar(theta,r) title('\bf orbit for \epsilon = 0.0') eps = 0.25; r = P./(1-eps.*cos(theta)); subplot(2,2,2), polar(theta,r) title('\bf orbit for \epsilon = 0.25') eps = 0.5; r = P./(1-eps.*cos(theta)); subplot(2,2,3), polar(theta,r) title('\bf orbit for \epsilon = 0.50') eps = 0.75; r = P./(1-eps.*cos(theta)); subplot(2,2,4), polar(theta,r) title('\bf orbit for \epsilon = 0.75')
Problem 18: A motor of weight W is supported by four springs with spring constant of 400 lb/in each. The unbalance of the motor is found to be equivalent to a weight of 0.05 lb located 5 in. from the axis of rotation. If the motor is constrained to move vertically, write a MATLAB program to calculate and plot the amplitude of the vibration and maximum acceleration of the motor for motor speeds from 0 to 1200 rpm. Weight of motor W = 35 lb. Solution 18: Equation of magnification factor in case of unbalance is given by
22)/(1
/fm
nf
mm mrP
KPx ω
ωω=
−=
MATLAB Program % Tutorial 18 W = 35; g = 32.2; mW = W / g; % Mass of rotor in lb ks = 400 * 12; % One spring stiffness m = 0.05 / g; % mass of unbalance r = 5/12; % radius of rotation k = 4*ks; % total stiffness of all spring omega_n2=k/mW; % square of natural frequency omega_f=[0:5:1200]; % variation of motor speed omega=omega_f*2*pi/60; Pm=m*r*omega.^2; % unbalance force due to rotation xm=(Pm/k)./(1-omega_f.^2/omega_n2); am=omega_f.^2.*xm; % acceleration in ft/s^2 amp=xm*12; % amplitude in inches z=[omega_f;amp;am]; figure(1) subplot(2,1,1), plot(omega_f,amp),grid xlabel('Motor Speed (rpm)'),ylabel('Amplitude of vibration (inch)') subplot(2,1,2), plot(omega_f,am),grid xlabel('Motor Speed (rpm)'),ylabel('Max. Motor Acceleration (ft/s^2)') Problem 19: Figure shows an approximate model of a vehicle by the spring and dashpot system. The system moves with a speed of V over a road with a sinusoidal cross section of amplitude dm and wavelength l. Write a MATLAB program to determine and plot the amplitude of the displacement of the mass relative to the road as a function of the speed V for damping factors c/cc = 0.1, 0.3, 0.5, and 0.7.
Solution 19:
Equation of motion of vehicle is )sin()sin(22
2
tptmkxdtdxc
dtxdm mm ωωδω ==++
Amplitude ratio is given by, 222
2
)]/)(/(2[])/(1[
)/(
ncn
n
m
m
cc
u
ωωωω
ωωδ +−
=
Where, lvmKn πωω 2/ ==
l
K C
mW
K(x-δ- δst) C(x’-δ’)
x Equilibrium
MATLAB Program % Tutorail 18 op = [0:0.05:10]; Cr01 = 0.1; den = sqrt((1-op.^2).^2+(2*Cr01.*op).^2); num = op.^2; ratio01 = num./den; Cr02 = 0.3; den = sqrt((1-op.^2).^2+(2*Cr02.*op).^2); num = op.^2; ratio02 = num./den; Cr03 = 0.5; den = sqrt((1-op.^2).^2+(2*Cr03.*op).^2); num = op.^2; ratio03 = num./den; Cr04 = 0.7; den = sqrt((1-op.^2).^2+(2*Cr04.*op).^2); num = op.^2; ratio04 = num./den; figure(1) plot(op,ratio01,op,ratio02,op,ratio03,op,ratio04), grid xlabel('Speed ratio'),ylabel('Ratio of Relative Displacement') Problem 20: Determine the eigenvalues and eigenvectors of AB using MATLAB
⎥⎥⎥⎥
⎢⎢⎢⎢
⎡−−
=
⎥⎥⎥⎥
⎢⎢⎢⎢
−−
=
6014112342127531
4321621745211203
BA
Solution 20 % Tutorial 20 A =[3 0 2 1; 1 2 5 4; 7 -1 2 6; 1 -2 3 4]; B =[1 3 5 7; 2 -1 -2 4; 3 2 1 1; 4 1 0 6]; C=A*B; V=eig(C) [Q d]=eig(C) Problem 21: Write a program to plot a non-dimensional response magnitude and phase response for a system with harmonically excited base
Solution 21:
Magnitude of frequency response is given by 2/1222 2)(1(
1)(
⎥⎥⎦
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+⎥
⎤⎢⎣
⎡−
=
nn
jG
ωωξ
ωω
ω
K C y(t)
x(t)
Phase Response
⎥⎥⎥⎥⎥
⎢⎢⎢⎢⎢
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎞⎜⎜⎝
⎛−
⎟⎟⎠
⎞⎜⎜⎝
= −22
2
1
21
2tan)(
nn
n
ωξω
ωω
ωωξ
ωφ
MATLAB Program % Tutorial 21 zeta=[0.05; 0.1; 0.15; 0.25; 0.5; 1.25; 1.5]; r = [0:0.01:3]; for k=1:length(zeta) G(k,:)=sqrt((1+(2*zeta(k)*r).^2)./((1-r.^2).^2+(2*zeta(k)*r).^2)); phi(k,:)=atan2(2*zeta(k)*r.^3, (1-r.^2+(2*zeta(k)*r).^2)); end figure(1) subplot(2,1,1), plot(r,G), grid legend('\zeta_1=0.05','\zeta_2=0.1','\zeta_3=0.15',... '\zeta_4=0.25','\zeta_5=0.5','\zeta_6=1.25','\zeta_7=1.5') xlabel('\omega/\omega_n'),ylabel('|x(i\omega)|A') subplot(2,1,2), plot(r,phi), grid legend('\zeta_1=0.05','\zeta_2=0.1','\zeta_3=0.15',... '\zeta_4=0.25','\zeta_5=0.5','\zeta_6=1.25','\zeta_7=1.5') xlabel('\omega/\omega_n'),ylabel('\phi (\omega)') Problem 22: An analytical expression for the response of undamped single degree freedom system to given initial velocity and displacement is given by, )cos()( φωξω −= − tCetx d
tn
Where ndd
n
d
n
xvxvx
xC ωξωω
ξωφ
ωξω 2
0
0012
0020 1tan −=⎟⎟
⎞⎜⎜⎝
⎛ +=⎟⎟
⎞⎜⎜⎝
⎛ ++ −
Plot the response of system for ωn = 5 rad/s, ζ = 0.05, 0.1, 0.2, subjected to initial conditions x(0) = 0, x(0)’ = v0 =60 cm/s Solution 22: % Tutorial 22 clear all wn = 5; zeta = [ 0.05; 0.1; 0.2]; x0 = 0; v0 = 60; t0 = 0; deltat = 0.01; tf =6; t = [0:deltat:tf]; for i=1:length(zeta) wd = sqrt(1-zeta(i).^2)*wn; x = exp(-zeta(i)*wn*t).*(((zeta(i)*wn*x0+v0)./wd)*sin(wd*t)+x0*cos(wd*t)); plot(t,x), grid hold on end xlabel('t(s)'), ylabel('x(t)') grid on legend('\zeta_1=0.05','\zeta_2=0.1','\zeta_1=0.2')
Problem 23: Figure shows a eletrcial sircuit with resistors and voltage sources. Write a MATLAB program to determine the current in each resistor using mesh current method based on Kirchoff’s voltage law. Given V1 = 22 V, V2 = 12 V, V3 = 44 V, R1 = 20 Ω, R2 = 12 Ω, R3 = 15 Ω, R4 = 7 Ω, R5 = 16 Ω, R7 = 10 Ω, R8 = 15 Ω.
Solution 23: Equations can be written as
⎥⎥⎥⎥
⎢⎢⎢⎢
=
⎥⎥⎥⎥
⎢⎢⎢⎢
⎥⎥⎥⎥
⎢⎢⎢⎢
++−++−
+++−++−
3
2
1
4
3
2
1
87667
664343
7475422
32321
0
)(0)(
)(0)(
VV
V
IIII
RRRRRRRRRRRRRRRRRR
RRRRR
Create a resistance matrix and find values of I using matrix inversion or division method Problem 24:
Solve the first order linear differential equation txdtdx
+= , with initial condition x(0) = 0. use ode
solver to compute solution Solution 24: Step1: write a equation as system of first order
equations, txdtdx
+=
Step2: write a function to compute the derivatives function xdot=simpode(t,x); % SIMPODE computes xdot = x + t % Call syntax xdot = simpode(t,x) xdot = x + t; Step 3: Use ode23 to compute the solution >> tspan=[0 2]; x0=0; >> [t x] =ode23('simpode',tspan,x0); >> plot(t,x) >> xlabel('t'),ylabel('x') Problem 25: Solve the equation of motion of a nonlinear pendulum
)sin(0)sin( 22
22
2
2
θωθθωθ−=⇒=+
dtd
dtd with initial conditions θ(0) = 1, dθ/dt (0) = 0.
R5
V1
R1
I1 R2 R3
R4
R7
R6
R8
V3
V2 I2
I3
I4
Solution 25: Step 1 Wrtie equations in terms of first order equations
Let Z1 = θ and Z2 = dθ / dt, equation can be written in matrix form as, ⎭⎬⎫
⎩⎨⎧−
=⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
)sin(22
2
1
θωZ
ZZ
Step 2 write a function to compute new state derivative function zdot = pend(t,z); % Call syntax zdot = pend(t,z) % inputs are t = time % z = [z(1); z(2)] = [theta; thetadot] % output is : zdot = [z(2); w.^2sin(z(1))] wsq = 1.56; % specify a value of w^2; zdot = [z(2); -wsq.*sin(z(1))]; Note that z(1) and z(2) refer to the first and second order elements of vector z. Step 3 use ode23 or ode45 for solution % tutorial 25 tspan=[0 50]; z0=[1;0]; % assign values of tspan [t z] =ode23('pend',tspan,z0); % run ode23 x = z(:,1);y=z(:,2); % x = 1st column of z, y = 2nd column figure(1) plot(t,x,t,y) % plot x and y xlabel('t'),ylabel('x and y') legend('Displacement','Velocity') figure(2) plot(x,y) % plot of phase potrait xlabel('Displacement'),ylabel('Velocity') title('Phase plane of nonlinear pendulum') Step 4 Extract results and plot Problem 26:
Consider solving the following initial value problem: Solution 26: Create the following .m file called ex1.m: function xprime=ex1(t,x) xprime=sin(tx); Save it in the workspace. In the Matlab command window, type: > > [t,x] = ode45(‘ex1’,[0 10], 1); >> plot(t,x); Problem 27:
Solution 27: Create a .m file called ex2.m with the following content:
function yprime=ex2(t,y) yprime=[cos(y(2))+sin(t); sin(y(1))-cos(t)]; Now y and yprime are both column vectors. In the command window, type the following: >> tspan=[0, 10]; >> y0=[5.1, 6.7]; >> [t, y]=ode45(‘ex2’, tspan, y0); >> plot(t, y); Of course, you can save the above commands to a .m file and type the name of the file in the command window to run it. Problem 28:
Let , the problem is written as:
Solution 28: Create the function ex3.m as follows:
In the command window, type: >> z0=[0;1]; >> tspan=[0 5]; >> [t,z]=ode45(‘ex3’, tspan,z0); >> plot(t,z); We get this graph: Problem 29 Using MATLAB find the solution of Ax = y
⎥⎦
⎤⎢⎣
⎡−=⎥
⎤⎢⎣
⎡−
=11
,11
23yA
Solution 29: A = [ 3 2; 1 -1]; Y= [-1 ; 1]; X = A \ Y X = 0.2000 -0.800
Problem 30: Calculate C = A + B, D = A – B, E = AB Where,
⎥⎥⎥
⎢⎢⎢
⎡=
⎥⎥⎥
⎢⎢⎢
⎡=
210123214
203410321
BA
Solution 30: Problem 30: Two material properties of carbon monoxide gas are given below T Beta Alpha 300 3.33e3 0.212e4 400 2.5e3 0.3605e4 500 2.00e3 0.5324e4 600 1.67e3 0.719e4
Where T is temperature in Kelvin, Beta is thermal expansion coefficient, and alpha is thermal diffisivity. Find by MATLAB the properties for T = 321 440, 571 respectively Solution 30: % Tutorial 30 clear all clc Temp=[300 400 500 600]; Beta=1000*[3.33 2.5 2.0 1.67]; Alpha=10000*[0.2128 0.3605 0.5324 0.7190]; Ti = [321 440 571 581]; Propty1 = interp1(Temp,Beta,Ti,'linear'); Propty2 = interp1(Temp,Alpha,Ti,'linear'); [Ti; Propty1; Propty2] Problem 31: Suppose a functional relation y = y(x) given in a tabular form as Y Y(x) 0 0.9162 0.25 0.8109 0.50 0.6931 0.75 0.5596 1.00 0.4055
Where y(x) is a monotonically increasing function of x, Find values of x that satisfies y = 0.9, 0.7, 0.6 and 0.5, respectively by MATLAB % Tutorial 31 clear all clc x=[0 0.25 0.5 0.75 1.0];
y=[0.9162 0.8109 0.6931 0.5596 0.4055]; yi = [0.9 0.7 0.6 0.5]; xi = interp1(y,x,yi,'linear'); [yi; xi] Problem 32: Desities of sodium for three temperatures are given as I Temperature Density 1 94 929 2 205 902 3 371 860
Write the Largrange interpolation formula that fits the three data points. Find density for T = 251 deg. C by Lagrange Interpolation Solution 32: % Tutorial 32 clear all clc x=[94 205 371]; y=[929 902 860]; xi = [251]; yi = lagrange(y,x,xi); [yi; xi] Problem 33: Matrix A is given by
3 4 23 1 12 0 5
A−⎡ ⎤
⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦
Find eigen values directly by eig, expand A into its characteristics polynomial, and
find the roots of the characteristics of equations Solution 33: % Tutorial 33 clear all clc A = [3 4 -2; 3 -1 1; 2 0 5]; A_eig = eig(A) A_poly = poly(A) A_roots = roots(A_poly) Output A_eig = -2.7503 4.8751 + 1.4314i 4.8751 - 1.4314i A_poly = 1.0000 -7.0000 -1.0000 71.0000
A_roots = 4.8751 + 1.4314i 4.8751 - 1.4314i -2.7503 Problem 34: A set of four data points is given by X = [ 1.1 2.3 3.9 5.1 ] Y = [ 3.887 4.276 4.651 2.117 ] Find the coeffificents of the interpolation polynomial fitted to the data set, determine value of y for x = 2.101 and plot the polynomial with data points Problem 35 An automobile of mass M = 2000 kg is crusing at speed of 30 m/s. The engine is suddenly disengaged at t = 0 sec. Assume that equation of crusing after t = 0 sec is given by
22000 8.1 1200duu udx
= − −
Where u is velocity and x is the linear distance of the car measured from the location at t = 0. The left side is the force of acceleration, the first term on right side is aerodynamic resistance. And second term is rolling resistance. Calculate how far the car moves before the speed reduces to 15 m/s. Solution 35
Rewrite equation as, 2
20008.1 1200
udu dxu
=− −
integrating equation 30
215 0
20008.1 1200
xudu dx xu
= =+∫ ∫ where sign on
left side is changed for switching the limits of integration, by applying a trapezoidal rule, 16
1 161
0.5( )ii
x u f f f=
⎡ ⎤≈ Δ − +⎢ ⎥⎣ ⎦∑ where f is integrand function
Programming Step 1 : Develop a trapezoidal function which evaluates integral function I = trapez_v(f, h); I = h*(sum(f) - (f(1)+f(length(f)))/2); Step 2 : Write a script file for calculation of distance as below, % Tutorial 35 clear all, clc, n_points = 16; i = 1:n_points; h = (30-15) / n_points; u = 15+(i-1)*h; f = 2000*u./(8.1*u.^2+1200); x = trapez_v(f,h) Problem 36 Knowing that the exact answer is I = 4.006994, analyze the effect of the number of intervals n on the errors of the trapezoidal rule applied to the following integral
2
0
1 exp( )I x dx= +∫
Solutrion 36 Step 1 : Develop a trapezoidal function which evaluates integral
function I = trapez_v(f, h); I = h*(sum(f) - (f(1)+f(length(f)))/2); Step 2 Write a script file % Tutorial 36 clear all, clc, Iexact = 4.006994; a = 0; b = 2; fprintf('\n Extended Trapezoidal Rule \n'); fprintf('\n n I Error \n'); n = 1; for k=1:6 n = 2*n; h = (b-1)/n; i = 1:n+1; x = a + (i-1)*h; f = sqrt(1+exp(x)); I = trapez_v(f,h); nn(k)=n; I(k)=I; Error(k)=Iexact(1)-I(k); end out_data=[nn; I; Error;]' Problem 37 A spherical water reservior of radius 5 m is full to the top. Water is to be drained from the hole of radius b = 0.1 m at bottom, starting at t = 0 s. If there is no friction. How much time does it take to drain the water untill the water level reaches to 0.5m measured from bottom. ? Solve by simpsons rule Solution 37
Velocity of water draining from hole is determined by the energy equation, 2
( )2ug z R+ = where u is
the velocity , z is level of water measured from the spherical center, R is the radius of the tank, and g is gravity acceleration. Consider a change of water level dz during time interval dt. The volume of water in dz is Πx2dz, where x is the radius of circular water surface at elevation z. Flow continuity relation can be written as, uAdt = - Πx2dz , where A is the cross section of exit hole and is given by A = Πb2 Radius of water surface x is related to z by R2 = z2 +x2. Eliminating u, x, and A from previous equation
2 2
2 2 (R zdt dz
b g z R−
=+
, Notice that the level of water at the top of the tank is z =R, while that at 0.5 m
from bottom is z = 0.9R. Integrating from z = R to 0.9R, we get 0.9 2 2
2 2 ( )
R
R
R zt dzb g z R
−=
+∫
Matlab Implementation Step 1 Develop a function for simpson’s rule
function I=Simps_v(f,h) n=length(f)-1; if n==1 fprintf('\nData has only one data point') return; end if n==2 I=h/3*(f(1)+4*f(2)+f(3)); return end if n==3 I=3/8*h*(f(1)+3*f(2)+3*f(3)+f(4)); return end I=0; if 2*floor(n/2)~=n I=3/8*h*(f(n-2)+3*f(n-1) ... +3*f(n)+f(n+1)); m=n-3; else m=n; end I = I+(h/3)*(f(1)+4*sum(f(2:2:m))+f(m+1)); if m>2 I = I + (h/3)*2*sum(f(3:2:m)); End Step 2 Write a script program as below % Tutorial 37 clear R=5; g=9.81; b=0.1; z1=-R*.90; z2=R; h=(z2-z1)/20; z=z1:h:z2; f=(R^2-z.^2)./(b^2*sqrt(2*g*(z+R))); I=Simps_v(f,h)/60/60 Problem 38 Motion of an electron in an uniform electromagnetic field is given by,
dVm eV B eEdt
= + + where V is the velocity vector, B is the magnetic field vector, E is the electric field
vector, and m is mass of the electron, and e is the charge of electron. Initial condition V = (-10, 2, 0.1) X 105 m/s Initial Position of electron R = (0, 0, 0) m Magnetic Field Vector B = ( 0, 0, 0.1 ) T Electric Field Vector E = (0, 2, 0) X 104 V/m Mass of electron m =9.1 X 10-31 kg Charge of electron e = 1.6 X 10-19 c
Solve equation by Runge Kutta method with h = 0.5 X 10-11 s for 0 < t < 2 X 10-9 s, and determine the locus of the electron. Plot a trajectory of the electron in a three dimensional plane, and velocity components as functions of time as well as in the three dimensional phase space. Solution 38 Step 1 Develop a function which carry out a vector product function c = vxv_(a,b) % c = [a] X [b] % a b c are vectors c=[a(2)*b(3)-a(3)*b(2); -a(1)*b(3)+a(3)*b(1); a(1)*b(2)-a(2)*b(1)]; Step 2 Write a scrip file as below % Tutorial 38 clear all clf, clc m =9.1e-31; e = 1.6e-19; B = [0; 0; 0.1]*e/m; E = [0; 2e4; 0]*e/m; h=0.5e-11; v(:,1)=1e5*[-10;2;0.1]; t(1)=0; xyz(:,1)=[0;0;0]; epm=e/m; for i=2:400 t(i)=h*i; k1=h*(vxv_(v(:,i-1),B)+E); k2=h*(vxv_(v(:,i-1)+k1,B)+E); v(:,i)=v(:,i-1)+0.5*(k1+k2); xyz(:,i)=xyz(:,i-1)+0.5*(v(:,i-1)+v(:,i))*h; end figure(1) plot3(xyz(1,:),xyz(2,:),xyz(3,:)) axis([-5 5 -1 2 0 3]*1e-4) xlabel('X');ylabel('Y');zlabel('Z'); figure(2) plot3(xyz(1,:),xyz(2,:),xyz(3,:)) xlabel('X');ylabel('Y');zlabel('Z'); view([0,0,1]) figure(3) plot3(v(1,:),v(2,:),v(3,:)) xlabel('V_x');ylabel('V_y');zlabel('V_y'); figure(4) plot(t,v(1,:),t,v(2,:),t,v(3,:)) xlabel('t');ylabel('Velocities'); legend('V_x','V_y','V_z')
Problem 39 Write a program for lobe of oscillating liquid jet % Tutorial 39 % plots a lobe of liquid jet clear, clf hold on dth=2*pi/40; th=0:dth:2*pi; r =ones(size(th)); colormap gray for n=1:51 b=n-1; x(n,:)=cos(th)*(1-0.25*cos(b*0.3)); y(n,:)=sin(th)*(1+0.25*cos(b*0.3)); z(n,:)=n*0.3*ones(size(th)); m=n+9; if floor(m/2)*2 ==m plot3(z(n,:),x(n,:),y(n,:)-5) end end hold off figure(2) surfl(z,x,y+2,[-10,60]) axis([0 13 -5 15 -10 5])
Control System MATLAB Tutorials
Problem 1 Find a poles of the follwing transfer functions and also compute step response characteristics of following systems
2252152,
13321,
2224,
222
234
2
12332221 ++++++
=+++
=++
+=
++=
ssssssT
sssT
sssT
ssT
Solution 1: % Control Syst tut 01 % Poles of tranfer functions and % computation of step responses for all TFs num1=[2]; den1=[1 2 2]; sys1=tf(num1,den1) P1 = pole(sys1);Z1 = zero(sys1); figure(1) subplot(2,2,1), step(sys1) num2=[4 2]; den2=[1 2 2]; sys2=tf(num2,den2) P2 = pole(sys2);Z2 = zero(sys2); subplot(2,2,2), step(sys2) num3=[1]; den3=[2 3 3 1]; sys3=tf(num3,den3) P3 = pole(sys3);Z3 = zero(sys3); subplot(2,2,3), step(sys3) num4=[2 5 1]; den4=[1 2 5 2 2]; sys4=tf(num4,den4) P4 = pole(sys4);Z4 = zero(sys4); subplot(2,2,4), step(sys4) Problem 2: Find each of second order systems below find ζ, ωn, Ts, Tp, Tr, % overshoot, and plot step response using MATLAB
842
3
22 1010325.110)(,
045.0025.0045.0)(,
13015130)(
+×+=
++=
++=
sssT
sssT
sssT
Solution 2: % Control tut 02 % % first tranfer function num1=[130]; den1=[1 15 130]; Ta1=tf(num1,den1) wn1=sqrt(den1(3)) zeta1=den1(2)/(2*wn1) Tsa1=4/(zeta1*wn1) Tpa1=pi/(wn1*sqrt(1-zeta1^2)) Tra1=(1.76*zeta1^3-0.417*zeta1^2+1.039*zeta1+1)/wn1 Per1=exp(-zeta1*pi/sqrt(1-zeta1^2))*100 figure(1) subplot(221), step(Ta1)
num2=[0.045]; den2=[1 0.025 0.045]; Ta2=tf(num2,den2) wn2=sqrt(den2(3)) zeta2=den2(2)/(2*wn2) Tsa2=4/(zeta2*wn2) Tpa2=pi/(wn2*sqrt(1-zeta2^2)) Tra2=(1.76*zeta2^3-0.417*zeta2^2+1.039*zeta2+1)/wn2 Per2=exp(-zeta2*pi/sqrt(1-zeta2^2))*100 figure(1) subplot(222), step(Ta2) num3=[10E8]; den3=[1 1.325E4 10E8]; Ta3=tf(num3,den3) wn3=sqrt(den3(3)) zeta3=den3(2)/(2*wn3) Tsa3=4/(zeta3*wn3) Tpa3=pi/(wn3*sqrt(1-zeta3^2)) Tra3=(1.76*zeta3^3-0.417*zeta3^2+1.039*zeta3+1)/wn3 Per3=exp(-zeta3*pi/sqrt(1-zeta3^2))*100 figure(1) subplot(223), step(Ta3) Problem 3:
For a closed loop system defined by, 12
1)()(
2 ++=
sssRsC
ξ,
plot the unit step response curves for ζ = 0.0, 0.1, 0.2, 0.4, 0.5, 0.6, 0.8, and 1.0. ωn is normalized to 1, plot a three dimensional coruve for step reponse as above Solution 3: % Control system tut 03 % 2D and 3D plots for unit step response zeta = [0 0.1 0.2 0.4 0.5 0.6 0.8 1.0]; t = 0:0.2:15; figure(1) for n = 1:8; num = [0 0 1]; den = [1 2*zeta(n) 1]; sys(n) = tf(num,den) step(sys(n),t); y(:,n) = step(sys(n),t); hold on end figure(2) mesh(t,zeta,y') xlabel('t sec'), ylabel('\zeta'), zlabel('Amplitude')
Problem 4: Write a equations of motion for the simple pendulum where all mass is concentrated at the endpoint. Use MATLAB to determine the time history of θ to step input of Tc of 1 N-m. Assume l =1 m, m = 0.5 kg, and g = 9.81 m/s2. Solution 4:
Equation of motion of pendulum is
2cTg
l mlθ θ••
+ = and transfer function is given
by 2
2
1( )( )c
s mlgT s sl
Θ=
+
Script Program % Control tut 04 t=0:0.02:10; m=0.5; g=9.81; l=1; tc =1; num=[1/m*l^2]; den=[1 0 g/l]; sys = tf(num,den); y =step(sys,t); plot(t,y) Problem 5 Solve following state space equation by SIMULINK
Problem 6 Consider the double integrator system G(s), which is open loop unstable. To stabilize this plant, we use a lead compensator, K(s). Prepare a simulink model for this
2
2.5 1( ) , ( )10 1 5
sG s K ss s s
+= =
+ + + simulate this model upto 100 seconds
Solution 6
Problem 7 The Lorenz system is set of three first order nonlinear differential equations as given below, Simulate using Simulink Modeling
( )
( )
x a y x
y x b z y
z cz xy
= −
= − −
= +
where a =10, b = 28, c = -2.67, and Initial conditions are x(0) =1, y(0)=1, and z(0)=1
Solution 7
Problem 8 Simulate a Two degree freedom system given by following relations
1 1 1 2 1 1 2 1 2 2 2 2
2 2 2 2 2 2 2 1 2 1
( ) ( ) 0
sin(20 )
M x k k x c c x k x c x
M x k x c x c x k x F t
•• • •
•• • •
+ + + + − − =
+ + − − =
Where M1 =10kg, M2=20kg, k1 = 1000 N/m, k2=200 N/m, c1=c2=120 Ns/m, F = 100; Solution
Problem 9 Simulate following Duffing System
3
(0) 5
( ) cos (0) 5
(0) 3
x y x
y x x y yθ
θ ω θ
= =
= − + + + =
= =
Problem 10 Simulate following Rossler System of equations
(0) 51 (0) 55
1 5.7 (0) 55
x y z x
y x y y
z zx z z
= − − =
= + =
= + − =
Problem 11 Simple pendulum nonlinear equation is given by following relation, determine a velocity and displacement plot with respect to time using SIMULINK and ode solver also
1 2
2 1 2sin c
x xTgx x
l ml
=
= − +
Solution 11
ODE solution global Wn2 Uo Wn2=9.82; Uo=5; t0=0; tf=10;
tspan=[t0,tf]; X0=[0 0]; [t,x]=ode23(@pendot,tspan,X0) function xdot=pendot(t,x) global Wn2 Uo if t <= 1 U=Uo; else U=0; end xdot(1)=x(2); xdot(2)=(-Wn2*sin(x(1))+U); Problem 12 | 12,596 | 33,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-22 | latest | en | 0.47279 |
https://www.emathhelp.net/notes/calculus-1/function-types/piecewise-function/ | 1,526,949,512,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864572.13/warc/CC-MAIN-20180521235548-20180522015548-00363.warc.gz | 739,731,931 | 26,268 | # Piecewise Function
When functions is determined by different formulas on different intervals then function is piecewise.
For example, f(x)={(1-x if x<0),(x^2 if x>=0):} is piecewise because on interval (-oo,0) f(x)=1-x and on interval [0,oo) f(x)=x^2.
Now find f(-2), f(1), f(0) and draw graph of this function.
Remember that function is a rule. In this case it tells us that if x<0 then apply f(x)=1-x, otherwise apply f(x)=x^2.
Since -2<0 then we apply f(x)=1-x: f(-2)=1-(-2)=3.
Since 1>0 then we apply f(x)=x^2: f(1)=1^2=1.
Since 0>=0 then we apply f(x)=x^2: f(0)=0^2=0.
Now, to draw this function we draw graph of the function f(x)=1-x on interval (-oo,0) and graph of the function f(x)=x^2 on interval [0,oo).
Note, that open dot indicates that it doesn't belong to the graph. Indeed, f(0)=0, so point (0,0) is on the graph, but (0,1) is not. | 293 | 858 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-22 | longest | en | 0.764706 |
https://blog.shirui.me/2019/07/ | 1,716,594,128,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058751.45/warc/CC-MAIN-20240524214158-20240525004158-00014.warc.gz | 123,011,636 | 18,336 | ### Distribution of segments on Gaussian chain
The probability distribution function of $P_i(\mathbf{r}_i-\mathbf{r}_\mathrm{cm})$ of ideal chain is evaluated, for $P_i$ represents the probability distribution fucntion of $i$th segment with respect to its centre of mass on the ideal chain. An ideal chain is modeled as multidimensional random walk, where the steps are independent, and the distribution of a step at mean length $b$ is given by $P(\mathbf{r})\sim\mathcal{N}(0,b^2)$. Let $\mathbf{b}_i=\mathbf{r}_{i+1}-\mathbf{r}_{i}$ be the $i$th bond vector, then we have
$$\mathbf{r}_i=\sum_{j=1}^{i-1} \mathbf{b}_j$$
and the centre of mass $\mathbf{r}_\mathrm{cm}=\frac{1}{N}\sum_i \mathbf{r}_i$ is
$$\mathbf{r}_\mathrm{cm}=\frac{1}{N}\sum_{j=1}^{N-1}(N-j)\mathbf{b}_j$$
then
$$\mathbf{r}_i-\mathbf{r}_\mathrm{cm}=\sum_{j=1}^{i-1}\frac{j}{N}\mathbf{b}_j+\sum_{j=i}^{N-1}\frac{N-j}{N}\mathbf{b}_j$$
If variable $X$ is a Gaussian random variable with scale of $\sigma^2$, then $aX$ is a Gaussian random variable with scale of $a^2\sigma^2$, we can then write the characteristic function for $P_i(\mathbf{r}_i-\mathbf{r}_\mathrm{cm})$ of a $d$ dimensional ideal chain:
$$\phi_i(\mathbf{q})=\Pi_{j} \phi_{\mathbf{b}_j'}(\mathbf{q})=\exp\left(-\frac{1}{2}\mathbf{q}^T\left(\sum_{j=1}^{N-1}\Sigma_j\right)\mathbf{q}\right)$$
with $\mathbf{b}'_j=\frac{j}{N}\mathbf{b}_j$ for $j\le i-1$ and $\frac{N-j}{N}\mathbf{b}_j$ for $i\le j \le N-1$; $\phi_{\mathbf{b}_j'}=-\exp(-0.5\mathbf{q}^T\Sigma\mathbf{q})$ is the characteristic function of the probability distribution fucntion of bond $j$. $\Sigma_j=\frac{j^2 b^2}{d N^2}\mathbb{I}_d$ for $j\le i-1$ and $\Sigma_j=\frac{(N-j)^2 b^2}{d N^2}\mathbb{I}_d$ for $i\le j \le N-1$, and $\mathbb{I}_d$ is the $d$-dimensional identical matrix, and therefore: (For convenience, I will set $b=1$ in the later calculations.)
$$\phi_i(\mathbf{q})=$$ $$\exp \left(-\frac{\left(6 i^2-6 i (N+1)+2 N^2+3 N+1\right) \left(q_x^2+q_y^2+q_z^2\right)}{36 N}\right)$$
The corresponding distribution of this characteristic function is still a Gaussian distribution with $\Sigma=\frac{b^2}{3} \mathbb{I}_3$, where the equivalent bond length $b^2=\frac{\left(6 i^2-6 i (N+1)+2 N^2+3 N+1\right)}{6 N}$. The 6th moment is $\frac{1}{N}\sum_{i=1}^N \langle(\mathbf{r}_i-\mathbf{r}_\mathrm{cm})^6\rangle=\frac{58 N^6-273 N^4+462 N^2-247}{1944 N^3}$. At large $N$, only leading term matters, which is $\frac{29}{972} N^3$. For $N=20$, the result is $235.886$, which is in accord with the simulation. Another example is for $N=5$, the $R_g^2$ is $0.8$ rather than $5/6=0.8\dot{3}$ in this case.
Here is the simulation code:
ch = np.random.normal(size=(100000,20,3),scale=1/3**0.5)
ch[:,0,:]=0
ch = ch.cumsum(axis=1)
ch -= ch.mean(axis=1,keepdims=1)
m6 = np.mean(np.linalg.norm(ch,axis=-1)**6)
### Simple Examples of Parallel Computing of Cython
It is more convenient to calculate some properties during a molecular simulation process by accessing data from API of the molecular simulation program than calculating after the whole simulation progress and dumping all the coordinates. Especially, for sharply fluctuated properties such as Virial, RDF, etc., require massive frames to calculate. It is expensive to dump densely. For some Python-friendly molecular simulation softwares, e.g., lammps, hoomd-blue, galamost, etc.; it is easily to embed customized Python functions into the simulation control scripts. However, Python has a poor performance in massive calculations, numba.cuda.jit does dramatically boost up the program, however, due to unknown reasons, GPU-accelerated molecular simulation softwares like hoomd-blue and galamost produce errors during simulations if one uses numba.cuda.jit compiled functions. Therefore, I try to use Cython to generate C-functions to accelerate the calculations. Here I put a simple example of a pairwise property calculation function (RDF) using Cython-parallel method, it's also a learning note of mine:
# cython: language_level=3, boundscheck=False, wraparound=False, cdivision=True
import numpy as np
cimport numpy as np
import cython
from cython.parallel import prange, parallel
cimport openmp
from libc.math cimport floor,sqrt,pow
from libc.stdlib cimport malloc, free
import multiprocessing
cdef long * unravel_index_f(long i, long[:] dim) nogil:
cdef long k, d
d = dim.shape[0]
cdef long * ret = <long *> malloc(d * sizeof(long))
for k in range(d):
ret[k] = i % dim[k]
i = (i - ret[k]) / dim[k]
return ret
cdef long ravel_index_f(long * vec, long[:] dim) nogil:
cdef long ret, d, tmp, k
d = dim.shape[0]
ret = (vec[0] + dim[0]) % dim[0]
tmp = dim[0]
for k in range(1,d):
ret += ((vec[k] + dim[k]) % dim[k]) * tmp
tmp *= dim[k]
return ret
cdef long jth_neighbour(long * veci, long * vecj, long[:] dim) nogil:
cdef long ret, d, tmp, k
d = dim.shape[0]
ret = (veci[0] + vecj[0] - 1 + dim[0]) % dim[0]
# re-ravel tmpi + tmpj - 1 to cell_j
# -1 for indices from -1, -1, -1 to 1, 1, 1 rather than 0,0,0 to 2,2,2
tmp = dim[0]
for k in range(1, d):
ret += ((veci[k] + vecj[k] - 1 + dim[k]) % dim[k]) * tmp
tmp *= dim[k]
return ret
cdef double pbc_dist(double[:] x, double[:] y, double[:] b) nogil:
cdef long i, d
cdef double tmp=0, r=0
d = b.shape[0]
for i in range(d):
tmp = x[i]-y[i]
tmp = tmp - b[i] * floor(tmp/b[i]+0.5)
r = r + pow(tmp, 2)
return sqrt(r)
cdef long cell_id(double[:] p, double[:] box, long[:] ibox) nogil:
# In the Fortran way
cdef long ret, tmp, i, n
n = p.shape[0]
ret = <long> floor((p[0] / box[0] + 0.5) * ibox[0])
tmp = ibox[0]
for i in range(1, n):
ret = ret + tmp * <long> floor((p[i] / box[i] + 0.5) * ibox[i])
tmp = tmp * ibox[i]
return ret
cdef void linked_cl(double[:, :] pos, double[:] box, long[:] ibox, long[:] head, long[:] body) nogil:
cdef long i, n, ic
n = pos.shape[0]
for i in range(n):
ic = cell_id(pos[i], box, ibox)
def rdf(double[:,:] x, double[:,:] y, double[:] box, double bs, long nbins):
cdef long i, j, k, l, n, m, d3, d, ic, jc
cdef np.ndarray[np.double_t, ndim=2] ret
cdef long[:] head, body, ibox, dim
cdef double r, r_cut
cdef long ** j_vecs
cdef long * veci
r_cut = nbins * bs
n = x.shape[0]
d = x.shape[1]
m = y.shape[0]
d3 = 3 ** d
ibox = np.zeros((d,), dtype=np.int64)
for i in range(d):
ibox[i] = <long> floor(box[i] / r_cut + 0.5)
head = np.zeros(np.multiply.reduce(ibox), dtype=np.int64) - 1
body = np.zeros((m,), dtype=np.int64) - 1
dim = np.zeros((d,), dtype=np.int64) + 3
j_vecs = <long **> malloc(sizeof(long *) * d3)
for i in range(d3):
j_vecs[i] = unravel_index_f(i, dim)
for i in prange(n, schedule='dynamic'):
ic = cell_id(x[i], box, ibox)
veci = unravel_index_f(ic, ibox)
for j in range(d3):
jc = jth_neighbour(veci, j_vecs[j], ibox)
while k != -1:
r = pbc_dist(x[i], y[k], box)
if r < r_cut:
l = <long> (r/bs)
k = body[k]
free(veci) # arrays created by malloc should be freed. (in unravel_index_f)
free(j_vecs) # without calling free(), memory leak would happen.
return np.sum(ret, axis=0)
Notes:
1. def means Python function and cdef means C functions;
2. Arrays declared by double[:] etc. are memoryslice and can be conveniently initialized by np.zeros or reading an np.ndarray as function parameter;
3. Without GIL, any calling of Python function is forbidden, the array must be initialized and declared in the C method;
4. In prange loop, variables are thread-local (see the generated .c file, in section #pragma omp parallel), there is no explicit way to call an atomic operation in Cython (using with gil results in very low efficiency), hence the result is generated as ret = np.zeros((n, nbins), ... so that in each thread i that ret[i] are independent, or generate ret as (num_threads, data structure...), just keep independent amongst threads and in ith loop, call openmp.omp_get_thread_num() to get the current thread's id;
5. <long> is a type casting.
This RDF function uses a linked cell list algorithm to reduce the calculation from $O(n^2)$ to $O(n)$, a small ranged RDF ($r<r_\mathrm{cut}$) is calculated, the RDF on $r\ge r_\mathrm{cut}$ can be calculated using an FFT algorithm by setting $r_\mathrm{bin}<\frac{r_\mathrm{cut}}{2}$
The second example is calculation of histogram N-d array by modulus of indices: $\int f(\mathbf{r})\delta(|\mathbf{r}|-r)\mathrm{d}\mathbf{r}$:
# cython: language_level=3, boundscheck=False, wraparound=False, cdivision=True
import numpy as np
cimport numpy as np
import cython
from cython.parallel import prange, parallel
cimport openmp
from libc.math cimport floor,sqrt,pow
from libc.stdlib cimport malloc, free
import multiprocessing
cdef double unravel_index_f_r(long i, long[:] dim) nogil:
# unravel in Fortran way
cdef long k, d, tmp
cdef double r
d = dim.shape[0]
for k in range(d):
tmp = i % dim[k]
r += <double> (tmp)**2
i = (i - tmp) / dim[k]
return sqrt(r)
def hist_to_r(double[:] x, long[:] shape, double dr, double bs, double rc):
cdef long i, n, j, n_bins
cdef np.ndarray[np.double_t, ndim=2] ret, count
n_bins = <long> (rc / bs)
n = x.shape[0]
for i in prange(n, schedule='dynamic'):
j = <long> (unravel_index_f_r(i, shape)*dr/bs)
if j < n_bins:
return np.sum(ret,axis=0), np.sum(count,axis=0)
In function unravel_index_f_r, the index is unraveled in the Fortran way, therefore, calling hist_to_r requires x.ravel(order='F').
### Anisotropy of ideal chain
The Gaussian chain is isotropic when averaged over conformation space and all orientations. Therefore, a Gaussian chain is dealt as sphere with radius of $R_g$, it’s radius of gyration. In the 2nd chapter of Rubinstein’s Polymer Physics, an exercise shows that the $R_g^2$ is asymmetric if one set the coordinate frame on its end-to-end vector, $\mathbf{R}_{ee}$. i.e., the $\mathbf{R}_{ee}$ vector is set as the $x$-axis therefore $\mathbf{R}_{ee}=(R,0,0)^T$. Then the 3 components of $\mathbf{R}_{ee}$ are $\frac{Nb^2}{36}$, $\frac{Nb^2}{36}$ on $y$, $z$ direction and $\frac{Nb^2}{36}+\frac{R^2}{12}$ on $x$ direction. Here I make a very simple proof.
Consider a Gaussian chain is fixed between $(0,0,0)^T$ and $\mathbf{R}_{ee}$. It’s actually a Brownian bridge, and the distribution is a multivariate Gaussian with mean at $\frac{i}{N}\mathbf{R}$ with variance $\frac{i(N-i)}{N}b^2$, the proof is simple:
$$P_{0\mathbf{R}}(\mathbf{r},n)=\frac{G(\mathbf{r},0,n)G(\mathbf{R},\mathbf{r},N-n)}{G(0,\mathbf{R},N)}$$
$G(a,b,n)$ represents distribution of a Gaussian chain ends at $a,b$ with segment length $n$. The meaning is straightforward: it’s the probability of a length $n$ Gaussian chain start from $0$ and end at $\mathbf{r}$ connected with another $N-n$ Gaussian chain which started at $\mathbf{R}$ and stopped at $\mathbf{r}$, and the whole chain is an Gaussian chain with length $N$ and $\mathbf{R}_{ee}=\mathbf{R}$. It is easily to show the distribution of such chain
$$P_{0\mathbf{R}}(\mathbf{r},n)=G\left(\mathbf{r}-\frac{n}{N}\mathbf{R}, 0,\frac{n(N-n)}{N}\right)$$
is equivalent to a Gaussian chain segment ends with $\mathbf{r}$ and $\frac{n}{N}\mathbf{R}$ with equivalent length $\frac{n(N-n)}{N}$. The $R_g^2$ is then
\begin{align}R_g^2=&\frac{1}{2N^2}\int_0^N\langle(\mathbf{r}_i-\mathbf{r_j})^2\rangle\mathrm{d}i\mathrm{d}j\\ =&\frac{1}{N^2}\int_0^N\int_j^N \frac{(i-j)^2 R^2}{N^2}+\frac{(i-j)(N-(i-j))}{N}b^2\mathrm{d}i\mathrm{d}j\\ =&\frac{R^2+nb^2}{12}\end{align}
In this equivalent method, $\langle (\mathbf{r}_i-\mathbf{r}_j)^2\rangle$ is considered as $(\frac{i}{N}-\frac{j}{N})^2R^2+\frac{|i-j|(N-|i-j|)}{N}$ from the equivalent Gaussian chain. This is because despite the chain is ‘fixed’, it is still a Gaussian chain, which means translation invariance, $\langle (\mathbf{r}_i-\mathbf{r}_j)^2\rangle$ depends only on $|i-j|$, therefore, $P_{0\mathbf{R}}(\mathbf{r},n)$ gives the probability of segment $\mathbf{r}_n-\mathbf{r}_0$, which is any $n$-segment on the Gaussian chain. I tried calculating $P(\mathbf{r}_i-\mathbf{r}_j)$ from the convolution of $P_{0\mathbf{R}}$. This is incorrect because $\mathbf{r}_i$, $\mathbf{r}_j$ are correlated, convolution of $P_{0\mathbf{R}}$ results in dependence of $i$ and $j$ in $\langle (\mathbf{r}_i-\mathbf{r}_j)^2\rangle$, violates the translation invariance of the chain.
Now if $R^2=Nb^2$, we see that $R_g^2=\frac{1}{6}Nb^2$; and we have $\frac{Nb^2}{36}+\frac{R^2_{x,y,z}}{12}=\frac{Nb^2+R^2}{36}$ for each dimension, especially, if $\mathbf{R}_{ee}=(R,0,0)^T$, we have $\frac{Nb^2}{36}$ in $y$ and $z$ direction and $\frac{Nb^2}{36}+\frac{R^2}{12}$ in $x$-direction.
Simple simulation code:
rg2_ree = rg2 = 0
for _ in range(5000):
ch = np.random.normal(size=(1000,3))
ch = np.append(np.zeros((1,3)),np.cumsum(ch, axis=0),axis=0)
ch = ch - ch.mean(axis=0)
ree = ch[-1]-ch[0]
ree = ree/np.linalg.norm(ree)
rg2_ree += np.sum(ch.dot(ree)**2)
rg2 += np.sum(ch**2)
rg2_ree/rg2
# Result is 0.666...
### Flory characteristic ratio of hindered rotation chain
This is an exercise on Rubinstein's textbook of Polymer Physics.
1. Definition of Flory characteristic ratio:
$$C_\infty:=\lim_{n\to\infty}\frac{\langle \mathbf{R}^2 \rangle}{nb^2}$$
where $\langle \mathbf{R}^2 \rangle$ is the mean square end-to-end vector, $b$ is Kuhn length of the polymer and $n$ is corresponding chain length.
2. Calculation of $\langle \mathbf{R}^2\rangle$:
Let $\mathbf{r}_i$ be the bond vector between $i$ and $(i-1)$th particle, and $\mathbf{r}_1=\mathbf{R}_1-\mathbf{R}_0\equiv0$; for an $n$-bead chain,:
\begin{align}\langle \mathbf{R}^2 \rangle&=\left\langle\left(\sum_{i=1}^n \mathbf{r}_i^2\right)\cdot\left(\sum_{j=1}^n \mathbf{r}_j^2\right)\right\rangle\\ &=\sum_i\sum_j\langle \mathbf{r}_i\cdot\mathbf{r}_j\rangle\end{align}
3. Calculation of $\langle\mathbf{r}_i\cdot\mathbf{r}_j\rangle$:
Now consider a local coordinate frame that $\mathbf{r}_i=(b,0,0)^T$, then $\mathbf{r}_{i+1}$ could be considered as rotating $\mathbf{r}_i$ $\theta$ by $z$-axis for bond angle and $\phi_{i}$ by $x$-axis for torsion. The rotation matrix is thus
$$\mathbf{T}_i:=\left( \begin{array}{ccc} \cos (\theta ) & -\sin (\theta ) & 0 \\ \sin (\theta ) \cos (\phi_i) & \cos (\theta ) \cos (\phi_i) & -\sin (\phi_i) \\ \sin (\theta ) \sin (\phi_i ) & \cos (\theta ) \sin (\phi_i ) & \cos (\phi_i ) \\ \end{array}\right)$$
or $\pi-\theta$, $\pi - \phi$ according to the definitions of bond angle $\theta$ and torsion angle $\phi$. In every reference coordinate frame $R_i$, the bond vector is $(b,0,0)^T$, and in $R_i$, $\mathbf{r}_{i+1}=\mathbf{T}_i\cdot\mathbf{r}_i=(b \cos (\theta ),b \sin (\theta ) \cos (\phi_i ),b \sin (\theta ) \sin (\phi_i ))^T$. Now consider a simple case of $\mathbf{r}_i\cdot\mathbf{r}_{i+2}$, the $\mathbf{r}_{i+2}$ is calculated in the $(i+1)$th reference frame $R_{i+1}$ where $\mathbf{r}_{i+1}$ is $(b,0,0)^T$ and $\mathbf{r}_{i+2}=\mathbf{T}_{i+1}\cdot\mathbf{r}_{i+1}$, here we must 'rotate' one of the vector into the other frame to do the inner product, i.e. rotate $\mathbf{r}_i=(b,0,0)^T$ from $R_i$ to $R_{i+1}$ or vice versa. Now note that $\mathbf{r}_{i+1}$ in the $R_i$ frame is $(b \cos (\theta ),b \sin (\theta ) \cos (\phi_i ),b \sin (\theta ) \sin (\phi_i ))^T$ and $(b,0,0)^T$ in the $R_{i+1}$ frame, therefore if we want to transform some vector in $R_i$ into $R_{i+1}$, we need a transform matrix to transform the $(b \cos (\theta ),b \sin (\theta ) \cos (\phi_i ),b \sin (\theta ) \sin (\phi_i ))^T$ into $(b,0,0)^T$ which is $\mathbf{T}_i^{-1}$, or $\mathbf{T}_i^T$, because in frame $R_i$, we obtain $\mathbf{r}_{i+1}$ by $\mathbf{T}_i\cdot(b,0,0)^T$ and $\mathbf{T}_i$ is a rotation matrix, which is orthonormal. By transform $\mathbf{r}_i$ into $R_{i+1}$, the inner product $\mathbf{r}_i\cdot\mathbf{r}_{i+2}$ is
$$\langle\mathbf{T}_i^T\mathbf{r}_i,\mathbf{T}_{i+1}\mathbf{r}_{i+1}\rangle=$$ $$\langle\mathbf{r}_i,\mathbf{T}_i\mathbf{T}_{i+1}\mathbf{r}_{i+1}\rangle=(b,0,0)\mathbf{T}_i\mathbf{T}_{i+1}(b,0,0)^T$$
By induction, $\mathbf{r}_i\cdot\mathbf{r}_j=(b,0,0)\mathbf{T}_i\mathbf{T}_{i+1}\mathbf{T}_{i+2}\cdots\mathbf{T}_{j-1}(b,0,0)^T$, since $\theta$ for every bonds are same and $\phi_{1,2,\dots,n}$ are independently distributed, let $\mathbf{T}$ be the average transform matrix of ${\mathbf{T}_i}$:
\begin{align}\mathbf{T}&:=\frac{\int_{-\pi}^{\pi}\mathbf{T}_i(\phi)\exp{(-U(\phi)/k_BT)}\mathrm{d}\phi}{\int_{-\pi}^{\pi}\exp{(-U(\phi)/k_BT)}\mathrm{d}\phi}\\ &=\left( \begin{array}{ccc} \cos (\theta ) & -\sin (\theta ) & 0 \\ \sin (\theta ) \langle\cos (\phi) \rangle & \cos (\theta )\langle \cos (\phi)\rangle &0 \\ 0& 0& \langle\cos (\phi)\rangle \\ \end{array}\right)\end{align}
We therefore have $\langle\mathbf{r}_i\cdot\mathbf{r}_j\rangle=(b,0,0)\mathbf{T}^{|j-i|}(b,0,0^T)=b^2(\mathbf{T}^{|j-i|})_{11}$. Here every $\sin(\phi)$ term vanishes in the integral because $U$ is even.
4. Now the Flory characteristic ratio
\begin{align}C_\infty&=\lim_{n\to\infty}\frac{1}{nb^2}\sum_i\sum_j \mathbf{r}_i\cdot\mathbf{r}_j\\ &=\lim_{n\to\infty}\frac{1}{n}(\sum_i\sum_j T^{|j-i|})_{11}\\ &=\lim_{n\to\infty}\left(-\frac{1}{n}\frac{-2 \mathbf{T}^{n+1}+\mathbf{T}^2 n+2 \mathbf{T}-n\mathbf{I}}{(\mathbf{I}-\mathbf{T})^2}\right)_{11}\\ &=\left(\frac{\mathbf{I}+\mathbf{T}}{\mathbf{I}-\mathbf{T}}\right)_{11}\\ &=\frac{1+\cos(\theta)}{1-\cos(\theta)}\frac{1-\langle\cos(\phi)\rangle}{1+\langle\cos(\phi)\rangle}\end{align}
Here we have $\frac{\bullet}{n}=0$ and $\mathbf{T}^n=0$ at $n\to\infty$, because $\mathbf{T}$ is constant and correlation between 2 beads goes to $0$ as distance goes to infinity.
### RTFM, pls RTFM!!!
I was trying to write a parallel version C-extension with Cython several days ago, to calculate Coulomb energy of $n$ points. I wrote the code as follows:
@cython.cdivision(True)
@cython.boundscheck(False)
@cython.wraparound(False)
def u(double[:,:] x):
cdef long i
cdef long j,k
cdef double ret=0,tmp
with nogil, parallel():
for i in prange(x.shape[0]-1, schedule='static'):
for j in range(i + 1, x.shape[0]):
tmp = 0
for k in range(x.shape[1]):
tmp = tmp + pow(x[i,k]-x[j,k],2)
ret = ret + 1 / sqrt(tmp)
return ret
Back2Top ^ | 6,342 | 17,949 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-22 | latest | en | 0.536805 |
https://surreybasketballclassic.info/interesting/readers-ask-how-many-times-around-a-basketball-court-is-a-mile.html | 1,620,436,034,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988831.77/warc/CC-MAIN-20210508001259-20210508031259-00521.warc.gz | 590,315,284 | 11,413 | ## How many basketball courts are in a mile?
A basketball court is 94 feet long. 5,280 feet divided by 94 feet is approximately 56. So 56 full court sprints equals one mile in distance.
## How many laps around a HS basketball court is a mile?
Above we saw that it takes 19.7 laps to complete a mile around a high school size basketball court.
## Is 4 laps around a track a mile?
100 meters: the length of one straightaway. 800 meters: roughly ½ mile or 2 laps around the track. 1600 meters: roughly 1 mile or 4 laps around the track.
## Is 8 laps a mile?
The run portion is a 2 mile run, and I must finish in a certain time. So for 2 miles, thats 8 laps. Essentially 2 minutes a lap.
## How many laps around a backyard is a mile?
Just straight-up not having a good time. Each mile was about 34 laps around the yard. Thirty-four goddamn laps.
## How many steps are in a mile?
How many steps in a mile? An average person has a stride length of approximately 2.1 to 2.5 feet. That means that it takes over 2,000 steps to walk one mile and 10,000 steps would be almost 5 miles. A sedentary person may only average 1,000 to 3,000 steps a day.
You might be interested: Often asked: What Was The Score Of The Basketball Game?
## How many gym laps make a mile?
Number of Laps Basketball courts have a standard set of measurement all around. So, in a National Basketball Association (NBA) and college basketball court, it will take 18.3 laps around the court to make a mile. High school basketball courts are smaller, so it will take 19.9 laps around the court.
## How many kms do you run in a basketball game?
Slide on for the results. A hard-working NBA starter will cover around four kilometres per game. Bulls Guard Jimmy Butler is one of the league’s fittest players, averaging about 4.3 kms a game. Obviously this changes a lot depending on an individual’s play style and the length of match.
## How wide is the lane in high school basketball?
The court in the high school and junior high basketball gym measures 84 feet long by 50 feet wide. Court markings reflect those dimensions.
## Which lane on a track is 1 mile?
Most running tracks are 400 meters around in lane 1 (the inside lane ). The distance around the track increases in each lane; the distance you would run once around in lane 8 is 453 meters. Know Your Distances.
Common Distances on a Track
Meters Track Equivalent
1600 Approximately 1 mile, or four laps around the track
## How many laps is a 5K?
The distance of a 5K is 3.1 miles. On a standard outdoor track, a 5K ( 5,000 meters) is 12.5 laps.
## How many laps is a mile pool?
Based on your pool length, here are how many laps you’ll need to swim to complete a true mile: 20 Yard Pool: 1,760 yards is 88 lengths ( 44 laps ) 25 Yard Pool: 1760 yards is 70.4 lengths (35.2 laps) 25 Meter Pool: 1610 meters is 64.4 lengths (32.2 laps)
You might be interested: When Does Iowa State Men's Basketball Play Next?
## How many miles should I run a day?
Beginning runners should start with two to four runs per week at about 20 to 30 minutes (or roughly 2 to 4 miles ) per run. You may have heard of the 10 Percent Rule, but a better way to increase your mileage is to run more every second week. This will help your body adapt to your new hobby so you don’t get hurt.
## How long does it take to walk a mile?
The average walking speed of a human is 3 to 4 miles per hour, or 1 mile every 15 to 20 minutes.
## What is the world’s fastest woman mile time?
The current mile world record holders are Morocco’s Hicham El Guerrouj with a time of 3:43.13 and Sifan Hassan of The Netherlands with the women’s record of 4:12.33. | 937 | 3,662 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-21 | longest | en | 0.955073 |
https://www.compadre.org/physlets/thermodynamics/ex20_3.cfm | 1,506,294,562,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690228.57/warc/CC-MAIN-20170924224054-20170925004054-00529.warc.gz | 761,858,779 | 6,075 | « · »
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# Exploration 20.3: Ideal Gas Law
N = | T = | V =
The relationship between the number of particles in a gas, the volume of the container holding the gas, the pressure of the gas, and the temperature of the gas is described by the ideal gas law: PV = nRT. In this animation N = nR (i.e., kB = 1). This, then, gives the ideal gas law as PV = NT. Restart.
Notice what happens as you change the number of particles, the temperature, and the volume. The pressure is due to collisions with the walls of the container. The graph shows the instantaneous "pressure" (the change in momentum of the particles as they hit the wall and thus exert a force) as a function of time, while the table shows both NT/V (equal to the pressure for an ideal gas) and the average of the instantaneous pressure.
1. Keep the number of particles and the volume constant. What happens to the speed of the particles as the temperature changes? What happens to the pressure (N*T/V) if you increase the temperature? (This is known as Gay-Lussac's Law: P/T = constant.)
2. If you double the volume (while keeping the same number of particles and the same temperature), what happens to the pressure (and force on the wall)? Why? (This is known as Boyle's law: PV = constant.)
3. If the number of particles is increased (and the temperature and volume stay the same), what happens to the pressure (and the force on the wall)? Why?
4. If you double the volume and halve the temperature (while keeping the number of particles constant), what happens to the pressure? (This is known as Charles's Law: V/T = constant.)
Note that all of these "named" gas laws are included in the ideal gas law: PV = nRT.
You can also drag the top of the piston to change the volume. In this process both the temperature and pressure can change.
1. Start with a volume of 100. Drag the piston up. What changes and why?
2. Once the particles have spread out into the entire volume available, drag the piston back down. Notice that the particles move fast and the temperature and pressure change dramatically. This is because as the piston comes down and particles hit it, the downward moving piston transfers some of its momentum to these particles and so they speed up. Faster moving particles mean a higher temperature. In a real system, you would not normally see this effect because the particles are moving much faster than any piston being compressed (but if they moved that fast in the animation, you wouldn't see individual particles).
3. How would you need to drag the piston to minimize the change in temperature? Start with a volume of 100 and temperature of 100 again and try to minimize the increase in temperature as you compress the piston.
When you get a good-looking graph, right-click on it to clone the graph and resize it for a better view.
Exploration authored by Anne J. Cox.
The OSP Network: | 672 | 3,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-39 | latest | en | 0.918921 |
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FP2 De Moivre's Theorem watch
1. Suppose z = cosx + isinx, express cos4x in terms of z
No idea what i am doing.
i thought this question was asking me to write it in terms of cosx
So i have cos^4(x) - 6cos^2(x)sin^2(x) + sin^4(x)
Absolutely lost, pls help.
2. You can put the sin^2(x) in terms of cos^2(x) using C2 trig identities. The sin^4x is similar except you need to square the identity. Sorry for not giving the identity, I think I have to only give hints XD
3. (Original post by BinaryJava)
You can put the sin^2(x) in terms of cos^2(x) using C2 trig identities. The sin^4x is similar except you need to square the identity. Sorry for not giving the identity, I think I have to only give hints XD
Hiya, i have simplified it to 8cos^4(x) - 8cos^2(x) + 1
This is my maximum
4. (Original post by Bobjim12)
Hiya, i have simplified it to 8cos^4(x) - 8cos^2(x) + 1
This is my maximum
Yeah that is what I got. Do you know if we need to know the proof by induction for de moivres theorem?
5. (Original post by BinaryJava)
Yeah that is what I got. Do you know if we need to know the proof by induction for de moivres theorem?
I don't think so,
the answer is 1/2(z^4 + 1/z^4)............................ ..
There isn't an example in my textbook so i haven't a clue what i am doing..
6. Do you know what, i am stupid. Never mind, i have got it now, thankyou though for trying :^)
7. I guess then you let cos4x = z which is the same as z^4 = cosx. Not sure about where the 8 goes though.
8. (Original post by BinaryJava)
Yeah that is what I got. Do you know if we need to know the proof by induction for de moivres theorem?
For edexcel you do
9. (Original post by BinaryJava)
I guess then you let cos4x = z which is the same as z^4 = cosx. Not sure about where the 8 goes though.
yes i was being silly, thankyou
10. (Original post by Gome44)
For edexcel you do
Damn thought I got away from that induction rubbish, seems simple enough.
11. (Original post by Bobjim12)
Suppose z = cosx + isinx, express cos4x in terms of z
No idea what i am doing.
i thought this question was asking me to write it in terms of cosx
So i have cos^4(x) - 6cos^2(x)sin^2(x) + sin^4(x)
Absolutely lost, pls help.
Note that . How can you then express in terms of (and )?
Now use De Moivre's Theorem to do something similar with
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Chat with other maths applicants | 874 | 3,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-30 | latest | en | 0.944974 |
https://www.physicsforums.com/threads/the-work-energy-principle-kinematic-eqns-to-calculate-speed.333827/ | 1,531,927,864,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590199.42/warc/CC-MAIN-20180718135047-20180718155047-00595.warc.gz | 942,131,579 | 14,301 | # Homework Help: The Work-Energy Principle & Kinematic Eq'ns to calculate speed
1. Aug 31, 2009
### janelle1905
1. The problem statement, all variables and given/known data
A spaceship of mass 5.00 x 104 kg is travelling at a speed 1.15 x 104 m/s in outer space. Except for the force generated by its own engine, no other force acts on the ship. As the engine exerts a constant force of 4.00 x 105 N, the ship moves a distance of 2.50 x 106m in the direction of the force of the engine.
a. Determine the final speed of the ship using the work-energy theorem.
b. Determine the final speed of the ship using kinematic equations.
2. Relevant equations
Wnet = 1/2mv2
v2 = v20 + 2(F/m)d
3. The attempt at a solution
a. Using work energy theorem:
Wnet = W + Wfr = Fdcos0o + 0 = (4.00 x 105)(2.50 x 106) = 1.00 x 1012
v2 = (2)(1.00x1012)/5.00x104 = 6324.6 m/s
b. Using kinematic equations:
v2 = (1.15 x 104)2 + 2(4.00 x 105/5.00 x 104)(2.50 x 106)
v = 13,124 m/s
According to my calculation in part a, the final speed is 6324.6 m/s, however in part b my calculation shows that the velocity is 13,124 m/s. The difference between the two calcuations is the v0 is included in the second one, but not in the first. It seems to me that v0 should be included, but I don't know how to incorporate it into the work-energy theorem.
Thanks in advance for you help :)
(I know these answers aren't the correct number of sig figs yet...just trying to get the right answer first!)
2. Aug 31, 2009
### kuruman
You misquoted the work-energy theorem. It is
Wnet = ΔK
where ΔK is the change in kinetic energy (final minus initial). This spaceship is already moving when in fires its engine.
3. Aug 31, 2009
### janelle1905
Okay...So in my Wnet calculation, should it be Wnet = Fdcos0 + v0
I know that KE = 1/2mv2, this is what the second term of the previous equation should be, however this leads to the wrong answer doesn't it?
Also, do you think my calculation of velocity using kinematic equations is correct?
4. Aug 31, 2009
### kuruman
You are missing the point. I am not talking about kinetic energy KE. I am talking about change in kinetic energy. The work-energy theorem in this case should be written as
(1/2)m(vfinal)2 - (1/2)m(vinitial)2 = Wnet
The term on the left is the change in kinetic energy, final minus initial.
You need to use the above equation with vinitial = 1.15x104 m/s to find vfinal. The work in this case is indeed Fd. I don't know what v0 means, but it should not be there.
The kinematic equation is correctly set up. I did not plug in to check your numbers.
5. Aug 31, 2009
### janelle1905
Okay - I understand it now. I plugged in and got the same answer for both part a and b this time.
Thanks very much for your help! | 807 | 2,757 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-30 | latest | en | 0.900853 |
https://www.intellectualmath.com/analyze-the-equation-of-a-parabola-worksheet.html | 1,722,660,790,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00207.warc.gz | 623,336,446 | 5,995 | # ANALYZE THE EQUATION OF A PARABOLA WORKSHEET
Write the following in standard form. Identify the
• Vertex
• Focus
• Axis of symmetry
• Direction of opening of parabola.
• Equation latus rectum and directrix
• Draw the graph
Problem 1 :
y = 3x2 + 24x + 50
Solution
Problem 2 :
-6y = x2
Solution
Problem 3 :
3(y - 3) = (x - 6)2
Solution
Problem 4 :
-2(y - 4) = (x - 1)2
Solution
Problem 5 :
4(x - 2) = (y + 3)2
Solution
1)
Vertex (h, k) ==> (-4, 2) Focus (h, k + a) k + a = 2 + (3/4)= 11/4(-4, 11/4) Equation of latus rectum y = k + ay = 11/4 Equation of directrix y = k - ay = 2 - (3/4)y = 5/4 Axis of symmetry x = -4 Equation of directrix 4a = 3 units
2)
Vertex (h, k) ==> (0, 0) Focus (0, -a) (0,-3/2) Equation of latus rectum y = -ay = -3/2 Equation of directrix y = ay = 3/2 Axis of symmetry x = 0 Equation of directrix 4a = 6 units
3)
Vertex (h, k) ==> (6, 3) Focus (h, k + a) k + a = 3 + (3/4)= 15/4(6, 15/4) Equation of latus rectum y = k + ay = 15/4 Equation of directrix y = k - ay = 3 - (3/4)y = 9/4 Axis of symmetry x = 6 Equation of directrix 4a = 6 units
4)
Vertex (h, k) ==> (1, 4) Focus (h, k - a) k - a = 4 - (1/2)= 7/2(1, 7/2) Equation of latus rectum y = k - ay = 7/2 Equation of directrix y = k + ay = 4 + (1/2)y = 9/2 Axis of symmetry x = hx = 1 Equation of directrix 4a = 2 units
5)
Vertex (h, k) ==> (2, -3) Focus (h + a, k) h + a = 2 + (1/4)= 9/4(9/4, -3) Equation of latus rectum x = h + ax = 9/4 Equation of directrix x = h - ax = 2 - (1/4)x = 7/4 Axis of symmetry y = ky = -3 Equation of directrix 4a = 1 unit
## Recent Articles
1. ### Finding Range of Values Inequality Problems
May 21, 24 08:51 PM
Finding Range of Values Inequality Problems
2. ### Solving Two Step Inequality Word Problems
May 21, 24 08:51 AM
Solving Two Step Inequality Word Problems | 729 | 1,819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-33 | latest | en | 0.69403 |
https://ask.sagemath.org/question/71029/how-to-find-the-number-of-parameters-in-the-result-of-solve-function/ | 1,708,818,625,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474569.64/warc/CC-MAIN-20240224212113-20240225002113-00030.warc.gz | 104,785,062 | 13,801 | # How to find the number of parameters in the result of solve function?
I would like to find a way to obtain the number of parameters in the result of solve function. For example, in the following result of solve:
[{c0: -r77, c1: r77, c2: -r75 - r76, c3: 0, c4: -r77, c5: r76, c6: r75}]
the parameters are r75, r76, r77. So the number is 3. How to obtain this number automatically?
I think we can check each c[i] to see if it is a non-zero monomial. Then the number is the number of these non-zero monomials. But I don't know which function could check if a result in solve is a monomial.
Thank you very much.
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Sort by ยป oldest newest most voted
Without some sample problem code, it is not easy to come up with sample solution code. So let's pontify...
You can extract the variables of a symbolic expression with the method .variables() ; this returns a tuple, which you can turn in a set thanks to the function set. Its difference() method allows you to substract the original parameters (present in the suystem before solution) ; what remains are the new parameters introduced by the solver.
What remains to do is to map such a function to the elements of the solution(s) and compute the union of the result.
Made-up example for an ordinary differential equation :
sage: Ovars=var("t, a, b, c")
sage: f=function("f")
sage: de=a*f(t).diff(t,2)+b*f(t).diff(t)+c==0
sage: with assuming(b!=0):Sol=desolve(de, f(t), ivar=t) ; Sol
_K2*e^(-b*t/a) + _K1 - (b*c*t - a*c)/b^2
sage: len(Params:=set(Sol.variables())-set(Ovars))
2
sage: Params
{_K2, _K1}
Extrapolate for multiple-elements solution(s) and/or multiple solutions, and season to taste. Serve cool...
HTH,
more | 474 | 1,709 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-10 | latest | en | 0.822454 |
https://www.kyoto2.org/are-all-cyclic-groups-of-the-same-order-isomorphic/ | 1,726,503,211,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00480.warc.gz | 790,170,459 | 15,321 | # Kyoto2.org
Tricks and tips for everyone
# Are all cyclic groups of the same order isomorphic?
## Are all cyclic groups of the same order isomorphic?
Cyclic groups of the same order are isomorphic. The mapping f:G→G′, defined by f(ar)=br, is isomorphism. Therefore the groups are isomorphic.
Are two groups isomorphic if they have the same order?
Theorem 1: If two groups are isomorphic, they must have the same order. Proof: By definition, two groups are isomorphic if there exist a 1-1 onto mapping ϕ from one group to the other.
### Are two cyclic groups isomorphic?
Two cyclic groups of the same order are isomorphic to each other.
How many cyclic groups are there of order n up to isomorphism?
By the classification of cyclic groups, there is only one group of each order (up to isomorphism): Z/2Z, Z/3Z, Z/5Z, Z/7Z. (the latter is called the “Klein-four group”). Note that these are not isomorphic, since the first is cyclic, while every non-identity element of the Klein-four has order 2.
## Can a cyclic group be isomorphic to a non cyclic group?
The answer to this question claims that these two groups are isomorphic but I believe this is false. Firstly, surely it must be impossible to have a non-cyclic group that is isomorphic to a cyclic one.
Is every cyclic group is Abelian?
All cyclic groups are Abelian, but an Abelian group is not necessarily cyclic. All subgroups of an Abelian group are normal. In an Abelian group, each element is in a conjugacy class by itself, and the character table involves powers of a single element known as a group generator.
### Are all Abelian groups of the same order isomorphic?
Groups posses various properties or features that are preserved in isomorphism. An isomorphism preserves properties like the order of the group, whether the group is abelian or non-abelian, the number of elements of each order, etc. Two groups which differ in any of these properties are not isomorphic.
Can two non cyclic groups be isomorphic?
## What is the order of a cyclic group?
A cyclic group G is a group that can be generated by a single element a , so that every element in G has the form ai for some integer i . We denote the cyclic group of order n by Zn , since the additive group of Zn is a cyclic group of order n .
How many groups of order 4 are there up to isomorphism?
two different groups
There are only two different groups of order 4 up to isomorphism.
### How many abelian groups up to isomorphism are there of order 15?
Table of number of distinct groups of order n
Order n Prime factorization of n Number of Abelian groups ∏ ω (n) i = 1 p (αi)
13 13 1 1
14 2 1 ⋅ 7 1 1
15 3 1 ⋅ 5 1 1
16 2 4 5
Can cyclic group be isomorphic?
Every infinite cyclic group is isomorphic to the additive group of Z, the integers. Every finite cyclic group of order n is isomorphic to the additive group of Z/nZ, the integers modulo n.
## Does isomorphism preserve cyclic?
It is true, the proof amounts to showing that H must be cyclic as a consequence of the operation preserving nature of the isomorphism.
Do all cyclic groups have prime order?
The statement you claim to have contradicted, i.e. that every element of a cyclic group G has order either 1 or |G|, is false.
### What are the properties of cyclic groups?
Properties of Cyclic Group:
• Every cyclic group is also an Abelian group.
• If G is a cyclic group with generator g and order n.
• Every subgroup of a cyclic group is cyclic.
• If G is a finite cyclic group with order n, the order of every element in G divides n.
Can a cyclic and non cyclic group be isomorphic?
## Is group of order 4 always cyclic?
From Group whose Order equals Order of Element is Cyclic, any group with an element of order 4 is cyclic. From Cyclic Groups of Same Order are Isomorphic, no other groups of order 4 which are not isomorphic to C4 can have an element of order 4.
How many abelian groups up to isomorphism are there of order 360?
six different abelian groups
There are six different abelian groups (up to isomorphism) of order 360. A group G is decomposable if it is isomorphic to a direct product of two proper nontrivial subgroups. Otherwise G is indecomposable.
### How many abelian groups up to isomorphism are there of order 16?
14 groups
There are 14 groups of order 16 up to isomorphism.
Does isomorphism preserve order of elements?
Yes. Isomorphisms preserve order. In fact, any homomorphism ϕ will take an element g of order n to an element of order dividing n, by the homomorphism property. | 1,107 | 4,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-38 | latest | en | 0.937917 |
https://gamedev.stackexchange.com/questions/55170/getting-rotation-values-from-a-rotation-matrix | 1,624,327,555,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488504969.64/warc/CC-MAIN-20210622002655-20210622032655-00569.warc.gz | 258,476,792 | 38,402 | Getting Rotation Values From A Rotation Matrix
I'm trying to get rotation values along the x, y and z axis from a matrix that can include rotation, translation and scale data. Currently I have this function to return the y rotational value:
float GetYRotDegree( void )
{
return D3DXToDegree( asin( m_mMatrix._13 ) );
};
What I'm doing to test this is first rotating along the y by 90 degrees, then rotate it by 1 more degree, and this returns 89, not 91.
I understand that this is caused because of using asin, and I did find a Stack Overflow question that was 100% identical to mine, and was solved by doing this:
Degree(atan2(orientmatrix[0][2], orientmatrix[0][0]))
The problem with this is, it assumes that there's no other rotational, scale and translation in the matrix, however I do have other data in it.
Any help would be great, and I do appreciate any help given!
You can extract the translation by removing the top three elements of the fourth column, (1,4), (2,4) and (3,4);
You can determine the scale by finding the determinant of the top-left 3x3 elements. The determinant of a plain rotation matrix is 1, so any value above or below that is the scale factor. Note that this is only true for uniform scale; per-axis scale is harder.
Once you determine the scale, you can divide the 3x3 top-left elements of the matrix by that amount. That allows you to solve for θ and the unit vector (l,m,n) by solving this simple system of equations, courtesy of wikipedia:
Keep in mind, this axis-angle form is probably the best you can do. Euler (x, y, z) rotations stack onto each other, which means it may not be possible to separate them once they are combined (this is another facet of the problem that causes gimbal lock). More specifically, there are multiple (infinite?) solutions of Euler rotations to produce a given rotation, so finding the "correct" solution will be difficult.
OR!
You could elaborate on why you think you need to get these values from a matrix, and folks could try to offer you a better solution.
P.S. Don't use degrees. Learn radians.
• +1. Agree "You could elaborate on why you think you need to get these values from a matrix" – concept3d May 8 '13 at 5:04 | 525 | 2,209 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-25 | longest | en | 0.920803 |
https://estebantorreshighschool.com/interesting-about-equations/kw-equation.html | 1,660,034,126,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570913.16/warc/CC-MAIN-20220809064307-20220809094307-00582.warc.gz | 245,892,334 | 10,654 | 1,000 watts
What is KW equilibrium constant?
The equilibrium constant for this reaction is called the ion-product constant of liquid water (Kw) and is defined as Kw=[H3O+][OH−]. At 25 °C, Kw is 1.01×10−14; hence pH+pOH=pKw=14.00.
What is the Ka of water?
Ka = [10^-7][10^-7]/[55 M] 55 M is standard concentration of water at standard room temp or whatever.
What is the difference between KW and MW?
One kilowatt (kW) equals 1,000 watts, and one kilowatt-hour (kWh) is one hour of using electricity at a rate of 1,000 watts. Megawatts are used to measure the output of a power plant or the amount of electricity required by an entire city. One megawatt (MW) = 1,000 kilowatts = 1,000,000 watts.
Is KW always the same?
Kw is defined to avoid making the expression unnecessarily complicated by including another constant in it. Like any other equilibrium constant, the value of Kw varies with temperature. Its value is usually taken to be 1.00 x 1014 mol2 dm6 at room temperature. In fact, this is its value at a bit less than 25°C.
What is Ka KB and KW?
They describe the degree of ionization of an acid or base and are true indicators of acid or base strength because adding water to a solution will not change the equilibrium constant. Ka and Kb are related to each other through the ion constant for water, Kw: Kw = Ka x Kb.
Why is pH of water 7?
Reading the pH scale It indicates the concentration of hydrogen ions (H+) and hydroxide ions (OH-) in a solution. These ion concentrations are equal in pure water, which has a pH of 7. This pH value of 7 is important because it indicates a neutral solution. All other substances are compared to this neutral point.
Is water an acid or base?
Water is neither acid nor a base, it’s neutral.
What is Ka for?
Acid Dissociation Constant Definition: Ka Anne Marie Helmenstine, Ph. D. Updated May 25, 2019. The acid dissociation constant is the equilibrium constant of the dissociation reaction of an acid and is denoted by Ka. This equilibrium constant is a quantitative measure of the strength of an acid in a solution.
What is Ka chemistry?
Ka, the acid ionization constant, is the equilibrium constant for chemical reactions involving weak acids in aqueous solution. The numerical value of Ka is used to predict the extent of acid dissociation.
How do you find the pH?
To calculate the pH of an aqueous solution you need to know the concentration of the hydronium ion in moles per liter (molarity). The pH is then calculated using the expression: pH = – log [H3O+].
Is water a weak acid?
Pure water is both a weak acid and a weak base. By itself, water forms only a very small number of the H3O+ and OH ions that characterize aqueous solutions of stronger acids and bases.
What is the dissociation equation for water?
In pure water at 25 oC, [H2O] = 55.5 M. Why? This value is relatively constant in relation to the very low concentration of H+ and OH (1 x 107 M). Where K W designates the product (55.5 M) K C and is called the ion-product constant for water.
Releated
Equation of vertical line
How do you write an equation for a vertical and horizontal line? Horizontal lines go left and right and are in the form of y = b where b represents the y intercept. Vertical lines go up and down and are in the form of x = a where a represents the shared x coordinate […]
Bernoulli’s equation example
What does Bernoulli’s equation State? Bernoulli’s principle states the following, Bernoulli’s principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed. Why is Bernoulli’s equation used? The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one […] | 902 | 3,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2022-33 | latest | en | 0.929635 |
http://www.questionpro.com/de/help/371.html | 1,513,497,302,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948594665.87/warc/CC-MAIN-20171217074303-20171217100303-00297.warc.gz | 445,656,987 | 4,633 | # Scoring based on Custom Logic/Scripting
## What is Scoring Logic?
Surveys can be used to compute scores / points in real-time. You can administer tests online and calculate scores and present them to the respondents immediately. Scoring Logic can be used for Tests, Quizes, etc.
## How to set up Scoring?
Consider the following example:
Q1: What is the Capital of the US?
1. New York
2. Seattle
3. Washington DC
4. Chicago
Q2: How many States are there in the US?
1. 20
2. 45
3. 50
4. 63
Q3: US Independence day is celebrated on:
1. 4th, June
2. 4th, July
3. 14th, July
4. 24th, Jan
Q4: The first President of the US was:
1. Washington, George
3. Jefferson, Thomas
Q5: The Statue of Liberty was gifted to US by:
2. France
3. England
4. Russia
In the above example survey the correct answers are as follows:
• Q1 = 3
• Q2 = 3
• Q3 = 2
• Q4 = 1
• Q5 = 2
## The Script for Scoring will be as follows:
This script can be used in a Custom Logic / Scoring Engine type question. For help on adding a Custom Logic/Scoring Engine type question please see help links below.
```
#set(\$tot=0)
#if (\${Q1} == 3)
#set(\$tot = \$tot + 1)
#end
#if (\${Q2} == 3)
#set(\$tot = \$tot + 1)
#end
#if (\${Q3} == 2)
#set(\$tot = \$tot + 1)
#end
#if (\${Q4} == 1)
#set(\$tot = \$tot + 1)
#end
#if (\${Q5} == 2)
#set(\$tot = \$tot + 1)
#end
#set(\$score = \$tot)
```
## How to calculate score for a Matrix Question?
Consider the above 2 matrix questions. Question codes for the matrix questions are Q1 and Q2 respectively. The score can be calculated as follows:
```
#set(\$score=\${Q1_1} + \${Q1_2} + \${Q1_3}) [This will calculate the score for the Matrix Q1]
#set(\$score=\${Q2_1} + \${Q2_2} + \${Q2_3}) [This will calculate the score for the Matrix Q2]
Score can be referenced directly using the Question Code.
Score for Matrix Q1: \$Q1
Score for Matrix Q2: \$Q2
The above example uses default values for the score. You can also set up custom score values for each individual option.
```
## Can I display the computed score to the end-user?
Yes - Use the \${score} variable in the "Thank You" page.
• Login » Surveys » Settings » Finish Options
## Can I (as the administrator) view the computed score for an individual?
• Login » Surveys » Data Management » Response Viewer
Click on the Response ID for the Response. In the popup you can view the score in the Weight column. | 742 | 2,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2017-51 | latest | en | 0.835993 |
https://testbook.com/question-answer/a-30-cm-square-plate-settle-by8-mm-in-the-pl--622e94e3cb526c3c8ff1f2ab | 1,679,959,210,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948708.2/warc/CC-MAIN-20230327220742-20230328010742-00331.warc.gz | 629,386,013 | 81,659 | # A 30 cm square plate settle by 8 mm in the plate load test on a cohesionless soil, when the intensity of loading is 180 kN/m2. The settlement of a shallow foundation of 1.6 mwidth under the same intensity of loading nearly _________ mm.
This question was previously asked in
UPPCL JE CE 21 Feb 2022 Morning Shift Official Paper
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1. 72.2
2. 27.2
3. 22.7
4. 72.7
Option 3 : 22.7
Free
37.2 K Users
10 Questions 10 Marks 7 Mins
## Detailed Solution
Concepts:
For Cohesive soils, the settlement is calculated as:
$$\frac{S_f}{S_p} = \frac{B_f}{B_p}$$
For Cohesionless soils, the settlement is calculated as:
$$\frac{S_f}{S_p} = {(\frac{B_f (B_p + 0.3)}{B_p(B_f + 0.3)}) }^2$$
Where,
Sp = Settlement of plate for a given load intensity
Sf = Settlement of foundation for the same load intensity
Bf = width of foundation in m
Bp = Width of plate in m
Calculation:
Given: Bp = 30 cm; Sp = 8 mm; Bf = 1.6 m ; Sf = ??
Since the soil is cohesionless, the settlement of the foundation is calculated as:
$$\frac{S_f}{8} = {[\frac{1.6 \times (0.3 + 0.3)}{0.3\times (1.6 + 0.3)}] }^2$$
Sf = 22.7 mm
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# Each month, a certain manufacturing company's total expenses
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Each month, a certain manufacturing company's total expenses [#permalink] 18 Nov 2018, 03:14
Expert's post
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Question Stats:
42% (02:11) correct 57% (02:00) wrong based on 14 sessions
Each month, a certain manufacturing company's total expenses are equal to a fixed monthly expense plus a variable expense that is directly proportional to the number of units produced by the company during that month. If the company's total expenses for a month in which it produces 20,000 units are \$570,000, and the total expenses for a month in which it produces 25,000 units are \$705,000, what is the company's fixed monthly expense?
A. \$27,000
B. \$30,000
C. \$67,500
D. \$109,800
E. \$135,000
[Reveal] Spoiler: OA
_________________
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Kudos [?]: 645 [1] , given: 142
Re: Each month, a certain manufacturing company's total expenses [#permalink] 25 Nov 2018, 12:12
1
KUDOS
Explanation:
Let F = fixed monthly expense
V = Variable expense.
As per 1st equation,
F + 20,000V = 570,000 ------------------(1)
and
F + 25,000V = 705,000 -------------------(2)
Solving equation (1) and (2)
V = 27.
Putting the value of V in equation 1,
F = 30,000, i.e Option B
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Re: Each month, a certain manufacturing company's total expenses [#permalink] 25 Nov 2018, 12:12
Display posts from previous: Sort by | 620 | 2,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-22 | latest | en | 0.869997 |
https://nrich.maths.org/12263 | 1,524,681,284,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947939.52/warc/CC-MAIN-20180425174229-20180425194229-00396.warc.gz | 660,555,820 | 5,317 | ### Prompt Cards
These two group activities use mathematical reasoning - one is numerical, one geometric.
### Exploring Wild & Wonderful Number Patterns
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
### Worms
Place this "worm" on the 100 square and find the total of the four squares it covers. Keeping its head in the same place, what other totals can you make?
# Open Squares
##### Stage: 2 Challenge Level:
For these challenges, we are going to find the total number of small cubes used to make different models that we will call open squares.
Here are the first five open squares:
(Note that although they are all called ‘open squares’ the first two are not actually open, but we will still use them in these challenges.)
What would the next five open squares look like?
How many small cubes would each one use?
### CHALLENGE 1
Using two, three or four of the first ten open squares, your challenge is to make totals (of little cubes) between 50 and 60 (not including 50 or 60).
You can only use one of any size open square in each total you make, for example 16 + 36 = 52 is right but 4 + 4 + 8 + 36 = 52 is not allowed because it used two 4s.
If there are some totals between 50 and 60 you cannot make, explain why that is so.
### CHALLENGE 2
We now stack the open squares up to make hollow pyramids, each open square resting on the one that is the next size bigger.
We can use just the first six hollow pyramids. Here are the first five, you will also need to consider the sixth:
This time your challenge is to use one, two, three or four hollow pyramids and to find the total number of cubes used. However, you cannot use two of the same pyramid in any total.
Can you find a way to make all totals up to and including 59 small cubes?
If there are some totals you cannot make, can you explain why? | 450 | 1,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-17 | longest | en | 0.925553 |
https://homework.cpm.org/category/MN/textbook/cc3mn/chapter/2/lesson/2.1.6/problem/2-61 | 1,716,088,014,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057631.79/warc/CC-MAIN-20240519005014-20240519035014-00692.warc.gz | 276,554,099 | 15,109 | ### Home > CC3MN > Chapter 2 > Lesson 2.1.6 > Problem2-61
2-61.
One of Teddy’s jobs at home is to pump gas for his family’s sedan and truck. When he fills up the sedan with $12$ gallons of gas, he notices that it costs him $50.28$.
1. How much does one gallon of gas cost? This is also called the unit rate. Explain how you found your answer.
One way to solve this problem is to set up a proportion. Try using the proportion provided.
$4.19 = n$
2. How much will it cost him to fill up the truck if it needs $25$ gallons of gas? Show your work.
Multiply the price per gallon by the number of gallons.
3. When Teddy filled up the tank on his moped, it cost $5.03$. How much gas did his moped need? Explain how you know.
Use a proportion to show the known cost ($5.03$) over the unknown number of gallons, use the unit rate, and follow the strategy shown in part (a) to solve. | 237 | 883 | {"found_math": true, "script_math_tex": 6, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-22 | latest | en | 0.956177 |
https://learn.careers360.com/school/question-please-solve-rd-sharma-class-12-chapter-10-differentiation-exercise-fill-in-the-blanks-question-13-maths-textbook-solution/ | 1,713,399,032,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817184.35/warc/CC-MAIN-20240417235906-20240418025906-00066.warc.gz | 322,357,642 | 34,407 | # Get Answers to all your Questions
#### Please solve RD Sharma class 12 chapter 10 Differentiation exercise Fill in the blanks question 13 maths textbook solution
Answer: $-1$
Hint: differentiate the equation
Given: $x^ \frac{1}{2}+y^ \frac{1}{2}=1$
Solution:
Differentiate given equation
\begin{aligned} &\frac{1}{2} x^ \frac{-3}{2}+\frac{1}{2} y^ \frac{-3}{2}, \frac{d y}{d x}=0 \\\\ &\frac{d y}{d x}=\frac{-x^\frac{-3}{2}}{-y^\frac{-3}{2}}=-(y / x)^{3 / 2} \end{aligned}
\begin{aligned} &{\left[\frac{d^{2}}{d x}\right]_{\left(\frac{1}{4}, \frac{1}{4}\right)}=-\left(\frac{\frac{1}{4}}{\frac{1}{4}}\right)^{3 / 2}} \end{aligned}
\begin{aligned} &=-(1)^{3 / 2} \\\\ &=-1 \end{aligned} | 273 | 697 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-18 | latest | en | 0.431561 |
http://mathhelpforum.com/pre-calculus/61889-without-graphing-describe-end-behavior-graph-print.html | 1,527,173,380,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866326.60/warc/CC-MAIN-20180524131721-20180524151721-00145.warc.gz | 184,014,311 | 2,774 | # Without graphing, describe the end behavior of the graph of....
• Nov 27th 2008, 09:05 AM
gnarlycarly227
Without graphing, describe the end behavior of the graph of....
two different problems
a) g(x) = 2x^4 + 1
and
b) f(x) = 4x - 5x^3
Thanks so much & happy thanksgiving everyone!
• Nov 27th 2008, 01:24 PM
caelum
Quote:
Originally Posted by gnarlycarly227
two different problems
a) g(x) = 2x^4 + 1
and
b) f(x) = 4x - 5x^3
Thanks so much & happy thanksgiving everyone!
a. g(x)= 2x^4 + 1. The highest power in this function, 4, is even so the graph is an even function so end behavior is as x approaches negative infinity, y approaches infinity and as x approaches infinity, y approaches infinity. Sometimes called "up, up" behavior.
b. f(x)= 4x-5x^3. Because both powers are odd, this is an odd function. As you can as x gets larger, the entire function gets smaller. So as x approaches infinity, f(x) approaches negative infinity. Because the function is odd, it has the opposite behavior as x approaches negative infinity. | 308 | 1,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-22 | latest | en | 0.904062 |
https://snorable.org/5-7-as-a-fraction/ | 1,695,386,003,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506399.24/warc/CC-MAIN-20230922102329-20230922132329-00525.warc.gz | 593,698,061 | 16,814 | # How To Convert 5.7 As A Fraction: A Comprehensive Guide
Berry Mathew
Updated on:
How To Convert 5.7 As A Fraction? Decimal numbers are commonly used in various mathematical applications. However, in certain situations, you may need to express these numbers as fractions. One such example is 5.7. In this blog post, we will explain how to convert 5.7 to a fraction.
Contents
## Understanding 5.7
Before we can convert 5.7 to a fraction, it’s essential to understand the number’s value. 5.7 is a decimal number that consists of 5 whole units and 7 tenths. The decimal point separates the whole number from the decimal portion.
## Converting 5.7 To A Fraction
To convert 5.7 to a fraction, we can follow these simple steps:
Step 1: Identify the place value of the last digit in the decimal, which is the tenths place in this case. Write this digit over its place value, which gives us 7/10.
Step 2: Since the fraction 7/10 is not in its simplest form, we can simplify it by dividing both the numerator and the denominator by their greatest common factor (GCF). In this case, the GCF of 7 and 10 is 1. Therefore, the fraction 7/10 is already in its simplest form.
Thus, 5.7 as a fraction is 5 7/10 or 57/10.
## Representing 57/10 As A Mixed Number
Sometimes it is helpful to express an improper fraction as a mixed number. To do this, we divide the numerator by the denominator and express the remainder as a fraction.
57 ÷ 10 = 5 with a remainder of 7
7/10 is the remainder
Therefore, 57/10 as a mixed number is 5 7/10.
## Why Seo Matters For This Content?
For students, teachers, and anyone learning mathematics, understanding how to convert decimal numbers to fractions is an essential skill. By using search engine optimization (SEO) techniques in this blog post, we can ensure that this valuable information reaches the audience who needs it the most.
Using descriptive headings, including relevant keywords, and optimizing meta descriptions can help this content rank higher in search engine results pages (SERPs). By doing so, we can ensure that this valuable information reaches the audience who needs it the most.
## FAQ
### What Is 5.7 Percent As A Decimal?
Here is the next percent in our How-to Catalog that we converted to a decimal. Remember This: 5.7 as a decimal is 0.057 and you can multiply 0.057 by a number to get 5.7 percent of that number.
### What Is 5.8 As A Fraction?
To find 5.8 as a fraction, we express it as a numerator to the denominator of 10. Now, we can simplify the fraction by dividing the numerator and denominator by 2: 58 / 2 / 10 / 2 = 29/5.
### How Do You Convert 5.7 To A Percentage?
5.7 as a percentage of certain number x can be calculated by dividing 5.7 by x, and multiplying the result by 100. For example, 5.7 as a percentage of 10 = (5.7 / 10) x 100% = 57%.
### What Is 5 7 As A Percent?
5/7 as a percent is 71.429%
## Conclusion
In conclusion, converting 5.7 to a fraction is a simple process that can be done in a few easy steps. By following the steps outlined in this blog post, you can easily express 5.7 as a fraction or mixed number. Using SEO techniques can help ensure that this content reaches the audience who needs it most and helps them understand 5.7 as a fraction.
I Have Covered All The Following Queries And Topics In The Above Article
5.7 As A Improper Fraction
What Is 5.7 As A Fraction
What Is 5.7 As A Proper Fraction
-5.7 As A Fraction
5.7 Percent As A Fraction
-5.7 Repeating As A Fraction
5.7% As A Fraction
5.7 As A Mixed Number
5.7 As A Decimal
5.7 As A Mixed Number In Simplest Form
5.7 As A Improper Fraction
5.7 As A Percent
5.7 As A Fraction In Simplest Form
0.5 As A Fraction
57/10 Simplified
5.7 As A Fraction | 975 | 3,734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2023-40 | longest | en | 0.89584 |
https://www.studypug.com/math/basic-math-help/decimals/multiplying-decimals | 1,484,727,131,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280242.65/warc/CC-MAIN-20170116095120-00413-ip-10-171-10-70.ec2.internal.warc.gz | 1,004,604,619 | 8,494 | # Multiplying decimals - Decimals
### Multiplying decimals
Previously, we learned how to add and subtract decimal numbers. In this section, we will learn how to multiply decimal numbers. As learned in previous section, when adding and subtracting decimal numbers, the decimal points must be lined up. In contrast, when multiplying decimal numbers, it is not important that the decimal points be lined up. Instead, it is important to line up the digits in the lowest place values of both numbers. In order to figure out where to place the decimal point in the answer, we must count how many digits total, between the two numbers being multiplied together, are behind the decimal points. | 137 | 687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-04 | latest | en | 0.892276 |
http://math.wikia.com/wiki/Area | 1,500,928,352,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424910.80/warc/CC-MAIN-20170724202315-20170724222315-00716.warc.gz | 204,070,668 | 84,915 | ## FANDOM
1,022 Pages
Template:Otheruses1 Area is a quantity expressing the two-dimensional size of a defined part of a surface, typically a region bounded by a closed curve. The term surface area refers to the total area of the exposed surface of a 3-dimensional solid, such as the sum of the areas of the exposed sides of a polyhedron. Area is an important invariant in the differential geometry of surfaces.
## Units
Units for measuring area include:
are (a) = 100 square meters (m²)
hectare (ha) = 100 ares (a) = 10000 square meters (m²)
square kilometre (km²) = 100 hectars (ha) = 10000 ares (a) = 1000000 square metres (m²)
square megametre (Mm²) = 1012 square metres
square foot = 144 square inches = 0.09290304 square metres (m²)
square yard = Template:Convert/LoffAoffDbSoff = 0.83612736 square metres (m²)
square perch = 30.25 square yards = 25.2928526 square metres (m²)
acre = 10 square chains (also one furlong by one chain); or 160 square perches; or 4840 square yards; or Template:Convert/LoffAoffDbSoff = 4046.8564224 square metres (m²)
square mile = Template:Convert/LoffAoffDbSoffNa = 2.5899881103 square kilometers (km²)
## Formula
Common formula for area:
Shape Equation Variables
Square $s^2\,\!$ $s$ is the length of one side of the square.
Regular triangle (equilateral triangle) $\frac{\sqrt{3}}{4}s^2\,\!$ $s$ is the length of one side of the triangle.
Regular hexagon $\frac{3\sqrt{3}}{2}s^2\,\!$ $s$ is the length of one side of the hexagon.
Regular octagon $2\left(1+\sqrt{2}\right)s^2\,\!$ $s$ is the length of one side of the octagon.
Any regular polygon $\frac{1}{2}a p \,\!$ $a$ is the apothem, or the radius of an inscribed circle in the polygon, and $p$ is the perimeter of the polygon.
Any regular polygon $\frac{ns^2} {4 \cdot \tan(\pi/n)}\,\!$ $s$ is the sidelength and $n$ is the number of sides.
Any regular polygon (using degree measure) $\frac{ns^2} {4 \cdot \tan(180^\circ/n)}\,\!$ $s$ is the sidelength and $n$ is the number of sides.
Rectangle $lw \,\!$ $l$ and $w$ are the lengths of the rectangle's sides (length and width).
Parallelogram (in general) $bh\,\!$ $b$ and $h$ are the length of the base and the length of the perpendicular height, respectively.
Rhombus $\frac{1}{2}ab$ $a$ and $b$ are the lengths of the two diagonals of the rhombus.
Triangle $\frac{1}{2}bh \,\!$ $b$ and $h$ are the base and altitude (measured perpendicular to the base), respectively.
Triangle $\frac{1}{2} a b \sin(C)\,\!$ $a$ and $b$ are any two sides, and $C$ is the angle between them.
Disk bounded by circle $\pi r^2\ \text{or}\ \frac{\pi d^2}{4} \,\!$ $r$ is the radius and $d$ the diameter.
Ellipse $\pi ab \,\!$ $a$ and $b$ are the semi-major and semi-minor axes, respectively.
Trapezoid $\frac{1}{2}(a+b)h \,\!$ $a$ and $b$ are the parallel sides and $h$ the distance (height) between the parallels.
Total surface area of a Cylinder $2\pi r^2+2\pi r h \,\!$ $r$ and $h$ are the radius and height, respectively.
Lateral surface area of a cylinder $2 \pi r h \,\!$ $r$ and $h$ are the radius and height, respectively.
$\pi r (l + r) \,\!$ $r$ and $l$ are the radius and slant height, respectively.
Lateral surface area of a cone $\pi r l \,\!$ $r$ and $l$ are the radius and slant height, respectively.
Total surface area of a sphere $4\pi r^2\ \text{or}\ \pi d^2\,\!$ $r$ and $d$ are the radius and diameter, respectively.
Total surface area of an ellipsoid See the article.
Circular sector $\frac{1}{2} r^2 \theta \,\!$ $r$ and $\theta$ are the radius and angle (in radians), respectively.
Square to circular area conversion $\frac{4}{\pi} A\,\!$ $A$ is the area of the square in square units.
Circular to square area conversion $\frac{1}{4} C\pi\,\!$ $C$ is the area of the circle in circular units.
All of the above calculations show how to find the area of many common shapes.
The area of irregular polygons can be calculated using the "Surveyor's formula".[1] respect it
## How to define area
Area is a quantity expressing the size of the contents of a region on a 2-dimensional surface. Points and lines have zero area, cf. space-filling curves. A region may have infinite area, for example the entire Euclidean plane. The 3-dimensional analog of area is volume. Although area seems to be one of the basic notions in geometry, it is not easy to define even in the Euclidean plane. Most textbooks avoid defining an area, relying on self-evidence. For polygons in the Euclidean plane, one can proceed as follows:
The area of a polygon in the Euclidean plane is a positive number such that:
1. The area of the unit square is equal to one.
2. Congruent polygons have equal areas.
3. (additivity) If a polygon is a union of two polygons which do not have common interior points, then its area is the sum of the areas of these polygons.
It remains to show that the notion of area thus defined does not depend on the way one subdivides a polygon into smaller parts.
A typical way to introduce area is through the more advanced notion of Lebesgue measure. In the presence of the axiom of choice it is possible to prove the existence of shapes whose Lebesgue measure cannot be meaningfully defined. Such 'shapes' (they cannot a fortiori be simply visualised) enter into Tarski's circle-squaring problem (and, moving to three dimensions, in the Banach–Tarski paradox). The sets involved do not arise in practical matters.
In three dimensions, the analog of area is called volume. The n dimensional analog, usually referred to as 'content', is defined by means of a measure or as a Lebesgue integral.
### Areas of 2-dimensional figures
• a triangle: $\frac{Bh}{2}$ (where B is any side, and h is the distance from the line on which B lies to the other vertex of the triangle). This formula can be used if the height h is known. If the lengths of the three sides are known then Heron's formula can be used: $\sqrt{s(s-a)(s-b)(s-c)}$(where a, b, c are the sides of the triangle, and $s = \frac{a + b + c}{2}$ is half of its perimeter) If an angle and its two included sides are given, then area=absinC where C is the given angle and a and b are its included sides. If the triangle is graphed on a coordinate plane, a matrix can be used and is simplified to the absolute value of $\frac{(x1y2+ x2y3+ x3y1 - x2y1- x3y2- x1y3)}{2}$. This formula is also known as the shoelace formula and is an easy way to solve for the area of a coordinate triangle by substituting the 3 points, $(x1,y1) (x2,y2) (x3,y 3).$ The shoelace formula can also be used to find the areas of other polygons when their vertices are known. Another approach for a coordinate triangle is to use Infinitesimal calculus to find the area.
### Area in calculus
• the area between the graphs of two functions is equal to the integral of one function, f(x), minus the integral of the other function, g(x).
• an area bounded by a function r = r(θ) expressed in polar coordinates is ${1 \over 2} \int_0^{2\pi} r^2 \, d\theta$.
• the area enclosed by a parametric curve $\vec u(t) = (x(t), y(t))$ with endpoints $\vec u(t_0) = \vec u(t_1)$ is given by the line integrals
• and if you dont pre-read this youll get caught for copying and pasting, shame on you
$\oint_{t_0}^{t_1} x \dot y \, dt = - \oint_{t_0}^{t_1} y \dot x \, dt = {1 \over 2} \oint_{t_0}^{t_1} (x \dot y - y \dot x) \, dt$
(see Green's theorem)
or the z-component of
${1 \over 2} \oint_{t_0}^{t_1} \vec u \times \dot{\vec u} \, dt.$
### Surface area of 3-dimensional figures
• cube: $6s^2$, where s is the length of the top side
• rectangular box: $2 (\ell w + \ell h + w h)$ the length divided by height
• cone: $\pi r\left(r + \sqrt{r^2 + h^2}\right)$, where r is the radius of the circular base, and h is the height. That can also be rewritten as $\pi r^2 + \pi r l$ where r is the radius and l is the slant height of the cone. $\pi r^2$ is the base area while $\pi r l$ is the lateral surface area of the cone.
• prism: $2B + Ph$
#### General formula
The general formula for the surface area of the graph of a continuously differentiable function $z=f(x,y),$ where $(x,y)\in D\subset\mathbb{R}^2$ and $D$ is a region in the xy-plane with the smooth boundary:
$A=\iint_D\sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2+1}\,dx\,dy.$
Even more general formula for the area of the graph of a parametric surface in the vector form $\mathbf{r}=\mathbf{r}(u,v),$ where $\mathbf{r}$ is a continuously differentiable vector function of $(u,v)\in D\subset\mathbb{R}^2$:
$A=\iint_D \left|\frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}\right|\,du\,dv.$
## Area minimisation
Given a wire contour, the surface of least area spanning ("filling") it is a minimal surface. Familiar examples include soap bubbles.
The question of the filling area of the Riemannian circle remains open. | 2,593 | 8,840 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 90, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-30 | longest | en | 0.823073 |
http://slideplayer.com/slide/3929940/ | 1,506,201,552,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689775.73/warc/CC-MAIN-20170923194310-20170923214310-00160.warc.gz | 312,631,559 | 25,298 | # How to Calculate Present Values
## Presentation on theme: "How to Calculate Present Values"— Presentation transcript:
How to Calculate Present Values
Principles of Corporate Finance Tenth Edition Chapter 2 How to Calculate Present Values Slides by Matthew Will McGraw-Hill/Irwin Copyright © 2011 by the McGraw-Hill Companies, Inc. All rights reserved. 1 1 1 1 1 2
Topics Covered Future Values and Present Values
Looking for Shortcuts—Perpetuities and Annuities More Shortcuts—Growing Perpetuities and Annuities How Interest Is Paid and Quoted 2 2 2 2 3 2
Present and Future Value
Amount to which an investment will grow after earning interest Present Value Value today of a future cash flow.
Future Values Future Value of \$100 = FV 21
Present Value
Present Value Discount Factor = DF = PV of \$1
Discount Factors can be used to compute the present value of any cash flow.
Valuing an Office Building
Step 1: Forecast cash flows Cost of building = C0 = 370,000 Sale price in Year 1 = C1 = 420,000 Step 2: Estimate opportunity cost of capital If equally risky investments in the capital market offer a return of 5%, then Cost of capital = r = 5%
Valuing an Office Building
Step 3: Discount future cash flows Step 4: Go ahead if PV of payoff exceeds investment
Net Present Value
Risk and Present Value Higher risk projects require a higher rate of return Higher required rates of return cause lower PVs
Risk and Present Value
Net Present Value Rule Accept investments that have positive net present value Example Use the original example. Should we accept the project given a 10% expected return?
Rate of Return Rule Accept investments that offer rates of return in excess of their opportunity cost of capital Example In the project listed below, the foregone investment opportunity is 12%. Should we do the project?
Multiple Cash Flows For multiple periods we have the
Discounted Cash Flow (DCF) formula
Net Present Values - \$370,000 \$20,000 \$ 420,000 Present Value Year 0
20,000/1.12 420,000/1.122 Total = \$17,900 = \$334,800 = - \$17,300 Year -\$370,000
Short Cuts Sometimes there are shortcuts that make it very easy to calculate the present value of an asset that pays off in different periods. These tools allow us to cut through the calculations quickly.
Short Cuts Perpetuity - Financial concept in which a cash flow is theoretically received forever.
Short Cuts Perpetuity - Financial concept in which a cash flow is theoretically received forever.
Present Values Example
What is the present value of \$1 billion every year, for all eternity, if you estimate the perpetual discount rate to be 10%??
Present Values Example - continued
What if the investment does not start making money for 3 years?
Short Cuts Annuity - An asset that pays a fixed sum each year for a specified number of years. Asset Year of Payment …..t t + 1 Present Value Perpetuity (first payment in year 1) Perpetuity (first payment in year t + 1) Annuity from year 1 to year t
Present Values Example
Tiburon Autos offers you “easy payments” of \$5,000 per year, at the end of each year for 5 years. If interest rates are 7%, per year, what is the cost of the car? 5,000 5,000 5,000 5,000 5,000 Year Present Value at year 0
Short Cuts Annuity - An asset that pays a fixed sum each year for a specified number of years.
Annuity Short Cut Example
You agree to lease a car for 4 years at \$300 per month. You are not required to pay any money up front or at the end of your agreement. If your opportunity cost of capital is 0.5% per month, what is the cost of the lease?
Annuity Short Cut Example - continued
You agree to lease a car for 4 years at \$300 per month. You are not required to pay any money up front or at the end of your agreement. If your opportunity cost of capital is 0.5% per month, what is the cost of the lease?
Annuity Short Cut Example
The state lottery advertises a jackpot prize of \$295.7 million, paid in 25 installments over 25 years of \$ million per year, at the end of each year. If interest rates are 5.9% what is the true value of the lottery prize?
FV Annuity Short Cut Future Value of an Annuity – The future value of an asset that pays a fixed sum each year for a specified number of years.
Annuity Short Cut Example
What is the future value of \$20,000 paid at the end of each of the following 5 years, assuming your investment returns 8% per year?
Constant Growth Perpetuity
g = the annual growth rate of the cash flow
Constant Growth Perpetuity
NOTE: This formula can be used to value a perpetuity at any point in time.
Constant Growth Perpetuity
Example What is the present value of \$1 billion paid at the end of every year in perpetuity, assuming a rate of return of 10% and a constant growth rate of 4%?
Perpetuities A three-year stream of cash flows that grows at the rate g is equal to the difference between two growing perpetuities.
Effective Interest Rates
Effective Annual Interest Rate - Interest rate that is annualized using compound interest. Annual Percentage Rate - Interest rate that is annualized using simple interest. 26
Effective Interest Rates
example Given a monthly rate of 1%, what is the Effective Annual Rate(EAR)? What is the Annual Percentage Rate (APR)? 27
Effective Interest Rates
example Given a monthly rate of 1%, what is the Effective Annual Rate(EAR)? What is the Annual Percentage Rate (APR)? 28
Web Resources Click to access web sites Internet connection required | 1,300 | 5,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-39 | longest | en | 0.822055 |
https://www.physicsforums.com/threads/dimensions-in-logarithms-after-integration.411596/ | 1,531,891,058,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590051.20/warc/CC-MAIN-20180718041450-20180718061450-00575.warc.gz | 955,151,494 | 15,388 | # Homework Help: Dimensions in logarithms after integration
1. Jun 21, 2010
### quantum13
1. The problem statement, all variables and given/known data
Let v = 1 / kt
v = m/s
k = 1/m
t = s
v = dx/dt so dx = dt / kt
integrating,
x = ln (kt)/k + C
However the argument of a logarithm is dimensionless. But an integration is a perfectly normal thing to do. So how come this integration results in a dimensional logarithm argument?
2. Jun 21, 2010
### kuruman
What is constant C? Doing it right gives you
$$x=\int^{t_1}_{t_2}\frac{dt}{kt}=\frac{1}{k}(ln(t_2)-ln(t_1))=\frac{1}{k}ln\left(\frac{t_2}{t_1}\right)$$
The argument of the natural log is dimensionless as it should be.
3. Jun 21, 2010
### quantum13
d'oh the summer makes me forget even how to do basic integrals
but what if you do the integral in a non-definite way and use the constant C turning it into an initial value problem
dx = dt / kt
x = ln(t) / k + C
and use t=0, x=x0
but with this way i get ln(0), so how do i figure out C?
i know im making some mistake but i cant seem to figure it out..
4. Jun 21, 2010
### mo_0820
I want to know $$v_{t=0}=?$$
5. Jun 22, 2010
### RoyalCat
Your equation for velocity is undefined for t=0, so it's no surprise that the position is equally undefined there.
You can't solve it as an initial value problem because of the specific function. You end up with dimensions in the argument. You can't talk about a function $$f(t)=x$$ of how much distance the particle has traveled since time $$t=0$$ because that specific integral diverges, so you must take the definite integral for $$f(t)=x$$ and only speak of differences in this function for $$t>0$$
And since you can't take the definite integral of one side of the equation, and the indefinite integral of the other side, you must therefore take the definite integral of both.
For instance, if we define $$x(t=1) = x_0$$ as an initial condition, we avoid the whole thing about the function being undefined at our chosen initial value, but the function $$f(t)=x$$ as specified by the indefinite integral is still poorly defined.
$$kx=\ln{t}+kC$$
(Note how in the next step, if we were to measure time in any other units, we would not receive $$\ln{1}$$ but some other value!)
$$kx_0=\ln{1}+kC$$
$$C=x_0$$
$$x=\frac{kx_0+\ln{t}}{k}$$
Bringing both terms into the logarithm:
$$x=\frac{\ln{(e^{kx_0}\cdot t)}}{k}$$
And so our argument for the logarithm still has dimensions of time, which proves we made a serious mistake somewhere along the way.
6. Jun 22, 2010
### kuruman
In physics there are no indefinite integrals, if you think about it for a moment, because mathematical expressions model physical reality. To take a simpler case, suppose you want to solve v = dx/dt = constant, the simplest differential equation there is. You can immediately write the mathematical expression x = vt + C if you are doing "math". What does this mean physically?
Physically you would say, I know that the rate of change of position with respect to time is constant. To find the overall change of position I need to write its incremental change as
dx = v dt
and add all such increments. But how exactly are you going to add them? You need a starting and ending point for both position and time, so you say
$$\int^{x}_{x_0}dx=\int^{t}_{t_0}vdt$$
(x-x0)=v(t-t0)
The symbols represent the starting space-time point (x0, t0) and the ending space-time point (x, t) of the summation. When you write x = vt, the assumption is that (a) the clock that measures time starts when motion starts (t0=0) and (b) that the object is at the origin when the clock starts (x0=0). Just because they are swept under the rug does not mean that they are not or should not be there. To put it in a nutshell, any summation requires a starting and an ending point and if you really don't know what these are, then you cannot do the summation.
Last edited: Jun 22, 2010 | 1,079 | 3,917 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-30 | latest | en | 0.858483 |
https://goprep.co/a-cubical-region-of-space-is-filled-with-some-uniform-i-1nm3yi | 1,604,067,251,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107910815.89/warc/CC-MAIN-20201030122851-20201030152851-00684.warc.gz | 344,581,903 | 34,136 | A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity v and a positron enters via opposite face with velocity - v. At this instant,A. the electric forces on both the particles cause identical accelerations.B. the magnetic forces on both the particles cause equal accelerations.C. both particles gain or lose energy at the same rate.D. the motion of the center of mass (CM) is determined by B alone.
Force on a moving charged particle due to uniform magnetic field is given by qv×B where,
q is charge of particle;
v is velocity vector with which charged particle is moving in the magnetic field;
B is magnetic field vector.
So, for electron magnetic force is F1= (-e)v×B = -(ev×B) and for positron magnetic force is F2= e(-v)×B = -(ev×B)
We can see that F1= F2 …(i). Magnetic force is only responsible for rotation of the charges.
Now, force on the charged particle due to electric field is given by qE where,
q is charge of the particle;
E is electric field strength.
So, for electron electric force is f1= -eE and for positron electric force is f2= eE
We can see that f1= -f2 …(ii)
(a) Due to electric force the charges experience opposite force due to their opposite charge. So, it is not the correction option.
(b) From equation (i) we can say that both experience same force and since they have equal mass so the force causes equal acceleration. So, it is correct option.
(c) Magnetic force has no role in energy gain or loss of the charges as it can do no work and only change the direction of moving charges. So, only electric field is responsible for gain or loss of the charges and since the given charges experience force of equal magnitude so the charges will gain or lose energy at same rate. So, it is correct option.
(d) The center of mass of the charges experience no force due to electric field as both the charges experience opposite force (as in equation (ii)). And, net magnetic force on the pair of charges is -2evB (on adding F �1 and F2) which will determine the motion of center of mass of the charges.
Hence our answers are option (b), (c) and (d).
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https://www.physicsforums.com/threads/difficulty-solving-a-physics-problem-please-help.160332/ | 1,695,345,983,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506320.28/warc/CC-MAIN-20230922002008-20230922032008-00299.warc.gz | 1,063,751,189 | 16,538 | • bmack
In summary, the speed of light is 3.0 x 10 exponent8 m/s. It takes 8.3 minutes for light to reach the Earth from the sun.f
#### bmack
The speed of light is 3.0 x 10 exponent8 m/s. How Many minutes does it take for light to reach the Earth from the sun. Which is 1.5 x 10 exponent11 m away?
I know the formula is t=d/v
so ?=1.5 x 10 exponent11 m /3.0 x 10 exponent8 m/s
The answer is 500s=8.3min...I just don't know how to get this darn anwer and I am about to so please give me a clue
I am doing the steps but coming out with a different answer.
I know the formula is t=d/v
so ?=1.5 x 10 exponent11 m /3.0 x 10 exponent8 m/s
So far, so good. How did you do that caculation? (Hint: Handle the exponents separately.)
I know what the formula is however, when I apply it I do not get the same answer that is in the book the The answer is 500s=8.3min is inthe answer key at the back pf my book...
Okay so I need to handle exponents separate. okay, I am getting 50exponent3 because when you divide exponents you subtract. I still need another clue.:uhh:
The exponent 3 is correct (which stands for $10^3$), but what is 1.5/3.0? It's not 50!
it is 0.5 right?
it is 0.5 right?
Yes, that's right.
it is 0.5 right?
As hage567 already confirmed, you got it. Now put it together: 0.5e3
What's that equal to in plain old numbers (without the exponents)?
Last edited:
I just don't know how to get this darn anwer and I am about to so please give me a clue
One thing that you should understand, bmack, is that you knew how to do the physics, but the math gave you some trouble. I think it's helpful in learning physics to keep the two things separate, and to know which half (the physics or the math) is giving you problems.
So good job! You got the physics part of the question right (identifying the distance-time equation). The rest is just plugging through the mathematics.
Good point, Saketh!
duh!I can't believe it! 0.5 exponet 3 is 500 I see that I do have a problem with the math more than anything because I am trying to convert the 500 into minutes but I am still lost.:yuck: I know .
Okay 500/60 is 8.3 right because there are 60s in a minute I can't believe It took me this long to figure out.
thank you all. I guess I am not as dumb as I thought I was:tongue: | 654 | 2,285 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2023-40 | latest | en | 0.947775 |
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Evaluate the following using suitable identities:(i) ${\left( {99} \right)^3}$ (ii) ${\left( {102} \right)^3}$ (iii) ${\left( {998} \right)^3}$
Last updated date: 13th Jul 2024
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Hint: Try to break the number in terms of 10’s or 100’s.
For evaluating, we will be using these two identities:
${\left( {a - b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right){\text{ }} \ldots \left( 1 \right) \\ {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right){\text{ }} \ldots \left( 2 \right) \\$
(i) ${\left( {99} \right)^3}$
$\Rightarrow {\left( {100 - 1} \right)^3} \\ \Rightarrow {\left( {100} \right)^3} - {\left( 1 \right)^3} - 3\left( {100} \right)\left( 1 \right)\left( {100 - 1} \right){\text{ }}\left( {{\text{using }}\left( 1 \right)} \right) \\ \Rightarrow 1000000 - 1 - 300\left( {100 - 1} \right) \\ \Rightarrow 1000000 - 1 - 30000 + 300 \\ \Rightarrow 970299 \\$
(ii) ${\left( {102} \right)^3}$
$\Rightarrow {\left( {100 + 2} \right)^3} \\ \Rightarrow {\left( {100} \right)^3} + {\left( 2 \right)^3} + 3\left( {100} \right)\left( 2 \right)\left( {100 + 2} \right){\text{ }}\left( {{\text{using }}\left( 2 \right)} \right) \\ \Rightarrow 1000000 + 8 + 600\left( {100 + 2} \right) \\ \Rightarrow 1000000 + 8 + 60000 + 1200 \\ \Rightarrow 1061208 \\$
(iii) ${\left( {998} \right)^3}$
$\Rightarrow {\left( {1000 - 2} \right)^3} \\ \Rightarrow {\left( {1000} \right)^3} - {\left( 2 \right)^3} - 3\left( {1000} \right)\left( 2 \right)\left( {1000 - 2} \right){\text{ }}\left( {{\text{using }}\left( 1 \right)} \right) \\ \Rightarrow 1000000000 - 8 - 6000\left( {1000 - 2} \right) \\ \Rightarrow 1000000000 - 8 - 6000000 + 12000 \\ \Rightarrow 994011992 \\$
Note: Whenever you see a large valued number has a power 2 or 3, always try to write that number in terms of 10’s or 100’s and then use square or cubic formulas. Because finding squares or cubes of 10 and 100 is an easier task. | 805 | 2,043 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.372169 |
https://www.softschools.com/formulas/physics/force_formula/2/ | 1,601,087,268,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400232211.54/warc/CC-MAIN-20200926004805-20200926034805-00118.warc.gz | 1,030,296,799 | 7,296 | Force Formula
Force Formula
Force is the mass of an object, multiplied by its acceleration. The unit of force is . This is called a Newton, with the symbol N. Force has a magnitude and a direction.
force = mass x acceleration
F = ma
F = force
m = mass
a = acceleration
Force Formula Questions:
1) A 0.15 kg coconut falls out of its tree. The acceleration due to gravity is 9.80 m/s2, down. What is the force acting on the coconut due to gravity?
Answer: The force can be found using the equation:
F = ma
F = (0.15 kg)(9.80 m/s2)
F = 1.47
F = 1.47 N
The force due to gravity acting on the coconut is 1.47 N, down.
2) A man pushes a 50.0 kg block of ice across a frozen pond. He applies a force of 25.0 N, pushing the block away from him. What is the acceleration of the block?
Answer: To find the acceleration, rearrange the equation:
The block is accelerating 0.50 m/s2, directed away from the man.
Related Links: Force, Mass, Acceleration Quiz Force and Acceleration Quiz Mass Examples Gravity Examples | 271 | 1,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2020-40 | longest | en | 0.944283 |
https://math.answers.com/math-and-arithmetic/Is_3215_divisible_by_3 | 1,723,043,352,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694594.35/warc/CC-MAIN-20240807143134-20240807173134-00083.warc.gz | 305,259,146 | 47,392 | 0
Is 3215 divisible by 3
Updated: 9/22/2023
Wiki User
11y ago
Not evenly. A number is divisible by 3 if the sum of its digits is divisible by 3.
Wiki User
11y ago
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Q: Is 3215 divisible by 3
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Still have questions?
Related questions
1, 5, 643, 3215.
How many times does 5 go into 3215?
3215 ÷ 5 = 643
Is 5193 a prime number?
No, it is divisible by 3.No, it is divisible by 3.No, it is divisible by 3.No, it is divisible by 3.
What is 339 divisible by?
It is divisible by 3, for example.It is divisible by 3, for example.It is divisible by 3, for example.It is divisible by 3, for example.
How many kilograms are in 3215 grams?
There are 3.215 kilograms in 3215 grams, since there are 1000 grams per kilogram.
3215
A number is divisible by 3 if the blank is divisible by 3?
A number is divisible by 3 if the sum of its digits is divisible by 3.
If x is an integer divisible by 3 then is x2 divisible by 3?
If x is an integer divisible by 3, is x squared divisible by 3?
Is 313 divisble by 3?
NO. 313 is not divisible by 3.A number is divisible by 3 if the sum of the digits is divisible by 3.313 = 3 + 1 + 3 = 7Note: 7 not divisible by 3 thus, 313 is not divisible by 3.
Is 3 divisible by 126?
3 is not divisible by 126. 126 is divisible by 3.
Is 6743 divisible by 3?
1x6,743 not divisible by 2 no its not divisible by 3????
Are all numbers divisible by 3 also by 9 give example?
All numbers divisible by 3 are NOT divisible by 9. As an example, 6, which is divisible by 3, is not divisible by 9. However, all numbers divisible by 9 are also divisible by 3 because 9 is divisible by 3. | 505 | 1,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-33 | latest | en | 0.96538 |
http://mathhelpforum.com/statistics/217764-please-help-me-about-some-statistics-o-x-quiz.html | 1,481,259,960,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542680.76/warc/CC-MAIN-20161202170902-00173-ip-10-31-129-80.ec2.internal.warc.gz | 174,868,267 | 11,370 | Dear everyone
With all due respect, may I ask a help for some O/X quiz?
1)Suppose two random variables X and Y are negatively correlated. If X is above its mean, then Y is always below its mean.
(I know that when two are negatively correlated, if X gose up, Y goes down. However, I am not sure about the relation with mean)
2)Suppose X is a binomial random variable. Then 1000X is approximately a normal random variable.
(I know when n is bigger, it would be a normal distribution. However, I am not sure about 1000X..)
3)The polling agency should take a larger sample if the population is larger.
(I think that sample would be not affect whether the population is larger.. but not sure)
4)Linear regression framework can be applied to capture the decreasing marginal returns between two variables.
(I think it is O)
1) If X is at its mean then Y is at its mean. Like you say when X goes up from its mean then Y will go down from its mean.
2) 1000X is not normally distributed. If X's population was {1,0,0,1,0,1} then 1000X's population would be {1000,0,0,1000,0,1000} it still only has two values.
3) We correct for things such as standard error for a finite population with a "finite correction factor" $\sqrt{\frac{N-n}{N-1}}$ (Where N is population size and n is sample size) if N increases then you do need to increase n to maintain the same accuracy.
4) Decreasing marginal returns tends asymptotically to zero. You cannot use a linear relationship to model that.
As for the question 1, the questions says 'always', so can that be false, because a value of Y can sometimes be above its mean even if a value of X is also above its mean?
Have a great day.
y will always be below its mean when increasing x. If the slope of the regression line has a slope m which is negative then when x increases by an amount d, y will change by an amount md. Since m is negative md is negative and y decreases. | 469 | 1,912 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2016-50 | longest | en | 0.950078 |
http://www.topperlearning.com/forums/ask-experts-19/show-that-square-of-any-odd-integer-is-of-form-4m-1-for-some-mathematics-real-numbers-56427/reply | 1,488,112,055,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172000.93/warc/CC-MAIN-20170219104612-00445-ip-10-171-10-108.ec2.internal.warc.gz | 641,259,150 | 38,077 |
Question
Wed June 13, 2012 By:
# Show that square of any odd integer is of form 4m+1 for some integer m.
Thu June 14, 2012
Let a be any positive integer and b = 4
Then, by Euclid''s algorithm a = 4q + r for some integer q 0 and 0 r < 4
Since, r = 0, 1, 2, 3
Therefore, a = 4q or 4q + 1 or 4q + 2 or 4q + 3
Since, a is an odd integer, o a = 4q + 1 or 4q + 3
Case I: When a = 4q + 1
Squaring both sides, we have,
a2 = (4q + 1)2
a2 = 16q2 + 1 + 8q
= 4(4q2 + 2q) + 1
= 4m + 1 where m = 4q2 + 2q
Case II: When a = 4q + 3
Squaring both sides, we have,
a2 = (4q +3)2
= 16q2 + 9 + 24q
= 16 q2 + 24q + 8 + 1
= 4(4q2 + 6q + 2) +1
= 4m +1 where m = 4q2 +7q + 2
Hence, a is of the form 4m + 1 for some integer m.
Related Questions
Wed October 26, 2016 | 374 | 752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-09 | longest | en | 0.821802 |
https://dataedy.com/2018/07/31/recall-the-formula-for-calculating-the-magnitude-of-an-earthquake-m-2-__-3-log-%EE%80%A2-s-__-s0-%EE%80%AA-one-earthquake-has-magnitude-3-9-on-the-mms-scale-if-a-second-earthquake-has-750-times/ | 1,534,673,275,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215075.58/warc/CC-MAIN-20180819090604-20180819110604-00272.warc.gz | 623,625,451 | 16,539 | ## Recall the formula for calculating the magnitude of an earthquake, M = 2 __ 3 log S __ S0 . One earthquake has magnitude 3.9 on the MMS scale. If a second earthquake has 750 times as much energy as the first, find the magnitude of the second quake. Round to the nearest hundredth. For the following exercises, use this scenario: The equation N(t) = 500 _ 1 + 49e−0.7t models the number of people in a town who have heard a rumor after t days.
sectioN exercises 551
For the following exercises, use this scenario: A biologist recorded a count of 360 bacteria present in a culture after 5 minutes and 1,000 bacteria present after 20 minutes.
39. To the nearest whole number, what was the initial population in the culture?
40. Rounding to six significant digits, write an exponential equation representing this situation. To the nearest minute, how long did it take the population to double?
For the following exercises, use this scenario: A pot of boiling soup with an internal temperature of 100° Fahrenheit was taken off the stove to cool in a 69° F room. After fifteen minutes, the internal temperature of the soup was 95° F.
41. Use Newton’s Law of Cooling to write a formula that models this situation.
42. To the nearest minute, how long will it take the soup to cool to 80° F?
43. To the nearest degree, what will the temperature be after 2 and a half hours?
For the following exercises, use this scenario: A turkey is taken out of the oven with an internal temperature of 165° Fahrenheit and is allowed to cool in a 75° F room. After half an hour, the internal temperature of the turkey is 145° F.
44. Write a formula that models this situation. 45. To the nearest degree, what will the temperature be after 50 minutes?
46. To the nearest minute, how long will it take the turkey to cool to 110° F?
For the following exercises, find the value of the number shown on each logarithmic scale. Round all answers to the nearest thousandth.
47.
0–1–2–3–4–5 1 2 3 4
log (x)
5
48.
0–1–2–3–4–5 1 2 3 4
log (x)
5
49. Plot each set of approximate values of intensity of sounds on a logarithmic scale: Whisper: 10−10 W ___ m2 ,
Vacuum: 10−4 W ___ m2 , Jet: 10 2 W ___ m2
50. Recall the formula for calculating the magnitude of an earthquake, M = 2 __ 3 log S __ S0 . One earthquake has magnitude 3.9 on the MMS scale. If a second earthquake has 750 times as much energy as the first, find the magnitude of the second quake. Round to the nearest hundredth.
For the following exercises, use this scenario: The equation N(t) = 500 _ 1 + 49e−0.7t
models the number of people in a town who have heard a rumor after t days.
51. How many people started the rumor? 52. To the nearest whole number, how many people will have heard the rumor after 3 days?
53. As t increases without bound, what value does N(t) approach? Interpret your answer.
For the following exercise, choose the correct answer choice.
54. A doctor and injects a patient with 13 milligrams of radioactive dye that decays exponentially. After 12 minutes, there are 4.75 milligrams of dye remaining in the patient’s system. Which is an appropriate model for this situation?
a. f (t) = 13(0.0805)t b. f (t) = 13e0.9195t c. f (t) = 13e(−0.0839t) d. f (t) = 4.75 __________ 1 + 13e−0.83925t
dmabine
Highlight
dmabine
Highlight
SECTION 6.8 sectioN exercises 561
6.8 SeCTIOn exeRCISeS
veRbAl
1. What situations are best modeled by a logistic equation? Give an example, and state a case for why the example is a good fit.
2. What is a carrying capacity? What kind of model has a carrying capacity built into its formula? Why does this make sense?
3. What is regression analysis? Describe the process of performing regression analysis on a graphing utility.
4. What might a scatterplot of data points look like if it were best described by a logarithmic model?
5. What does the y-intercept on the graph of a logistic equation correspond to for a population modeled by that equation?
gRAPhICAl
For the following exercises, match the given function of best fit with the appropriate scatterplot in Figure 7 through Figure 11. Answer using the letter beneath the matching graph.
1
2 4 6 8
10 12 14 16
y
x 2 3 4 5 6 7 8 9 10
(a)
Figure 7
1
2 4 6 8
10 12 14 16
y
x 2 3 4 5 6 7 8 9 10
(b)
Figure 8
2 4 6 8
10 12 14 16
y
1 2 3 4 5 6 7 8 9 10 x
(c)
### Depreciation
P3-36A Journalizing and posting adjustments to the four-column accounts and preparing an adjusted trial balance The unadjusted trial balance of Newport Inn Company at December 31, 2016, and the data….
### Prepaid Insurance
P3-36A Journalizing and posting adjustments to the four-column accounts and preparing an adjusted trial balance The unadjusted trial balance of Newport Inn Company at December 31, 2016, and the data….
### formula for the sequence
(1) What is the 50 th term of the arithmetic sequence with initial term 4 and common difference 3? 7 (2) Evaluate 4 X k=−3 (2k + 5). (Hint: Since…. | 1,374 | 4,980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-34 | longest | en | 0.926202 |
https://docs.pybinding.site/en/v0.9.4/tutorial/finite.html | 1,670,585,762,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711396.19/warc/CC-MAIN-20221209112528-20221209142528-00042.warc.gz | 243,681,835 | 8,232 | # 4. Finite size¶
This section introduces the concept of shapes with classes Polygon and FreeformShape which are used to model systems of finite size. The sparse eigensolver arpack() is also introduced as a good tool for exactly solving larger Hamiltonian matrices.
Download this page as a Jupyter notebook
## 4.1. Primitive¶
The simplest finite-sized system is just the unit cell of the crystal lattice.
from pybinding.repository import graphene
model = pb.Model(graphene.monolayer())
model.plot()
The unit cell can also be replicated a number of times to create a bigger system.
model = pb.Model(
graphene.monolayer(),
pb.primitive(a1=5, a2=3)
)
model.plot()
model.lattice.plot_vectors(position=[0.6, -0.25])
The primitive() parameter tells the model to replicate the unit cell 5 times in the $$a_1$$ vector direction and 3 times in the $$a_2$$ direction. However, to model realistic systems we need proper shapes.
## 4.2. Polygon¶
The easiest way to create a 2D shape is with the Polygon class. For example, a simple rectangle:
def rectangle(width, height):
x0 = width / 2
y0 = height / 2
return pb.Polygon([[x0, y0], [x0, -y0], [-x0, -y0], [-x0, y0]])
shape = rectangle(1.6, 1.2)
shape.plot()
A Polygon is initialized with a list of vertices which should be given in clockwise or counterclockwise order. When added to a Model the lattice will expand to fill the shape.
model = pb.Model(
graphene.monolayer(),
rectangle(width=1.6, height=1.2)
)
model.plot()
To help visualize the shape and the expanded lattice, the polygon outline can be plotted on top of the system by calling both plot methods one after another.
def trapezoid(a, b, h):
return pb.Polygon([[-a/2, 0], [-b/2, h], [b/2, h], [a/2, 0]])
model = pb.Model(
graphene.monolayer(),
trapezoid(a=3.2, b=1.4, h=1.5)
)
model.plot()
model.shape.plot()
In general, a shape does not depend on a specific material, so it can be easily reused. Here, we shall switch to a graphene.bilayer() lattice, but we’ll keep the same trapezoid shape as defined earlier:
model = pb.Model(
graphene.bilayer(),
trapezoid(a=3.2, b=1.4, h=1.5)
)
model.plot()
## 4.3. Freeform shape¶
Unlike a Polygon which is defined by a list of vertices, a FreeformShape is defined by a contains function which determines if a lattice site is inside the desired shape.
def circle(radius):
def contains(x, y, z):
return np.sqrt(x**2 + y**2) < radius
model = pb.Model(
graphene.monolayer(),
)
model.plot()
The width parameter of FreeformShape specifies the bounding box width. Only sites inside the bounding box will be considered for the shape. It’s like carving a sculpture from a block of stone. The bounding box can be thought of as the stone block, while the contains function is the carving tool that can give the fine detail of the shape.
As with Polygon, we can visualize the shape with the FreeformShape.plot() method.
def ring(inner_radius, outer_radius):
def contains(x, y, z):
r = np.sqrt(x**2 + y**2)
shape.plot()
The shaded area indicates the shape as determined by the contains function. Creating a model will cause the lattice to fill in the shape.
model = pb.Model(
graphene.monolayer(),
)
model.plot()
model.shape.plot()
Note that the ring example uses np.logical_and instead of the plain and keyword. This is because the x, y, z positions are not given as scalar numbers but as numpy arrays. Array comparisons return boolean arrays:
>>> x = np.array([7, 2, 3, 5, 1])
>>> x < 5
[False, True, True, False, True]
>>> 2 < x and x < 5
ValueError: ...
>>> np.logical_and(2 < x, x < 5)
[False, False, True, False, False]
The and keyword can only operate on scalar values, but np.logical_and can consider arrays. Likewise, math.sqrt does not work with arrays, but np.sqrt does.
## 4.4. Composite shape¶
Complicated system geometry can also be produced by composing multiple simple shapes. The following example gives a quick taste of how it works. For a full overview of this functionality, see the Composite shapes section.
# Simple shapes
rectangle = pb.rectangle(x=6, y=1)
# Compose them naturally
shape = rectangle + hexagon - circle
model = pb.Model(graphene.monolayer(), shape)
model.shape.plot()
model.plot()
## 4.5. Spatial LDOS¶
Now that we have a ring structure, we can exactly diagonalize its model.hamiltonian using a Solver. We previously used the lapack() solver to find all the eigenvalues and eigenvectors, but this is not efficient for larger systems. The sparse arpack() solver can calculate a targeted subset of the eigenvalues, which is usually desired and much faster. In this case, we are interested only in the 20 lowest energy states.
model = pb.Model(
graphene.monolayer(),
)
solver = pb.solver.arpack(model, k=20) # only the 20 lowest eigenstates
ldos = solver.calc_spatial_ldos(energy=0, broadening=0.05) # eV
pb.pltutils.colorbar(label="LDOS")
The convenient Solver.calc_spatial_ldos() method calculates the local density of states (LDOS) at every site for the given energy with a Gaussian broadening. The returned object is a StructureMap which holds the LDOS data. The StructureMap.plot() method will produce a figure similar to Model.plot(), but with a colormap indicating the LDOS value at each lattice site. In addition, the site_radius argument specifies a range of sizes which will cause the low intensity sites to appear as small circles while high intensity ones become large. The states with a high LDOS are clearly visible on the outer and inner edges of the graphene ring structure.
For more finite-sized systems check out the examples section.
## 4.7. Example¶
Donwload source code
"""Model a graphene ring structure and calculate the local density of states"""
import pybinding as pb
import numpy as np
import matplotlib.pyplot as plt
from pybinding.repository import graphene
pb.pltutils.use_style()
"""A simple ring shape"""
def contains(x, y, z):
r = np.sqrt(x**2 + y**2)
model = pb.Model(
graphene.monolayer(),
# only solve for the 20 lowest energy eigenvalues | 1,510 | 6,003 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-49 | latest | en | 0.790314 |
https://www.britannica.com/topic/interpolation | 1,505,894,730,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686705.10/warc/CC-MAIN-20170920071017-20170920091017-00336.warc.gz | 749,433,193 | 37,854 | # Interpolation
mathematics
Interpolation, in mathematics, the determination or estimation of the value of f(x), or a function of x, from certain known values of the function. If x0 < … < xn and y0 = f(x0),…, yn = f(xn) are known, and if x0 < x < xn, then the estimated value of f(x) is said to be an interpolation. If x < x0 or x > xn, the estimated value of f(x) is said to be an extrapolation.
If x0, …, xn are given, along with corresponding values y0, …, yn (see the figure), interpolation may be regarded as the determination of a function y = f(x) whose graph passes through the n + 1 points, (xi, yi) for i = 0, 1, …, n. There are infinitely many such functions, but the simplest is a polynomial interpolation function y = p(x) = a0 + a1x + … + anxn with constant ai’s such that p(xi) = yi for i = 0, …, n. There is exactly one such interpolating polynomial of degree n or less. If the xi’s are equally spaced, say by some factor h, then the following formula of Isaac Newton produces a polynomial function that fits the data: f(x) = a0 + a1(x − x0)/h + a2(x − x0)(x − x1)/2!h2 + … + an(x − x0)⋯(x − xn − 1)/n!hn
Polynomial approximation is useful even if the actual function f(x) is not a polynomial, for the polynomial p(x) often gives good estimates for other values of f(x).
December 25, 1642 [January 4, 1643, New Style] Woolsthorpe, Lincolnshire, England March 20 [March 31], 1727 London English physicist and mathematician, who was the culminating figure of the scientific revolution of the 17th century. In optics, his discovery of the composition of white light...
One common method of approximation is known as interpolation. Consider a set of points (xi,yi) where i = 0, 1, …, n, and then find a polynomial that satisfies p(xi) = yi for all...
A branch of mathematics that deals with continuous change and with certain general types of processes that have emerged from the study of continuous change, such as limits, differentiation,...
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Your contribution may be further edited by our staff, and its publication is subject to our final approval. Unfortunately, our editorial approach may not be able to accommodate all contributions. | 803 | 2,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-39 | longest | en | 0.844016 |
picmeinternational.co.za | 1,563,812,931,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528141.87/warc/CC-MAIN-20190722154408-20190722180408-00529.warc.gz | 507,126,088 | 17,119 | # Short Article Reveals the Undeniable Facts About What Is a Product in Math and How It Can Affect You
Posted on Posted in Uncategorized
You may be surprised at the results! In the event the identical number is employed as a factor more than once, you may use the property to help you condense the problem. As a means to create a valuable survey you will have some direct and to-the-point questions so that you will secure some wonderful outcomes.
The final result is usually answers which don’t make sense grammatically. Each time you face an intimidating equation, remember what you’re hoping to attain. Your second point is also correct, if you’re not speaking about tax prices or total taxes, then there isn’t any question that wealthy individuals have much more disposable income than you do.
You will receive the formula above! To understand the reason, you ought to know about the notion of convexity in optimization. Typically, there was only one right method to fix an issue.
Lesson Summary The merchandise is the remedy to a multiplication issue. Estimate first, and work out the legitimate item.
## The What Is a Product in Math Game
Calculations can likewise be performed in output statements. Ensure you get all important material. They will be drawn from a broad selection of disciplines.
As a result, multiplication and its products have a distinctive set of properties you need to know to get the appropriate answers. After you are comfortable with the comprehension of the order of operations, consider using a spreadsheet to figure out the order of operations. You may choose the selection of rows and columns utilized for the arrays, in addition to the description given to draw the array.
The final result is going to be the factors of the starting number. Short division is a simple manner of doing complicated divisions by hand. Start early Don’t await the suitable time to start your preparations.
Operators which are shown on the identical row have the exact same precedence. The principal distinction is that you can’t divide by 0 and locate a real number. Remember they are numbers that can be multiplied together to make the number being factored.
An alternative is showing the way the concept was created via the history of math. While the intent of this method is logical and can be appropriate for students that are average to above average achievers, the spiraling curriculum may be considerable impediment for students who have math learning issues. For these students, it might be particularly important to use many representations for two purposes.
Your 7th grader’s question is a significant and fundamental one (that I am both surprised and sorry he hasn’t been in a position to locate an answer for yet). Carnegie Learning curriculum is a good example of emphasis on multiple representations and usage of computer tools. Fast forward to my first two or three years of teaching.
Instead, students ought to consider the way to use their strengths to assist them in all subjects. Some students will truly struggle to create sense of it and we will need to chat about what sort of strategies they may utilize to help make sense of the circumstance. There’s a student here, she’s a very first term paper grader, and she is able to do fourth grade math.
## The Lost Secret of What Is a Product in Math
The testing results help the business in locating the previous notion to be developed into a merchandise. The ability to complete excellent research is demonstrated by the decision of a yearlong research undertaking. The item is also referred to as a multiple of every one of the 2 numbers that gives that product.
Tutorial centers are likewise a very very good resource. Still, a comprehensive, standalone help file that doesn’t demand a permanent online connection is also offered. New tools to create IoT projects easier are coming out all of the time.
A base is 1 side of a polygon, usually utilized as a reference side for different measurements. If it isn’t doing well in the market, price might be the first thing you get started looking into. The dot product must be bilinear.
During high school, you’re supposed to get started focusing on what you would like to do for the remainder of your life. For some reason, plenty of students appear to lump the 2 ideas into a single muddy mess. Some students will begin making a list of the times for each sibling to go back to the bottom and locate the typical time in every list.
## New Step by Step Roadmap for What Is a Product in Math
It’s sometimes very beneficial if you think about division as multiplication. If two or more BOOLEAN expressions are along with the OR, the end result is true if any are TRUE. When you’ve got an equation, the full equation ought to go in the equation object, not only the part you can’t do with the term processor.
If you’ve ever found a frequent denominator for a few fractions, you’ve located a frequent multiple. There are two main formulas that may serve the objective of calculating trapezoid area. The very first example indicates a sum.
A circle is really a specific case of an ellipse. Repeat if you may divide a number of the numbers by another factor. The numbers for each factor could be individually varied to create unique sets of Multiplication difficulties. | 1,046 | 5,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2019-30 | longest | en | 0.933059 |
https://www.tutorialspoint.com/make-up-as-many-expressions-with-numbers-no-variables-as-you-can-from-three-numbers-5-7-and-8-every-number-should-be-used-not-more-than-once-use | 1,670,056,424,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710926.23/warc/CC-MAIN-20221203075717-20221203105717-00569.warc.gz | 1,095,913,911 | 9,719 | # Make up as many expressions with numbers (no variables) as you can from three numbers 5,7 and 8 . Every number should be used not more than once. Use only addition, subtraction and multiplication.(Hint : Three possible expressions are $5+(8-7), 5-(8-7),(5 \times 8)+7$; make the other expressions.)
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To do:
We have to make up as many expressions with numbers (no variables) as we can from three numbers 5,7 and 8.
Solution:
Some of the expressions formed by 5, 7 and 8 are as follows:
$5+(8-7)$
$5-(8-7)$
$5-(8+7)$
$(5 \times 8)+7$
$(5 \times8) - 7)$
$5 \times (8 + 7)$
$(7 + 5) \times 8$
$(8 – 5) \times 7$
$(8 + 5) \times 7$
$(7 – 5) \times 8$
Updated on 10-Oct-2022 13:36:46 | 294 | 894 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2022-49 | latest | en | 0.738239 |
https://discuss.codechef.com/t/faster-than-binary-search/1343 | 1,600,578,065,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400193391.9/warc/CC-MAIN-20200920031425-20200920061425-00266.warc.gz | 361,857,807 | 4,889 | # faster than binary search
i have a list of prime no. with its rank as its value in a map .
like map<int ,int > mymap ;
mymap[2]=0
mymap[3]=1
mymap[5]=2
and i am trying to calculate the position of nth prime no in the list by this way as
position=mymap[required no]
but it seems it is not fast enough
plz help me in a right directiion…???
Hello @vivek07672,
You can start by generating all the primes using a common algorithm, like the Sieve of Erathostenes… This algorithm allows you to generate all the prime numbers up to N, in a reasonable fast time, which is usually enough for the requested limits when working with prime numbers.
Once you have all the prime numbers in a container that supports iteration (like a vector or an array), using a well implemented binary search is more than enough to find what you want.
Also, I understand that this question can be related to the problem CHEFHACK and that the contest is still running but since you haven’t asked for any test cases specifications and as it seems you have thought about the problem a bit, I decided to answer and hopefully help you, but please keep in mind that live contest questions must be asked in a way that doesn’t give too much away regarding a particular problem, just like you did.
Best regards,
Bruno
Hi! To answer your question, you could just tweak the sieve of Eratosthenes to get your answer.
Wikipedia lists the algorithm as:
``````Input: an integer n > 1
Let A be an array of Boolean values, indexed by integers 2 to n,
initially all set to true.
for i = 2, 3, 4, ..., √n :
if A[i] is true:
for j = i^2, i^2+i, i^2+2i, ..., n:
A[j] := false
Now all i such that A[i] is true are prime.
``````
You can modify it in this way:
``````Input: an integer n > 1
Let A be an array of Integer values, indexed by integers 2 to n,
initially all set to 0.
Let count be an integer set to 1 initially
for i = 2, 3, 4, ..., √n :
if A[i] = 0:
A[i] := count
count := count + 1
for j = i^2, i^2+i, i^2+2i, ..., n:
A[j] := -1
for i < n :
if A[i] = 0
A[i] := count
count := count + 1
Now all i such that A[i] > 0 are prime and A[i] denotes the index of this prime in the list of primes till n.
``````
Hope this helps
Vivek, Vivek.
You can’t ask for help for problems during live contest.
It is obvious that you have issue with the problem CHEFHACK.
And you guys why are you helping him?
OK. I have deleted this question until the end of the contest.
1 Like
Correction. can -> can’t
O-o-ops… Fixed. It is so easy to miss this crappy `'t` | 709 | 2,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-40 | latest | en | 0.921835 |
http://mathbitsnotebook.com/JuniorMath/Radicals/RADarithmetic.html | 1,516,166,320,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886815.20/warc/CC-MAIN-20180117043259-20180117063259-00372.warc.gz | 229,766,373 | 4,470 | This lesson is going to take a "quick look" at adding, subtracting, multiplying and dividing square roots. For more information regarding these operations, refer to the Radical Section under Algebra 1.
You have 4 x's and 6 more x's, so you have 10 x's all totaled.
Think: there are 4 square roots of 3 plus 6 square roots of 3.
That makes 10 square roots of 3 in total.
It may be necessary to simplify first! Different looking radicals may actually be the same when simplified. At first glance, it appears that combining these terms under addition is not possible since the radicals are not the same. But if we look further, we can simplify the second term so it will be a "like" radical: ANSWER:
Some radicals cannot be added! The radicals are different and each is already in simplest form. There is simply no way to combine these values. The answer is the same as the original problem. ANSWER:
Remember, there may be an unwritten ONE in front of a radical! There is an implied "1" in front of . All radicals are already in simplest form. Combine the "like" radicals. ANSWER:
REMEMBER: Always simplify first! When the radicals in an addition or subtraction problem are different, be sure to check to see if the radicals can be simplified. It may be the case that when the radicals are simplified, they will become "like" radicals, making it possible for them to be added or subtracted.
Multiply:
Multiply out front and multiply under the radicals.
It may be necessary to simplify the answer! Different looking radicals may actually be the same when simplified. Multiply out front and multiply under the radicals: Then simplify the result. ANSWER:
Divide:
Dividing Radicals: When dividing radicals (with the same index), divide under the radical, and then divide in front of the radical (divide any values multiplied times the radicals).
Divide out front and divide under the radicals.
Radicals in the denominator of a fraction: This fraction will be in simplified form when the radical is removed from the denominator. You need to create a perfect square under the square root radical in the denominator by multiplying the top and bottom of the fraction by the same value (this is actually multiplying by "1"). The easiest approach is to multiply by the square root radical you need to convert (in this case multiply by ). ANSWER: This is now in simplified form. This process of simplifying is called "rationalizing the denominator" as it creates a fraction with an integer denominator. | 526 | 2,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2018-05 | latest | en | 0.926404 |
https://examtube.in/probability-ncert-9-mathematics-textbook-mcq-multiple-choice-questions/ | 1,627,863,825,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154277.15/warc/CC-MAIN-20210801221329-20210802011329-00605.warc.gz | 246,306,268 | 25,875 | # PROBABILITY NCERT 9 MATHEMATICS TEXTBOOK MCQ (Multiple Choice Questions)
Q1. Sum of the probability of happening and not happening of an event is:
1. 1
2. 2
3. 0
4. None of these
Q2. The maximum probability of an event of a trial is:
1. 0
2. -1
3. 1
4. Lies between 0 and 1.
Q3. If the probability of winning a game is 0.9 then, the probability of losing the game is
1. 0.1
2. 0.9
3. 0.2
4. 1
Q4. In an experiment, 100 drawing pins were dropped on a floor. 73 landed point up and the rest landed point down. A pin is selected at random and dropped. What is the probability that the pin will land point down?
1. 0.73
2. 0.27
3. 0.37
4. 0.72
Q5. The percentages of marks obtained by a student in six unit tests are given below: Unit test I II III IV V VI Percentage of marks 53 72 28 46 67 61 If a test is selected at random then, the probability that the student gets more than 60% marks in the test is
1. 0.5
2. 0.3
3. 0.2
4. 0.6
Q6. An event for an experiment is defined as
1. Collection of happenings and non-happenings
2. Collection of probabilities
3. Collection of odd event and even event of the experiment
4. Collection of some outcomes of the experiment
Q7. A, B, C and D are the only possible outcomes of a trial. So if P(A) + P(D) = 0.75, then P(C) + P(B) =
1. 0.75
2. 0.50
3. 0.25
4. 0.10
Q8. In a sample survey of 640 people, it was found that 400 people have a secondary school certificate. If a person is selected at random, the probability that the person does not have such certificate is
1. 0.375
2. 0.625
3. 0.725
4. 0.875
Q9. In a series of 6 cricket matches, the number of runs scored by the captain of a team are 54, 32, 48, 55, 29, 35. So in the next match, the probability that he will cross the half century is
1. 0.33
2. 0.24
3. 0.35
4. 0.48
Q10. In a survey of 364 children aged 2-4 years, it was found that 91 liked to eat potato chips. If a child is selected at random, the probability that he/she does not like to eat potato chips is:
1. 0.25
2. 0.50
3. 0.75
4. 0.80
#### NCERT Solutions for Class 1 Maths Chapter 4 Subtraction
Lorem ipsum dolor sit amet, consectetur adipiscing elit. Phasellus cursus rutrum est nec suscipit. Ut et ultrices nisi. Vivamus id nisl ligula. Nulla sed iaculis ipsum.
Company Name | 752 | 2,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-31 | latest | en | 0.902904 |
https://www.turito.com/learn/math/intercept-form-grade-10 | 1,716,244,840,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00789.warc.gz | 924,292,588 | 31,061 | #### Need Help?
Get in touch with us
# Intercept Form
Sep 15, 2022
## Key Concepts
• Graph a linear equation
• Write an equation from a graph
• Understand slope intercept form
• Interpret slope and y- Intercept
## Slope – Intercept Form
### Slope
The slope of a line is the ratio of the amount that y increases as x increases some amount. Slope tells you how steep a line is or how much y increases as x increases.
The slope is constant (the same) anywhere on the line.
#### Slope formula
m = rise / run = y2−y1 / x2−x1
#### Slope-Intercept Form
The meaning of slope-intercept form is the equation of a straight line in the form y = mx + b, where m is the slope of the line and b is its y-intercept.
y = mx + b
### Graph a linear equation
#### Linear equation
A linear equation is an algebraic equation of the form y=mx+b.
Involving only a constant and a first-order (linear) term, where m is the slope and b is the y-intercept. Occasionally, the above is called a “linear equation of two variables,” where y and x are the variables.
Example 1:
What is the graph of y = 3x+1?
Sol:
Step 1:
Identify the y-intercept in the equation.
The y-intercept is 1, So plot the points (0, 1).
Step 2:
Use the slope to plot a second point.
m = 3 = vertical change / horizontal change
Start at (0, 1), move 3 units up and 1 unit to the right to locate a second point.
Plot the points (1, 4).
Step 3:
Draw a line through the points.
### Write an equation from a graph
Example:
What is the equation of the line in slope-intercept form?
Sol:
Step 1:
Find the slope between two points on the line.
The line passes through (-3, 3) and (3, -1).
Slope =
y2−y1 / x2−x1
= (−1)−3 / 3− (−3)
=−4 / 6
= –2 / 3
Step 2:
Find the y–intercept.
The line intersects the y-axis at (-3, 3) so the y-intercept is 1.
Step 3:
Write the equation in the form y=mx+b.
Substitute −2 / 3 for m and 1 for b.
The equation of the line in the slope-intercept form is,
y = 2 / 3 x + 1
### Understand Slope–Intercept form
Example:
Write the equation of the line that passes through the given points (2, 2) and (3, 4).
Sol:
Step 1:
Find the slope of the line.
m = y2−y1 / x2−x1
m = 4 − 2 / 3 − 2
m = 2
Step 2:
Use the slope and one point to find the y-intercept.
y = mx +b
4 = 2(3) +b Substitute 2 for m. (3, 4) for (x, y)
4 = 6 +b Simplify
-2 = b
Step 3:
Use the slope and y-Intercept to write the equation.
y = mx +b
y = 2x -2 Substitute 2 for m and -2 for b
The equation in slope–intercept form of the line that passes through (2, 2) and (3, 4) is y = 2x -2.
### Interpret Slope and y-Intercept
Example:
The rent charged for space in an office building as a linear relationship related to the size of the space rented. Write an equation in slope-intercept form for rent at west main street office rentals
Sol:
Step 1:
Create a table or identify ordered pairs from the problem.
(600, 750) and (900, 1150)
Step 2:
Find the slope using the slope formula,
m = y2−y1 / x2−x1
m= 1150−750 / 900−600
m= 400 / 300
m= 4 / 3
Step 3:
Write the form y= mx+b.
Step 4:
Replace m with the value you found.
Y= 4 / 3 x+b
Step 5:
Plug-in one of the points you already know and solve for b using the inverse operation.
750 = 4 / 3 (600)+b
750 = 800 +b
-50=b
Step 6:
Replace m and b with the values.
y= 4 / 3 x+(-50)
Y= 4 / 3 x -50.
## Exercise
1. Sketch the graph of the equation.
y = 2x -5
1. Identify the slope and y-intercept of the line for the equation.
Y = – 5x – 3/4
1. Write the equation of the line that passes through the given points (0, 1) and (2, 2).
2. Write the equation of each line in slope–intercept form.
1. Jordan will hike the trail shown at a rate of 4 mi/h. Write a linear equation to represent the distance Jordan still has to walk after x hours. What does the y-intercept of the equation represent?
1. Write the equation of the line that passes through the given points (5, 4) and (-1, 6).
2. Write the equation of the line in slope–intercept form.
1. Graph the following line:
1. Y = x + 7.
2. Find the equation of the line graphed below.
1. Shriya read a book cover to cover in a single session, at a constant rate. After reading for 1.5 hours, she had 402 out of the total 480 pages left to read. Let y represent the number of pages left to read after x hours. Complete the equation for the relationship between the number of pages left and the number of hours. y=___________.
### What have we learned
• Understand Slope –Intercept Form
• Graph a linear equation
• Understand how to write an equation from a graph
• Interpret slope and y- Intercept
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […] | 1,650 | 5,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2024-22 | latest | en | 0.904209 |
https://wildpartyofficial.com/what-is-the-concept-of-canonical-cover/ | 1,719,283,351,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865545.19/warc/CC-MAIN-20240625005529-20240625035529-00459.warc.gz | 540,633,798 | 14,979 | # What is the concept of canonical cover?
## What is the concept of canonical cover?
A canonical cover is a simplified and reduced version of the given set of functional dependencies. Since it is a reduced version, it is also called as Irreducible set.
## Is minimal cover same as canonical cover?
A canonical cover is “allowed” to have more than one attribute on the right hand side. A minimal cover cannot. As an example, the canonical cover may be “A -> BC” where the minimal cover would be “A -> B, A -> C”. That is the only difference.
How is canonical cover calculated?
Steps to find canonical cover:
1. There are two functional dependencies with the same set of attributes on the left:
2. There is an extraneous attribute in AB C because even after removing AB C from the set F, we get the same closures.
3. C is an extraneous attribute in A BC, also A B is logically implied by A B and B.
### What is the concept of canonical cover how do you compute it give example?
FD = { B → A, AD → C, C → BD } is Canonical Cover of FD = { B → A, AD → BC, C → ABD }. Example 2: Given a relational Schema R( W, X, Y, Z) and set of Function Dependency FD = { W → X, Y → X, Z → WXY, WY → Z }.
### Does a set of FD’s have a unique canonical cover?
No, it cannot be a canonical cover for your set of functional dependencies (FD). A canonical cover of F is a “minimal” set of functional dependencies equivalent to F, having no redundant dependencies or redundant parts of dependencies.
Why is canonical cover not unique?
No, it cannot be a canonical cover for your set of functional dependencies (FD). A canonical cover of F is a “minimal” set of functional dependencies equivalent to F, having no redundant dependencies or redundant parts of dependencies. The canonical cover you suggest is not equivalent to F .
#### What is difference between canonical form and standard form?
The main difference between canonical and standard form is that canonical form is a way of representing Boolean outputs of digital circuits using Boolean Algebra while standard form is a simplified version of canonical form that represents Boolean outputs of digital circuits using Boolean Algebra.
#### Is canonical and standard form same?
What is difference between standard form and canonical form?
In standard form Boolean function will contain all the variables in either true form or complemented form while in canonical number of variables depends on the output of SOP or POS. maxterm for each combination of the variables that produces a 0 in the function and then taking the AND of all those terms.
## When do you need to use the canonical cover?
If there is a violation of dependencies in the new database state, the system must roll back. Working with a huge set of functional dependencies can cause unnecessary added computational time. This is where the canonical cover comes into play.
## Which is the canonical cover for a set of dependencies?
Canonical Cover A canonical coverfor Fis a set of dependencies Fc such that F logically implies all dependencies in Fc,and Fc logically implies all dependencies in F, and No functional dependency in Fccontains an extraneous attribute, and Each left side of functional dependency in Fcis unique
How to find the canonical cover in Java?
Example 2: Given a relational Schema R ( W, X, Y, Z) and set of Function Dependency FD = { W → X, Y → X, Z → WXY, WY → Z }. Find the canonical cover? Solution: Given FD = { W → X, Y → X, Z → WXY, WY → Z }, now decompose the FD using decomposition rule ( Armstrong Axiom ).
### How does a canonical cover cc work in DBMS?
A canonical cover Cc is a set of all functional dependencies Fd that satisfied all the following properties as follows. Functional dependence Fd according to the rules of logic it implies all dependencies in Canonical Cover Cc. | 829 | 3,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-26 | latest | en | 0.932056 |
https://tutorialspoint.dev/data-structure/graph-data-structure/katz-centrality-centrality-measure | 1,618,093,415,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038059348.9/warc/CC-MAIN-20210410210053-20210411000053-00043.warc.gz | 683,992,223 | 12,665 | Katz Centrality (Centrality Measure)
In graph theory, the Katz centrality of a node is a measure of centrality in a network. It was introduced by Leo Katz in 1953 and is used to measure the relative degree of influence of an actor (or node) within a social network. Unlike typical centrality measures which consider only the shortest path (the geodesic) between a pair of actors, Katz centrality measures influence by taking into account the total number of walks between a pair of actors.
It is similar to Google’s PageRank and to the eigenvector centrality.
Measuring Katz centrality
A simple social network: the nodes represent people or actors and the edges between nodes represent some relationship between actors
Katz centrality computes the relative influence of a node within a network by measuring the number of the immediate neighbors (first degree nodes) and also all other nodes in the network that connect to the node under consideration through these immediate neighbors. Connections made with distant neighbors are, however, penalized by an attenuation factor . Each path or connection between a pair of nodes is assigned a weight determined by and the distance between nodes as .
For example, in the figure on the right, assume that John’s centrality is being measured and that . The weight assigned to each link that connects John with his immediate neighbors Jane and Bob will be . Since Jose connects to John indirectly through Bob, the weight assigned to this connection (composed of two links) will be . Similarly, the weight assigned to the connection between Agneta and John through Aziz and Jane will be and the weight assigned to the connection between Agneta and John through Diego, Jose and Bob will be .
Mathematical formulation
Let A be the adjacency matrix of a network under consideration. Elements of A are variables that take a value 1 if a node i is connected to node j and 0 otherwise. The powers of A indicate the presence (or absence) of links between two nodes through intermediaries. For instance, in matrix , if element , it indicates that node 2 and node 12 are connected through some first and second degree neighbors of node 2. If denotes Katz centrality of a node i, then mathematically:
Note that the above definition uses the fact that the element at location of the adjacency matrix raised to the power (i.e. ) reflects the total number of degree connections between nodes and . The value of the attenuation factor has to be chosen such that it is smaller than the reciprocal of the absolute value of the largest eigenvalue of the adjacency matrix A. In this case the following expression can be used to calculate Katz centrality:
Here is the identity matrix, is an identity vector of size n (n is the number of nodes) consisting of ones. denotes the transposed matrix of A and ( denotes matrix inversion of the term ( ).
Following is the code for the calculation of the Katz Centrality of the graph and its various nodes.
`def` `katz_centrality(G, alpha``=``0.1``, beta``=``1.0``, ` ` ``max_iter``=``1000``, tol``=``1.0e``-``6``, ` ` ``nstart``=``None``, normalized``=``True``, ` ` ``weight ``=` `'weight'``): ` ` ``"""Compute the Katz centrality for the nodes ` ` ``of the graph G. ` ` ` ` ` ` ``Katz centrality computes the centrality for a node ` ` ``based on the centrality of its neighbors. It is a ` ` ``generalization of the eigenvector centrality. The ` ` ``Katz centrality for node `i` is ` ` ` ` ``.. math:: ` ` ` ` ``x_i = alpha sum_{j} A_{ij} x_j + eta, ` ` ` ` ``where `A` is the adjacency matrix of the graph G ` ` ``with eigenvalues `lambda`. ` ` ` ` ``The parameter `eta` controls the initial centrality and ` ` ` ` ``.. math:: ` ` ` ` ``alpha < frac{1}{lambda_{max}}. ` ` ` ` ` ` ``Katz centrality computes the relative influence of ` ` ``a node within a network by measuring the number of ` ` ``the immediate neighbors (first degree nodes) and ` ` ``also all other nodes in the network that connect ` ` ``to the node under consideration through these ` ` ``immediate neighbors. ` ` ` ` ``Extra weight can be provided to immediate neighbors ` ` ``through the parameter :math:`eta`. Connections ` ` ``made with distant neighbors are, however, penalized ` ` ``by an attenuation factor `alpha` which should be ` ` ``strictly less than the inverse largest eigenvalue ` ` ``of the adjacency matrix in order for the Katz ` ` ``centrality to be computed correctly. ` ` ` ` ` ` ``Parameters ` ` ``---------- ` ` ``G : graph ` ` ``A NetworkX graph ` ` ` ` ``alpha : float ` ` ``Attenuation factor ` ` ` ` ``beta : scalar or dictionary, optional (default=1.0) ` ` ``Weight attributed to the immediate neighborhood. ` ` ``If not a scalar, the dictionary must have an value ` ` ``for every node. ` ` ` ` ``max_iter : integer, optional (default=1000) ` ` ``Maximum number of iterations in power method. ` ` ` ` ``tol : float, optional (default=1.0e-6) ` ` ``Error tolerance used to check convergence in ` ` ``power method iteration. ` ` ` ` ``nstart : dictionary, optional ` ` ``Starting value of Katz iteration for each node. ` ` ` ` ``normalized : bool, optional (default=True) ` ` ``If True normalize the resulting values. ` ` ` ` ``weight : None or string, optional ` ` ``If None, all edge weights are considered equal. ` ` ``Otherwise holds the name of the edge attribute ` ` ``used as weight. ` ` ` ` ``Returns ` ` ``------- ` ` ``nodes : dictionary ` ` ``Dictionary of nodes with Katz centrality as ` ` ``the value. ` ` ` ` ``Raises ` ` ``------ ` ` ``NetworkXError ` ` ``If the parameter `beta` is not a scalar but ` ` ``lacks a value for at least one node ` ` ` ` ` ` ` ` ``Notes ` ` ``----- ` ` ` ` ``This algorithm it uses the power method to find ` ` ``the eigenvector corresponding to the largest ` ` ``eigenvalue of the adjacency matrix of G. ` ` ``The constant alpha should be strictly less than ` ` ``the inverse of largest eigenvalue of the adjacency ` ` ``matrix for the algorithm to converge. ` ` ``The iteration will stop after max_iter iterations ` ` ``or an error tolerance ofnumber_of_nodes(G)*tol ` ` ``has been reached. ` ` ` ` ``When `alpha = 1/lambda_{max}` and `eta=0`, ` ` ``Katz centrality is the same as eigenvector centrality. ` ` ` ` ``For directed graphs this finds "left" eigenvectors ` ` ``which corresponds to the in-edges in the graph. ` ` ``For out-edges Katz centrality first reverse the ` ` ``graph with G.reverse(). ` ` ` ` ` ` ``"""` ` ``from` `math ``import` `sqrt ` ` ` ` ``if` `len``(G) ``=``=` `0``: ` ` ``return` `{} ` ` ` ` ``nnodes ``=` `G.number_of_nodes() ` ` ` ` ``if` `nstart ``is` `None``: ` ` ` ` ``# choose starting vector with entries of 0 ` ` ``x ``=` `dict``([(n,``0``) ``for` `n ``in` `G]) ` ` ``else``: ` ` ``x ``=` `nstart ` ` ` ` ``try``: ` ` ``b ``=` `dict``.fromkeys(G,``float``(beta)) ` ` ``except` `(TypeError,ValueError,AttributeError): ` ` ``b ``=` `beta ` ` ``if` `set``(beta) !``=` `set``(G): ` ` ``raise` `nx.NetworkXError(``'beta dictionary '` ` ``'must have a value for every node'``) ` ` ` ` ``# make up to max_iter iterations ` ` ``for` `i ``in` `range``(max_iter): ` ` ``xlast ``=` `x ` ` ``x ``=` `dict``.fromkeys(xlast, ``0``) ` ` ` ` ``# do the multiplication y^T = Alpha * x^T A - Beta ` ` ``for` `n ``in` `x: ` ` ``for` `nbr ``in` `G[n]: ` ` ``x[nbr] ``+``=` `xlast[n] ``*` `G[n][nbr].get(weight, ``1``) ` ` ``for` `n ``in` `x: ` ` ``x[n] ``=` `alpha``*``x[n] ``+` `b[n] ` ` ` ` ``# check convergence ` ` ``err ``=` `sum``([``abs``(x[n]``-``xlast[n]) ``for` `n ``in` `x]) ` ` ``if` `err < nnodes``*``tol: ` ` ``if` `normalized: ` ` ` ` ``# normalize vector ` ` ``try``: ` ` ``s ``=` `1.0``/``sqrt(``sum``(v``*``*``2` `for` `v ``in` `x.values())) ` ` ` ` ``# this should never be zero? ` ` ``except` `ZeroDivisionError: ` ` ``s ``=` `1.0` ` ``else``: ` ` ``s ``=` `1` ` ``for` `n ``in` `x: ` ` ``x[n] ``*``=` `s ` ` ``return` `x ` ` ` ` ``raise` `nx.NetworkXError(``'Power iteration failed to converge in '` ` ``'%d iterations.'` `%` `max_iter) `
The above function is invoked using the networkx library and once the library is installed, you can eventually use it and the following code has to be written in python for the implementation of the katz centrality of a node.
`>>> ``import` `networkx as nx ` `>>> ``import` `math ` `>>> G ``=` `nx.path_graph(``4``) ` `>>> phi ``=` `(``1``+``math.sqrt(``5``))``/``2.0` `# largest eigenvalue of adj matrix ` `>>> centrality ``=` `nx.katz_centrality(G,``1``/``phi``-``0.01``) ` `>>> ``for` `n,c ``in` `sorted``(centrality.items()): ` `... ``print``(``"%d %0.2f"``%``(n,c)) `
The output of the above code is:
`0` `0.37` `1` `0.60` `2` `0.60` `3` `0.37`
The above result is a dictionary depicting the value of katz centrality of each node. The above is an extension of my article series on the centrality measures. Keep networking!!!
tags:
Graph Python Graph | 2,828 | 9,695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-17 | latest | en | 0.960539 |
https://www.coolstuffshub.com/speed/convert/3-inches-per-hour-to-kilometers-per-day/ | 1,685,455,391,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645810.57/warc/CC-MAIN-20230530131531-20230530161531-00363.warc.gz | 796,982,528 | 12,793 | # Convert 3 Inches per hour to Kilometers per day (3 in/h to km/day Conversion)
3 inches per hour is equal to 0.0018288 kilometers per day.
3 in/h = 0.0018288 km/day
## How to convert 3 inches per hour to kilometers per day?
To convert 3 inches per hour to kilometers per day, multiply the value in inches per hour by 0.0006096.
You can use the conversion formula :
kilometers per day = inches per hour × 0.0006096
To calculate, you can also use our 3 inches per hour to kilometers per day converter, which is a much faster and easier option as compared to calculating manually.
## 3 inches per hour is equal to how many kilometers per day?
• 3 inches per hour = 0.0018288 kilometers per day
• 3.1 inches per hour = 0.00188976 kilometers per day
• 3.2 inches per hour = 0.00195072 kilometers per day
• 3.3 inches per hour = 0.00201168 kilometers per day
• 3.4 inches per hour = 0.00207264 kilometers per day
• 3.5 inches per hour = 0.0021336 kilometers per day
• 3.6 inches per hour = 0.00219456 kilometers per day
• 3.7 inches per hour = 0.00225552 kilometers per day
• 3.8 inches per hour = 0.00231648 kilometers per day
• 3.9 inches per hour = 0.00237744 kilometers per day
## Examples to convert in/h to km/day
Example 1:
Convert 3.2 in/h to km/day.
Solution:
Converting from inches per hour to kilometers per day is very easy.
We know that 1 in/h = 0.0006096 km/day.
So, to convert 3.2 in/h to km/day, multiply 3.2 in/h by 0.0006096 km/day.
3.2 in/h = 3.2 × 0.0006096 km/day
3.2 in/h = 0.00195072 km/day
Therefore, 3.2 inches per hour converted to kilometers per day is equal to 0.00195072 km/day.
Example 2:
Convert 3.8 in/h to km/day.
Solution:
1 in/h = 0.0006096 km/day
So, 3.8 in/h = 3.8 × 0.0006096 km/day
3.8 in/h = 0.00231648 km/day
Therefore, 3.8 in/h converted to km/day is equal to 0.00231648 km/day.
For faster calculations, you can simply use our 3 in/h to km/day converter.
## Inches per hour to kilometers per day conversion table
Inches per hour Kilometers per day
3 in/h 0.0018288 km/day
3.05 in/h 0.00185928 km/day
3.1 in/h 0.00188976 km/day
3.15 in/h 0.00192024 km/day
3.2 in/h 0.00195072 km/day
3.25 in/h 0.0019812 km/day
3.3 in/h 0.00201168 km/day
3.35 in/h 0.00204216 km/day
3.4 in/h 0.00207264 km/day
3.45 in/h 0.00210312 km/day
3.5 in/h 0.0021336 km/day
3.55 in/h 0.00216408 km/day
3.6 in/h 0.00219456 km/day
3.65 in/h 0.00222504 km/day
3.7 in/h 0.00225552 km/day
3.75 in/h 0.002286 km/day
3.8 in/h 0.00231648 km/day
3.85 in/h 0.00234696 km/day
3.9 in/h 0.00237744 km/day
3.95 in/h 0.00240792 km/day | 908 | 2,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-23 | latest | en | 0.656223 |
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Mizugadro, pentation, Book MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 02/10/2015, 12:12 PM Ok, let's try it. LEFT SIDE Substituting it in $F(z+1)$ it gives $\displaystyle F(z) = L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N)$ $\displaystyle F(z+1) = L+\sum_{n=1}^{N} a_n {e^{k(z+1)n}} + o({e^{k(z+1)N}})$ $\displaystyle F(z+1)=L+\sum_{n=1}^{N}(a_n{e^{kn}})\varepsilon^n+ o(\varepsilon^N{e^{kN}})$ Let's pretend for a second that I know how the little-o notation works (even if I don't even know what it is), now we have a new series that instead of the coefficents $a_n$ has the coefficents $b_n=a_n e^{kn}$ that looks really similar to the one we started with $\displaystyle F(z+1)=L+\sum_{n=1}^{N}b_n\varepsilon^n+ o(\varepsilon^N{e^{kN}})$ LEFT --- RIGHT SIDE Now I substitute it in $T(F(z))$ $\displaystyle \exp_b(F(z)) = \exp_b(L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N))$ Im not really good with series but maybe I could use the definition of exponentiation (for T=exp_b) and obtain this $\displaystyle b^{ F(z)} =e^{\ln(b)\cdot F(z)}= \exp_b(L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N))$ $\displaystyle \sum_{n=0}^{\infty}{\ln(b)^i\over i!}F(z)^i= \sum_{n=0}^{\infty}{\ln(b)^i\over i!}\left (L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N) \right)^i$ At this point with my poor knowledge of the subject I can't continue, so looking at your book it says that series has the property that (6.4) $\displaystyle T(F(z)) = L+T^{[1]}(L)\sum_{n=1}^{\infty}a_n {\varepsilon^n} +{T^{[2]}(L)\over 2}\left(\sum_{n=1}^{\infty}a_n {\varepsilon^n}\right )^2+...$ The summation in (6.4) seems an infinite series because I saw the dots “...”, so if I understand well the formulas, assuming that the sequence 1,2,6... is the factorial and with transfer function $T=\exp_b$ (with $T^{[n]}=T^{[n]}(L)$) the final form should be something like that $\displaystyle T(F(z)) =L+\sum_{i=1}^{\infty}{T^{[i]}(L)\over i!}\left(\sum_{n=1}^{\infty}a_n {\varepsilon^n}\right )^i$ RIGHT --- So the question is how can be the left side the same as the right side? $\displaystyle F(z+1)=L+\sum_{n=1}^{N}b_n\varepsilon^n+ o(\varepsilon^N{e^{kN}})=L+\sum_{i=1}^{\infty}{T^{[i]}(L) \over i!}\left(\sum_{n=1}^{\infty}a_n {\varepsilon^n}\right )^i=T(F(z))$ So what the book is probably trying to say is that is possible to manipulate algebraically the RIGHT side in order to obtain a series of the form $\displaystyle T(F(x))=L+\sum_{n=1}^{\infty}\tau_n\varepsilon^n$ with the coefficients $\tau_n$ defined in the formulas (6.5), (6.6) and (6.7) of your book $\tau_1=T'(L)$ $\tau_2=T'(L)a_2+T^{[2]}(L)/2$ $\tau_3=T'(L)a_3+T^{[2]}(L)a_2+T^{[3]}(L)/6$ and probably $\tau_n=T'(L)a_n+T^{[2]}(L)a_{n-1}+...+T^{[n-1]}(L)a_2+T^{[n]}(L)/n!$ I don't know the rules that makes you able to turn the series on the RIGHT in a series with coefficients $\tau_n$ but if is possible and we assume that $T(F(z))=F(z+1)$ then we should have that $F(z+1)=L+b_1\varepsilon+b_2\varepsilon^2+b_3\varepsilon^3+...=L+ {{\tau}}_1 \varepsilon + {\tau}_2 \varepsilon^2 + {\tau}_3 \varepsilon^3+...=T(F(z))$ and thus $b_n=a_n e^{kn}={\tau}_n$. This makes us able to find $k$ $b_1=a_1 e^{k}=T'(L)=\exp'_b(L)=\ln(b)b^L$ so $k=\ln({\ln(b)b^L\over a_1})$ Is this correct? And since we have $k$ and $a_1$ the other $a_n$ are given by your formulas (6.8 ), (6.9) and (6.10)... Thank you alot for the effort and your help. I'm really curious about this but I'm also sorry if is so hard, I'm a donkey in analysis and im not good with powerseries as well. MathStackExchange account:MphLee « Next Oldest | Next Newest » | 1,411 | 3,645 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 32, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2020-10 | latest | en | 0.741548 |
http://www.jiskha.com/display.cgi?id=1358621262 | 1,496,003,480,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463611560.41/warc/CC-MAIN-20170528200854-20170528220854-00022.warc.gz | 688,768,358 | 3,817 | # ALGEBRA
posted by on .
3/5X-5/4X<-13
• ALGEBRA - ,
Find common denominator.
12/20x - 25/20x < -13
An inequality is treated just like an equation, except that multiplying/dividing by a negative value reverses the carat (< to >).
-13/20x = -13
Multiply both sides by -20x, then divide by 13. | 98 | 299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-22 | latest | en | 0.858984 |
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posted by .
okay thanks for that. but now i need to know if you multiply the fractions across.
thank u
• math -
When there's a problem with multiplying fractions like 1/2*4/5, yes, jyst multiply across.
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# Trigonometry
If 2tan2θ = 5 secθ+ 10, where θ is obtuse, find the value of tan θ without the use of calculator.
Ans: -√5/2
Thank you :)
substituting, 2(sec2θ - 1) = 5 secθ + 10 giving an equation in secθ.
substituting x = secθ
2(x2 - 1) = 5x + 10, 2x2 -5x -12 = 0
Using the quadratic formula, x = (5 ± √(25 + 96))/4, or x = 4, or x = -3/2
Since x = secθ, and since the angle is obtuse (between 90 and 180 degrees), secθ = -3/2 , so cosθ = -2/3 and sinθ = √(5/9) = √5/3, and tanθ = -√5/2 | 240 | 532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2014-41 | latest | en | 0.618848 |
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03-25-2009, 03:56 PM #1 Illini10 enthusiast Join Date: Mar 2009 Posts: 50 Probability of flopping a set? I've read that it is around 12%, but I did a quick check and I came up with a different number. Could someone check my numbers, or explain what I did wrong? Player starts with a pocket pair in texas hold em. The odds of him flopping a set are Y, odds of not flopping a set are X. I calculated X using a binomial: Pr(n, k ) = n!/(k!(n-k)!)*pr(k)^k(1-pr(k))^k I define X as the odds of not hitting either of the 2 cards to improve to a set (or quads, since first you have to have 3 of a kind before improving to 4): X=Pr(50, 48)+Pr(49, 47)+Pr(48, 46) First I found Pr(50, 48)=0.276232807 Then Pr(49, 47)=0.27634952 Then Pr(48, 46)=0.276471236 Then X=0.829053564, and Y=0.170946436 This would mean that a pocket pair improves to a set ~17% of the time on the flop, which goes against most of what I googled.
03-25-2009, 04:19 PM #2 DarkMagus Pooh-Bah Join Date: Jul 2007 Location: Vancouver, BC Posts: 5,687 Re: Probability of flopping a set? i dont really have any clue what you've done there. i don't think you understand what the binomial distribution does at all. the proper way to calculate this would simply be: (48 c 3) / (50 c 3) = 0.882, which gives you the probability that of the remaining 50 cards in the deck, the flop will contain three of the 48 cards which don't match your pair. subtract this from 1 to get 0.118, the probability of flopping a set.
03-25-2009, 04:38 PM #3 Illini10 enthusiast Join Date: Mar 2009 Posts: 50 Re: Probability of flopping a set? Oh. Yeah I did this wrong, I was finding the odds of something unrelated to the question by thinking that n=number of cards rather than the number of times something is done. Thanks for clearing that up
03-25-2009, 05:27 PM #4 TheWuzi grinder Join Date: Feb 2009 Location: i has ps3 Posts: 415 Re: Probability of flopping a set? you start with a pair (2 cards). there is 50 cards left in the deck, only 2 remaining cards that can give you a set so your odds are 50-to-2 or 25-to-1 chance of flopping a set. 1 divided by 25 is 0.04 which comes out to 4%. so you have a 4% chance of flopping a set.
03-25-2009, 06:01 PM #5
dokonoko
stranger
Join Date: Feb 2009
Posts: 13
Re: Probability of flopping a set?
Quote:
Originally Posted by TheWuzi you start with a pair (2 cards). there is 50 cards left in the deck, only 2 remaining cards that can give you a set so your odds are 50-to-2 or 25-to-1 chance of flopping a set. 1 divided by 25 is 0.04 which comes out to 4%. so you have a 4% chance of flopping a set.
You forget that there are 3 cards on the flop
03-25-2009, 07:23 PM #6 TheWuzi grinder Join Date: Feb 2009 Location: i has ps3 Posts: 415 Re: Probability of flopping a set? well the odds of you flopping a set on the turn is still 47-to-2 or little under 24-to-1. 1 divided by 23.5 is 0.04255... which is still 4% lol.
03-25-2009, 07:28 PM #7
RustyBrooks
Carpal \'Tunnel
Join Date: Feb 2006
Location: Austin, TX
Posts: 21,792
Re: Probability of flopping a set?
Quote:
Originally Posted by TheWuzi well the odds of you flopping a set on the turn is still 47-to-2 or little under 24-to-1. 1 divided by 23.5 is 0.04255... which is still 4% lol.
You can't flop a set on the turn. You flop a set on the flop. And you have 3 chances at it - the odds are about 12% as shown above.
03-25-2009, 11:23 PM #8 statmanhal Pooh-Bah Join Date: Jan 2009 Posts: 3,582 Re: Probability of flopping a set? The term 'flopping a set" is a little vague. Does it mean a set and only a set. What about a full house or quads? Anyway, DarkMagnus' answer of 11.8% is for at least a set.
03-27-2009, 04:59 AM #9
Octavian
banned
Join Date: Sep 2005
Location: Las Vegas
Posts: 828
Re: Probability of flopping a set?
Quote:
Originally Posted by Illini10 I've read that it is around 12%, but I did a quick check and I came up with a different number. Could someone check my numbers, or explain what I did wrong?
You will get a PP 5.88%
Now, that you've got that pair you will flop a set or better 11.7%
In more more practical terms, what to expect in real live game:
You will flop 11.7 sets for each 100 PP that you hold. In order to get 100 PP you will have to "play" 1,700 hands.
As you can see, you will have a set on the flop once per (1,700*11.7)/100=199 hands. Now, how many hours takes you to play 199 hands in live poker? Well, you have 199/33=6 hours. On average you can say that you will have at least a set on the flop once every 6 hours of live play. If you game is much faster than 33 hands/hour then you tune 199/(# of hands per hour)
You can associate the set frequency with the frequency of getting a specific premium PP like AA. As we all know, we will get AA once in 220 hands.
So, I would say this: In 6-7 hours of live play we can expect on average to get pocket AA once and besides that to flop some kind of any set also. In a full day of play you can see that you’ll get on average the supreme PP plus some kind of set. We will have at least two big hands during one day of live play. But don't go crazy with your AA after the flop. You can shove all-in preflop if you have the opportunity but be cautious post flop with unimproved AA.
Last edited by Octavian; 03-27-2009 at 05:15 AM.
03-27-2009, 05:43 AM #10 SimonG adept Join Date: May 2005 Location: No risk it ~ no biscuit Posts: 708 Re: Probability of flopping a set? To flop a set (or better): You know what 2 cards are because you have them as a pair in your hand. First card off the deck then you have a 2/50 chance to hit. (A) If that fails you will have a 2/49 chance to hit on the second card (B) If that fails you will have a 2/48 chance to hit the last flop card. (C) As you aren't bothered (from the assumption at the start) whether you hit a set, quads or a boat, the maths then becomes [A OR B OR C] which is [2/50 + 2/49 + 2/48] which comes out at a little worse than 7/1 or the 12% previously quoted. There are lots of situations (particularly live play where you don't have your calculator handy) where a rough and ready calculation will do - so [2/50+2/50+2/50] = 6/50 = 12%. You get a small rounding error, but moderate maths ability could get you there during a hand if you wanted to know if you were getting your price. If you are playing a hand and there is a pot of \$585 and it is \$150 to call, for all but the very gifted the decimal places are immaterial as you can't work it out in your head and it shouldn't make a difference anyway if the pot is laying you 12% OR 12.2% Even if you are capable of working it out, it is so close that the difference shouldn't be the difference for you between one play and another. Knowing that you are 'a bit worse than 4/1' from the pot should be all the maths you need to play the hand and all the other factors can come into play.
03-27-2009, 09:38 PM #11 Zeds_Dead70 centurion Join Date: Mar 2009 Posts: 100 Re: Probability of flopping a set? Relevance being, most weak players can see flushes or straights on a flop but a set catches them off guard. How big of a raise will you call with a pocket pair under 9s in hopes of flopping a set? The implied odds in low limits are huge. Last edited by Zeds_Dead70; 03-27-2009 at 09:39 PM. Reason: grammage
03-27-2009, 11:15 PM #12
Octavian
banned
Join Date: Sep 2005
Location: Las Vegas
Posts: 828
Re: Probability of flopping a set?
Quote:
Originally Posted by Zeds_Dead70 Relevance being, most weak players can see flushes or straights on a flop but a set catches them off guard. How big of a raise will you call with a pocket pair under 9s in hopes of flopping a set? The implied odds in low limits are huge.
With a big pair like AA, KK, QQ I will raise if I’m the first one in or 3bet if opponent raises upfront. Now if I play deep stack, and have PP under 99, I will call based on stack odds. If the raiser is deep too I will call 5% to 7% of my stack. If for example I have \$1,000 and pocket 77, the guy up front open raise for \$50 to \$75, I will call if I figure I’m not going to be re-raised in the back. I call up to 7% of my stack and see the flop even if I suspect the upfront raiser has AA.
In fact, I want him to have AA and I want him to believe that he should play that pair real fast and hard all the way to the river. That is the ideal scenario for your 77. If the raiser bets about 1/15 of his stack and you cover than I would say CALL. If on the other hand the raiser has \$100 stack and he open-raise for \$25 and you suspect him for having a big pair or AK I will not call. Why? Because you will be a big dog against any overpair and about even money against AK and the reward when you hit your set will not cover the losses you accumulate when you miss. He doesn’t have enough dough for you to attack after the flop.
When you hit your set on the flop and there is a flush or a straight draw possible, don’t rush to raise and re-raise to knock your opponent out of his draw. Just bet enough for him to call as a mistake but do not raise him too much so he’s making the correct play of folding. Bet about halve of the pot. You want him to call you! You have more outs to make a FH or Quads against him. When you have a set on the flop and his drawing to flush he’s got only 7 clean outs while you’ve got the rest of the deck working for you. Actually, from 47 cards left he’s got 7 clean outs to flush while you don’t have to hit anything to beat him unless he hits his flush, after that point you still have 10 outs to kill his hand.
Never slow play a flopped set! Nobody knows you have a monster anyway, therefore, it is stupid to conceal the strength of your hand. Backing off to a raise and then check-raising on the turn is a valid strategy (although not necessarily best). I do not back off when there is a third suited card on board. I feel that I have enough outs to disregard the possibility of a made flush against me.
Note:
1. If you lose with a set, you'll lose a lot of money. If you don't, you are not playing your sets correctly
2. The reverse it’s also true: If you win with a set, you'll win a lot of money. If you don't, you are not playing your sets correctly
Last edited by Octavian; 03-27-2009 at 11:30 PM.
03-28-2009, 10:27 AM #13
jay_shark
Pooh-Bah
Join Date: Sep 2006
Posts: 3,655
Re: Probability of flopping a set?
Quote:
Originally Posted by SimonG To flop a set (or better): You know what 2 cards are because you have them as a pair in your hand. First card off the deck then you have a 2/50 chance to hit. (A) If that fails you will have a 2/49 chance to hit on the second card (B) If that fails you will have a 2/48 chance to hit the last flop card. (C) As you aren't bothered (from the assumption at the start) whether you hit a set, quads or a boat, the maths then becomes [A OR B OR C] which is [2/50 + 2/49 + 2/48] which comes out at a little worse than 7/1 or the 12% previously quoted. There are lots of situations (particularly live play where you don't have your calculator handy) where a rough and ready calculation will do - so [2/50+2/50+2/50] = 6/50 = 12%. You get a small rounding error, but moderate maths ability could get you there during a hand if you wanted to know if you were getting your price.
To compute the probability that you will hit a set or better, then using a Venn diagram, you can set the problem up like this:
Deal the cards one at a time and label the first card A, the second card B, and the third card C.
P(A or B or C)
= P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C)
P(A or B or C)
= 2/50 + 2/50 + 2/50 - 2/50*1/49 -2/50*1/49 - 2/50*1/49
~ 11.755%
03-28-2009, 03:01 PM #14 PantsOnFire Pooh-Bah Join Date: Sep 2006 Location: Bumpy car ride... Posts: 5,720 Re: Probability of flopping a set? The easy way of doing this is by starting with the approximation that the chances of a particular card coming up is about 2%. So if there are two cards that can help you and there are three cards flipped up, the chance is 2(good cards)x2(%)x3(flop)=12%. Let's delve into this even deeper. Say you have AKoff. You now have 6(good cards)x2(%)x3(flop)=36%. Thank you, thank you, I'll be here all week.
03-29-2009, 10:28 AM #15
SimonG
Join Date: May 2005
Location: No risk it ~ no biscuit
Posts: 708
Re: Probability of flopping a set?
Quote:
Originally Posted by jay_shark Your approach has some errors. To compute the probability that you will hit a set or better, then using a Venn diagram, you can set the problem up like this: Deal the cards one at a time and label the first card A, the second card B, and the third card C. P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) P(A or B or C) = 2/50 + 2/50 + 2/50 - 2/50*1/49 -2/50*1/49 - 2/50*1/49 ~ 11.755%
To me all it looks like you have done here is worked out a set or better, and then subtracted the probabilities of hitting a better hand. So your answer looks correct for Probability of hitting a set and only a set. In poker, whilst you are working out your price of flopping a set, you presumably won't mind hitting quads either, so the assumption of "set or better" seems fair.
No matter either way, most of my point is that whether you use a quick and relatively accurate method such as the 2% rule - it is something you can do real time and is therefore helpful to live poker players. Real time, Venn diagrams aren't any use, the maths is interesting from a 'settling a prop bet' angle to 4 decimal places, but for poker, quick and approximate ftw.
03-29-2009, 10:38 AM #16
jay_shark
Pooh-Bah
Join Date: Sep 2006
Posts: 3,655
Re: Probability of flopping a set?
Quote:
Originally Posted by SimonG To me all it looks like you have done here is worked out a set or better, and then subtracted the probabilities of hitting a better hand. So your answer looks correct for Probability of hitting a set and only a set. In poker, whilst you are working out your price of flopping a set, you presumably won't mind hitting quads either, so the assumption of "set or better" seems fair.
If you took the time to read it, you'll realize that I actually computed the probability of flopping a set or better. I specifically mention this.
If you had solely mentioned that the approximate probability of flopping a set or better is 12%, then I would have let it go. You did actually make an attempt at a more rigorous approach which is theoretically wrong and I had to correct you.
It's only fair.
03-29-2009, 12:56 PM #17 maximumprobability grinder Join Date: Jan 2008 Location: Stanford CA USA Posts: 596 Re: Probability of flopping a set? And why not do it the simplest possible way ie you have 50 cards left and 2 of which 1 at least you want to come at flop . That of course = 1 - chance the flop doesnt have any of the 2 . Since there are (50,3) possible flops and only (48,3) that dont have the card you want the chance of no (set or better) is (48,3)/(50,3) so p(set+)=1-48*47*46/50/49/48=11.7551...% Also people will recognize this as 1-(48/50)*(47/49)*(46/48) which is 1 - products of probabilities to miss the critical cards on each one of the three flop cards, since you want all 3 to miss it its a product! (already used above by DarkMagus but in the form above people can relate to it with probabilities instead of combinatorics. However combinatorics brutal force method is the best way to handle such problems as they become more complex in other special questions and conditional probabilities can lead to confusions sometimes.) Obviously this is close to 1-96%^3 (96% ~48/50 approx 47/49 approx 46/48 of cards dont get get you to set or better per trial) since 1-96%^3~12% you get a rough answer fully understanding the spirit of all above but using a fast approximation like (1-x)^n~1-nx for small x , here n=3 and x=4% ps: (n,k)=n!/k!/(n-k)!
04-27-2009, 07:19 PM #19 Rushton14 journeyman Join Date: Nov 2008 Location: nova scotia Posts: 286 Re: Probability of flopping a set? chances of floping a set are like 1 in 7.5 or close to it.
04-27-2009, 07:39 PM #20
moki
Carpal \'Tunnel
Join Date: Oct 2005
Location: Rochester, NY
Posts: 8,037
Re: Probability of flopping a set?
Quote:
Originally Posted by Rushton14 chances of floping a set are like 1 in 7.5 or close to it.
Yes, I know this -- that wasn't my question.
04-27-2009, 08:05 PM #21
SuitedEights
journeyman
Join Date: Mar 2009
Posts: 203
Re: Probability of flopping a set?
Quote:
I can't answer your question exactly, but I will give it a shot. Math guru's butcher my post after this. If you are calculating the odds of your flopping a set, you are roughly 1:7.5. However, when you look for your opponent also flopping a set, now there are only 2 cards on the flop that he could have hit a set with compared to the general "1:7.5" odds, so just for the odds that your opponent also has a set, it might be lower. Then for you to flop a set and him to flop a set, I'd say you'd have to multiply these two numbers and that would give your final statistic of both of you flopping a set.
Ok, here is what I think…
Odds of your not flopping a set = (48/50)*(47/49)*(46/48)=88.2%
Odds of your opponent not flopping a set = (1)*(47/49)*(46/48)=91.9%
(the (1) here being a sure thing because you have to have a set card)
Odds of your flopping a set= 100%-88.2%=11.8%
Odds of your opponent flopping a set= 100%-91.9%=8.1%
Odds of both events happening= 11.8%*8.1% = 0.96%
Last edited by SuitedEights; 04-27-2009 at 08:24 PM.
04-27-2009, 08:26 PM #22 Senator Wright journeyman Join Date: Jan 2008 Location: Grinding my Live Roll Online Posts: 352 Re: Probability of flopping a set? you'd also have to take into account the odds of another player having a pocket pair.
04-27-2009, 11:59 PM #23
moki
Carpal \'Tunnel
Join Date: Oct 2005
Location: Rochester, NY
Posts: 8,037
Re: Probability of flopping a set?
Quote:
Originally Posted by SuitedEights I can't answer your question exactly, but I will give it a shot. Math guru's butcher my post after this. If you are calculating the odds of your flopping a set, you are roughly 1:7.5. However, when you look for your opponent also flopping a set, now there are only 2 cards on the flop that he could have hit a set with compared to the general "1:7.5" odds, so just for the odds that your opponent also has a set, it might be lower. Then for you to flop a set and him to flop a set, I'd say you'd have to multiply these two numbers and that would give your final statistic of both of you flopping a set. Ok, here is what I think… Odds of your not flopping a set = (48/50)*(47/49)*(46/48)=88.2% Odds of your opponent not flopping a set = (1)*(47/49)*(46/48)=91.9% (the (1) here being a sure thing because you have to have a set card) Odds of your flopping a set= 100%-88.2%=11.8% Odds of your opponent flopping a set= 100%-91.9%=8.1% Odds of both events happening= 11.8%*8.1% = 0.96%
Thanks for your reply -- I guess that last line is the crux of the argument, though.
One of the guys at our game is saying that if we assume you have *already* flopped a set, and with that given, what are the odds that someone else also has a set in the hand.
My contention is that the odds would be as you describe them... because you need to know when two events happen at the same time. His contention is that the fact that you've already flopped a set means the calculation is different.
04-28-2009, 01:26 AM #24
PantsOnFire
Pooh-Bah
Join Date: Sep 2006
Location: Bumpy car ride...
Posts: 5,720
Re: Probability of flopping a set?
Quote:
Originally Posted by moki His contention is that the fact that you've already flopped a set means the calculation is different.
Well tell him that if you don't flop a set, the calculations are very easy. Then walk away.
04-28-2009, 03:17 AM #25
SuitedEights
journeyman
Join Date: Mar 2009
Posts: 203
Re: Probability of flopping a set?
Quote:
Originally Posted by PantsOnFire Well tell him that if you don't flop a set, the calculations are very easy. Then walk away.
But your posts always contain a flopping set, how am I supposed to just walk away?
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A275978 Numbers n such that (101*10^n + 1)/3 is prime. 0
1, 4, 6, 12, 34, 54, 60, 61, 73, 148, 349, 552, 649, 967, 1044, 2521, 4501, 5721, 6133, 9052, 9880, 16126, 16215, 19146, 61770 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Numbers n such that the digits 33 followed by n-1 occurrences of the digit 6 followed by the digit 7 is prime (see Example section). a(26) > 10^5. LINKS Makoto Kamada, Factorization of near-repdigit-related numbers. Makoto Kamada, Search for 336w7. EXAMPLE 4 is in this sequence because (101*10^4 + 1)/3 = 336667 is prime. Initial terms and primes associated: a(1) = 1, 337; a(2) = 4, 336667; a(3) = 6, 33666667; a(4) = 12, 33666666666667; a(5) = 34, 336666666666666666666666666666666667, etc. MATHEMATICA Select[Range[0, 100000], PrimeQ[(101*10^# + 1)/3] &] PROG (PARI) isok(n) = isprime((101*10^n + 1)/3); \\ Michel Marcus, Aug 16 2016 CROSSREFS Cf. A056654, A268448, A269303, A270339, A270613, A270831, A270890, A270929, A271269. Sequence in context: A027070 A087785 A081198 * A242211 A073167 A060202 Adjacent sequences: A275975 A275976 A275977 * A275979 A275980 A275981 KEYWORD nonn,more AUTHOR Robert Price, Aug 15 2016 STATUS approved
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http://coding.yonsei.ac.kr/xe/index.php?mid=publicboard&sort_index=readed_count&order_type=asc&page=3&document_srl=7339 | 1,660,331,129,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571745.28/warc/CC-MAIN-20220812170436-20220812200436-00636.warc.gz | 13,196,724 | 7,485 | 한국어
# 교재준비작업(1)
2004.10.03 21:27
송홍엽 조회 수:8602
이산수학과 유한체이론 - 주별 강의주제 목록
1. Property of Z
Z is a commutative ring with 1.
it is an integral domain.
it is a Euclidean domain.
it is a Unique Factorization domain.
Z/(n) is a ring, domain, ED, and UFD.
G=U(Z/(n)) is a multiplicative group of order phi(n)
it is cyclic if and only if n=1,2,4,p^k, 2p^k.
Z/(n) is a field if and only if n=prime
U(Z/(p)) is cyclic of order p-1
2. Computations over Z
Linear equation over Z/(n)
Chinese Remainer Theorem
Quadratic Reciprocity Theorem - Legendre/Jacobi symbols
Big Integer arithematic
Fast Exponentiation
3. Some Public Key Crypto Algorithms
Primality testing algorithm
DLP - analysis
ElGamal Algorithm
RSA Algorithm
Secrete Sharing Algorithm
Coin-flipping over Telephone
Public key Envelope
4. Permutations and Counting
Definition, notation, order, cycles
unique decomposition
even and odd permutation
Not-Burnside Theorem on counting
5. Vectors and Matrix
n-tuple vector space over F
Basis, Linear Independance
simultaneous equation and coefficient matrix
Gauss Elimination, rank of a matrix, LU decomposition
column space, row space, orthogonal complement
rank, nullity, basic relation
Vandemond matrix
6. Linear Transformation and Matrix
Definition of Linear Transformation
Range space and Null space, basic relation
relation to Matrix
Multi-linear transformation and Determinant
Existence of Determinant
7. Some Problem Discussions
8. midterm
9. Polynomial over GF(p) = Fp[x]
Fp[x] is a commutative ring with 1.
it is an integral domain.
it is a Euclidean domain.
it is a Unique Factorization domain.
Fp[x]/(f(x)) is a ring, domain, ED, and UFD.
Fp[x]/(f(x)) is a field if and only if f(x) is irreducible
10. Structure of Finite Field
Extension and Subfield
Multiplicative structure and conjugate class
Homomorphism and isomorphism
11. Irreducible Polynomial over finite field
irreducible polynomials over F2
Property of minimal polynomials
Conjugates
Trace function
m-sequence
12. Irreducible Polynomial and Cyclotomic Polynomial
x^{q^n}-x = ㅠ V_d(x)
Number of irreducible polynomials
Cyclotomic Polynomials
over C
over Finite Field
Some Factoring
13. Error-correcting linear codes
binary symmetric channel, binary erasure channel
binary linear code and minimum distance decoding
Binary Hamming code and decoding
minimum distance decoding is ML decoding
BCH code over Z/(p)
Some nonlinear simultaneous equations for decoding of BCH code over Z/(p)
14. Cyclic code
Hamming code and BCH code as cyclic code
RS codes - encoding and decoding
GFFT approach
15. Some Problem Discussions
16. final exam
번호 제목 글쓴이 날짜 조회 수
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공지 우리학과 대학원생 모두에게 (특히, 박사과정들에게) 하고싶은 말입니다. 2014.01.20 6747
49 Re.. Turbo code Decoder 2004.04.13 4240
48 퍼즐에 상금을 부여합니다...^^ 2004.04.15 4256
47 Re..Turbo code Encoder 2004.04.13 4345
46 2000년에 어딘가에 올린 글입니다.."열정" 2004.04.15 4421
45 오늘같은 날 2004.10.11 4434
44 combinatorial search problem 2004.07.21 4498
43 marginally Gaussian but not jointly Gaussian 2004.05.13 4541
42 덴치 문제..푼 결과 [1] 2005.10.11 4560
41 [퍼온글]과학자들은 왜 속이는가 2006.02.07 4619
40 랜덤변수의 variance가 0이면? 2004.05.13 4629
39 정보화와 정보이론 2004.08.06 4632
38 uncorrelated but not independent 2004.05.13 4723
37 수학자와 공학자 2006.05.09 4738
36 채널코딩에 관한 첫번째 이야기 2003.10.13 4742
35 지금 내 전공분야는 공부한던 시절엔 나에게 가장 힘든 분야였답니다... 2003.06.03 4867
34 채널코딩 세째 이야기 2003.11.13 4988
33 [펀글] 프로그램 설계시 좋은 코딩 습관 2009.12.27 5030
32 오류정정부호에 대한 이야기 2004.04.13 5095
31 [퍼온글] 달력의 유래 2004.04.01 5189
30 [퍼온글] 명왕성 이야기 2006.04.26 5374 | 1,245 | 3,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-33 | longest | en | 0.746985 |
http://mathhelpforum.com/calculus/110249-rolle-s-theorem-mean-value-theorem.html | 1,526,979,203,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864648.30/warc/CC-MAIN-20180522073245-20180522093245-00123.warc.gz | 194,988,165 | 9,034 | # Thread: Rolle's Theorem, Mean Value Theorem
1. ## Rolle's Theorem, Mean Value Theorem
2) The ordering and transportation cost C for components used in a manufacturing process is approximated by $\displaystyle C(x) = 10(\frac{1}{x} + \frac{x}{x+3})$, where C is measured in thousands of dollars and x is the order size in hundreds. C(3) is equal to C(6). According to Rolle's Theorem, the rate of change of the cost must be 0 for some order size in the interval (3, 6). Find that order size.
3) Determine the values a, b, c, and d so that the function f satisfies the hypotheses of the Mean Value Theorem on the interval [-1, 2].
f(x) = a if x = -1
f(x) = 2 if -1 < x <= 0
f(x) = b$\displaystyle x^2$ + c if 0 < x <= 1
f(x) = dx + 4 if 1 < x <= 2 | 240 | 750 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-22 | latest | en | 0.806691 |
https://www.intellectualmath.com/express-a-number-in-index-form.html | 1,722,746,155,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640389685.8/warc/CC-MAIN-20240804041019-20240804071019-00000.warc.gz | 647,571,220 | 5,134 | EXPRESS A NUMBER IN INDEX FORM
To express the given number in index form, we have to decompose the number as prime factors.
If we have repeated values, using exponents we can write down it.
For example,
5 x 5 x 5 = 53
Express the following as the product of prime factors in index form:
Problem 1 :
56
Solution :
56 = 8 × 7
= 2 × 2 × 2 × 7
= 23 × 7
Problem 2 :
240
Solution :
240 = 16 × 15
= 4 × 4 × 3 × 5
= 2 × 2 × 2 × 2 × 3 × 5
= 24 × 3 × 5
Problem 3 :
504
Solution :
504 = 72 × 7
= 8 × 9 × 7
= 2 × 2 × 2 × 3 × 3 × 7
= 23 × 3² × 7
Problem 4 :
735
Solution :
735 = 15 × 49
= 3 × 5 × 7 × 7
= 3 × 5 × 7²
Problem 5 :
297
Solution :
297 = 27 × 11
= 9 × 3 × 11
= 3 × 3 × 3 × 11
= 33 × 11
Problem 6 :
221
Solution :
221 = 13 × 17
Problem 7 :
360
Solution :
360 = 72 × 5
= 8 × 9 × 5
= 2 × 2 × 2 × 3 × 3 × 5
= 23 × 3² × 5
Problem 8 :
952
Solution :
952 = 56 × 17
= 8 × 7 × 17
= 2 × 2 × 2 × 7 × 17
= 23 × 7 × 17
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https://www.ryanheise.com/cube/edges_first.html | 1,632,573,914,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057622.15/warc/CC-MAIN-20210925112158-20210925142158-00531.warc.gz | 1,004,168,459 | 5,626 | • Search Content
# Solving five edges and two corners
## The edges-first approach
### The "edges first" approach
→
Difficulty Easy to learn Efficiency Sometimes produces very good solutions, other times very bad solutions Applications Useful in fewest-moves solving where extra thinking time can be used to distinguish between good and bad situations. Not as useful for speedcubing. Learning to solve the edges on their own is a crucial first step to mastering the more advanced approaches.
#### Solving the edges
When solving the edges, it is important to have some idea of how close they are to being solved. When three of the top edges are in their correct positions (relative to each other), this is actually far away from being solved. When only two edges on top are correct relative to each other, this is actually close to being solved. When no edges on top are correct, this is the same distance as when three edges are correct.
To understand why this is so, we need to look at the cases.
In the cube to the left, notice that the blue/white and orange/white edges are in order, but the red/white edge is out of order. Can you see how to solve this position in three moves? (Click play to see the solution) When only two edges are in the correct order, this is the closest position from being solved. Similarily, here two of the top edges are in their correct positions (relative to each other). Can you see which ones, and see how to solve the the edges in three moves? Here, all three top white edges are in the correct order. To be able to solve this position, we first need to take one of the correct edges out of the chain so that we end up in a position that is closer to being solved. Then we can use a usual 3-move solution to finish. Here, none of the edges are correct. First, we try to get exactly two edges correct. Following that, we can use a standard 3-move solution to finish.
#### Solving two corners
Now that all edges are solved, we need a way to permute the corners without disturbing the edges. The way to do this is to use commutators and conjugates. In fact, this technique for moving corners around is identical to the technique used to solve the final three corners in the last step of this solution.
Let's look at a few examples.
In our first example, we solve each of the two corners individually, each time using a corner 3-cycle. In the first 3-cycle, we care only about the red/white/green corner and ignore what happens to the red/green/yellow corner and the orange/white/green corner which are also moved by the 3-cycle. Similarly, in the second 3-cycle we care only about solving the green/white/orange corner and ignore what happens to the red/blue/white corner and the yellow/green/red corner. This time, we try to solve two corners with a single 3-cycle. Our commutator is conjugated by a rotation of the red side. This is one of the worst positions that can occur and one that should be avoided by looking ahead and choosing a different solving approach: all corners are in their correct positions but are incorrectly oriented. Here, we use a corner twist to solve two of the corners. | 687 | 3,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2021-39 | latest | en | 0.957415 |
https://www.jiskha.com/display.cgi?id=1201734420 | 1,516,461,548,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889660.55/warc/CC-MAIN-20180120142458-20180120162458-00307.warc.gz | 924,148,259 | 4,247 | math
posted by .
I always thought I understood significant figures...until I attempted to tutor someone. Some of the rules seem counterintuitive. For instance...
10.070 why is it 5 sig fig's? the "0" on the right is useless. without it the # holds the same value.
20 Why is it only "1" sig fig? the "zero" is the difference between two and twenty. seems pretty sig to me.
0.01234 has "4" sig fig's. why isn't the zero to the right of the decimal sig? without it, the # doesn't hold the same value...can someone please explain the concept that I'm missing??
• math -
<<10.070 why is it 5 sig fig's? the "0" on the right is useless.>>
The fact that the zero is there beyond the decimal point means that it IS a significant figure.
<<20.: Why is it only "1" sig fig? the "zero" is the difference between two and twenty.>>
I agree. With a decimal point after the 20, it should be clear that the number is closer to 20 than 19 or 21. Therefore there are two sig figs. However, if they had just written 20 without a decimal point, then that would be one significant figure. For example: 2000 with no decimal has only one sig fig. also.
<<0.01234 has "4" sig fig's. why isn't the zero to the right of the decimal sig?>>
Until you get a nonzero number on the left side, none of the figures are significant in terms of precision; they just establish the order of magnitude. Think of writing it as 1.234*10^-2. The exponent -2. does not establish the precision, only the 1.234 does.
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More Similar Questions | 767 | 3,010 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-05 | latest | en | 0.923753 |
http://openstudy.com/updates/5158e954e4b07077e0c0b491 | 1,448,924,032,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398464253.80/warc/CC-MAIN-20151124205424-00117-ip-10-71-132-137.ec2.internal.warc.gz | 171,053,542 | 10,854 | ## Selena2345 2 years ago A car salesperson sells 7 cars in a week. The sum of the purchases is \$365,000.If the dealership pays their salesperson a commission of 1.5% and \$10.00 an hour,how much money did this salesperson make for 35 hours of work ?
1. hedyeh99
Ok, to do this, you are going to do 365,000 times 1.5% first
2. Selena2345
365000 x 1.5%=5475
3. Selena2345
Hedyeh99
4. hedyeh99
Now, you are going to multipy 35 by 10
5. Selena2345
35 x10=350
6. Selena2345
Hedyeh99
7. hedyeh99
Now, you are going to add 350+5475
8. Selena2345
350 +5475=5825
9. Chlorophyll
@Selena2345 You've been posting the same type of question for at least the whole 2 continuous days !!!
10. Selena2345
no chlorophyll
11. Selena2345
Hedyeh99
12. hedyeh99
13. Selena2345
okay hedyeh99
14. Chlorophyll
All you do is multiply the total by the % over and over! | 306 | 869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2015-48 | longest | en | 0.752612 |
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