The second equation of motion is used to calculate the distance traveled by an object. To derive this equation, we consider an object that starts with an initial velocity u and accelerates for a time t until it reaches a final velocity v, covering a distance s.

The area under a velocity-time graph represents the distance traveled by the object. Using this concept, we can express the distance s as the area under the graph, given by:

s = ¹/₂(v + u)t

Substituting the first equation of motion, v = u + at, into the above equation, we get:

s = ¹/₂(v + u)t
s = ¹/₂([u + at] + u)t
s = ¹/₂(u + at + u)t
s = ¹/₂(2u + at)t
s = ¹/₂(2ut + at²)
s = ut + ¹/₂at²

Therefore, the second equation of motion is:

s = ut + ¹/₂at²

This equation can be used to solve problems involving distance, initial velocity, final velocity, acceleration, and time.

Example 1:
A bus starts from rest and accelerates at a rate of 2.5 m/s² for 50 s. Determine the distance traveled by the bus during the acceleration phase.

Given: u = 0 m/s, t = 50 s, a = 2.5 m/s²
Using the second equation of motion:
s = ut + ¹/₂at²
= 0*50 + 0.5*2.5*50²
= 0 + 1.25*2500
= 3125 m

Example 2:
An airplane traveling at 300 m/s lands on a 2000 m long track and decelerates for 25 s until it comes to rest. Calculate the deceleration needed to stop the plane.

Given: u = 300 m/s, t = 25 s, s = 2000 m
Using the second equation of motion:
s = ut + ¹/₂at²
Rearranging to solve for a:
a = 2(s - ut)/t²
= 2(2000 - 300*25)/25²
= -8.8 m/s²

To make t the subject of the formula, we can rearrange the equation as follows:
0.5at² + ut - s = 0
Using the quadratic formula, we get:
t = [-u ± √(u² - 4*0.5a*(-s))]/(2*0.5a)
t = [-u ± √(u² + 2as)]/a

This equation can be used to calculate the time taken by an object to travel a certain distance under constant acceleration.


# Python数据挖掘代写 | COMP809 Data Mining and Machine Learning Assignment 2

本次新西兰**代写**是Python数据挖掘的一个assignment。您需要探索三种聚类方法：K-Means、DBSCAN和Agglomerative，并将每种算法应用于三个不同的数据集：Seoul Bike Sharing Demand、Sales Transactions和Water Treatment Plant。

任务包括三个部分：

1. 对每个数据集进行数据探索、数据预处理和特征选择，然后应用K-Means、DBSCAN和Agglomerative算法。评估指标包括时间、Sum of Squares Errors (SSE)、Cluster Silhouette Measure (CSM)和Davis-Bouldin score。
2. 对每个算法，提交Python代码和结果，包括3x3表格和CSM图。
3. 比较每个数据集上三种算法的性能，确定哪种算法表现最佳，并解释其原因。

具体要求：

a) 运行K-Means算法，获得最佳K值，并计算SSE、CSM和时间。提交3x3表格和CSM图。[7分]

b) 重复a)中的步骤，使用DBSCAN算法。[7分]

c) 重复a)中的步骤，使用Agglomerative算法。[7分]

d) 对每个数据集，确定哪种算法表现最佳，并解释其原因。[7分]

e) 对每个数据集和获胜算法，解释为什么该算法产生了最佳CSM值。[7分]

f) 根据d)中的结果，确定哪种算法是总体最佳，并解释其原因。[5分]


To prove that x = –3, given the equation 7x + 2 = 2 + 5x – 6, we use the commutative property to rearrange the terms so that like terms are next to one another, resulting in 7x + 2 = 2 – 6 + 5x. Then, we apply the associative property to group the like terms, giving 7x + 2 = (2 – 6) + 5x. Next, we combine like terms to get 7x + 2 = –4 + 5x. Using the subtraction property of equality, we subtract 5x from both sides of the equation, yielding 2x + 2 = –4. Then, we subtract 2 from both sides, resulting in 2x = –6. Finally, we use the division property of equality to divide both sides by 2, giving a final solution of x = –3.

A more straightforward approach to solving this equation is to simply apply the rule that whatever is done to one side of the equation must be done to the other. Thus, subtracting 2 and 5x from both sides gives 7x - 5x = -6, simplifying to 2x = -6, and then dividing both sides by 2 yields x = -3.

The justification left out of the original paragraph proof is the associative property of addition, which allows us to group like terms.

Which justification was left out of the paragraph proof above?

A. Commutative Property of Addition
B. Associative Property of Addition
C. Distributive Property
D. Addition Property of Equality

The correct answer is B. Associative Property of Addition. This property is necessary for grouping like terms in the equation.


The formula to calculate the perimeter of a rectangle is given by Perimeter = 2l + 2w, where l = length and w = width. This can also be written as 2(l + w).

To find the perimeter, we need to know the length and width of the rectangle. For example, if the length and width are L and W, the perimeter is given by Perimeter = 2(L + W).

Given the perimeter and area of a rectangle, we can solve for the length and width. For instance, if the perimeter is 200 and the area is 2400, we can set up the equations:

Perimeter = 2(L + W) = 200
Area = L * W = 2400

Solving these equations, we get L + W = 100 and L = 100 - W. Substituting L into the area equation, we get (100 - W) * W = 2400, which simplifies to W^2 - 100W + 2400 = 0.

Solving this quadratic equation, we find W = 40 feet and L = 60 feet.

The area of a rectangle is given by Area = L * W, where L is the length and W is the width. For example, if L = 5 in and W = 2 in, the area is A = 5 * 2 = 10 in^2.

In some cases, we may be given a relationship between the length and width, such as L = 2W or W = L + 2. However, if the relationship is inconsistent, such as L = 2W and W = W + 2, it may not be possible to solve for the length and width.

To find the width given the length and area, we can use the formula W = A / L. For example, if A = 3x^2 + 9xy + 6y^2 and L = 3x + 6y, we can find W by dividing A by L.

The formula for the perimeter of a rectangle can also be applied to a square, where the length and width are equal. In this case, the perimeter is simply 4 times the length of one side.

Q: 2l plus 2w is used for what?
A: The perimeter of a rectangle.

Note: The perimeter of a rectangle is given by the formula P = 2L + 2W, which can also be written as P = L + L + W + W. This formula is used to calculate the distance around the outside of a rectangle.


**ReCal OIR: Reliability Calculator for Ordinal, Interval, and Ratio Data**

ReCal OIR is an online utility that computes intercoder/interrater reliability coefficients for nominal, ordinal, interval, and ratio data judged by two or more coders. As of 5/22/17, ReCal OIR can also be used to compute coefficients for incomplete nominal datasets.

**Key Features:**

* Calculates four reliability coefficients: Krippendorff's alpha for nominal, ordinal, interval, and ratio data
* Accepts any range of possible variable values, including decimal values and negative numbers
* Allows missing data (as of 5/22/17)
* Results are valid for nominal, ordinal, interval, or ratio data sets coded by two or more coders

**Documentation:**

ReCal OIR operates under the following requirements and assumptions:

* Data should be nominal, ordinal, interval, or ratio
* Each file should represent two or more coders working on a single variable
* Each column should represent a single coder's work on one variable
* Each row should represent a single unit of analysis
* All code values must be represented numerically
* Input file must be formatted properly
* All columns must contain the same number of units of analysis

**Formatting Data for ReCal OIR Analysis:**

* Make sure each code value is represented by a unique number or a hash mark (#) for missing data
* Create a new file in Excel, SPSS, or another spreadsheet-like program
* Enter a different coder's data into each column, one unit of analysis per row
* Save the file in comma-separated values (.csv), semicolon-separated values (.csv), or tab-separated values format (.tsv)

**Frequently Asked Questions:**

* Do all codes have to be represented to obtain an accurate alpha? No, the math of Krippendorff's alpha does not take account of variable values that do not actually appear in the data.
* How is Krippendorff's alpha interpreted? A value closest to 1 means the most reliable, while a value closest to 0 means complete reliability.
* Can ReCal OIR handle missing data? Yes, ReCal OIR allows missing data, which should be represented by a single hash mark (#) in the input file.
* What is the level of significance used for calculating the alpha coefficients? The level of significance is not specified, but it can be considered as part of the calculation.

**Example Use Cases:**

* Computing reliability between two coders for one variable with ordinal data
* Handling missing data in a coding scheme with multiple coders
* Interpreting Krippendorff's alpha values, including negative values and values closest to 0

**Troubleshooting:**

* Check the FAQ/troubleshooting page for common issues
* Ensure proper formatting of the input file
* Contact the developer for further assistance if needed

**User Feedback and Suggestions:**

* Users have found ReCal OIR to be a useful tool for calculating reliability coefficients
* Some users have requested additional features, such as interval estimations (confidence limits)
* Others have asked for clarification on interpreting Krippendorff's alpha values and handling missing data.


A complicated problem is presented, involving a spring, gravity, and motion. A frictionless table has a spring compressed at distance x from its equilibrium. An object of mass m is launched off the table and falls to the ground, with an initial height of h. The goal is to find the final velocity (Vf) of the object when it hits the ground.

The problem can be solved using the principle of conservation of energy, which states that the initial energy is equal to the final energy. Initially, the object has potential energy due to its height and elastic potential energy due to the compressed spring. When it falls to the ground, all the energy is converted to kinetic energy.

The equation for conservation of energy can be written as: initial energy = final energy. The initial energy consists of potential energy (mgh) and elastic potential energy (0.5kx^2), where m is the mass, g is the acceleration due to gravity, h is the initial height, k is the spring constant, and x is the compression distance. The final energy is all kinetic energy (0.5mVf^2).

By equating the initial and final energies, the equation becomes: 0.5mVf^2 = mgh + 0.5kx^2. Simplifying the equation, Vf can be solved for. Multiplying both sides by 2 and dividing by m gives: Vf^2 = 2gh + kx^2/m. Taking the square root of both sides gives: Vf = sqrt(2gh + kx^2/m).

However, the provided solution is Vf = sqrt(2gh + kx^2/m) which can be approximated to Vf = sqrt(2gh) when the spring's effect is negligible, but the exact solution provided in the text is not complete.

To find the correct solution, the complete equation should be used: Vf = sqrt(2gh + kx^2/m). This equation shows that both the height and the spring's compression affect the final velocity of the object.


To find a percentage of a number, convert the percentage to a decimal by dividing by 100. For example, 40% becomes 0.4. Then, multiply this decimal by the number.

What number is 10% of 150?
10% of 150 is 15.

What number is 40% of 250?
40% of 250 is 100.

What number is 35% of 80?
35% of 80 is 28.

What number is 40% of 150?
40% of 150 is 60.

What is 40% off?
40% off a price means the price is reduced by 40% of its original value.

How do you solve 40% of 200?
40% of 200 is 0.4 * 200 = 80.

How do you find 40% of 60?
40% of 60 is 0.4 * 60 = 24.

What number is 30% of 50?
30% of 50 is 0.3 * 50 = 15.

How much is 40% off 100?
40% off 100 is 100 - (0.4 * 100) = 60.

What number is 40% of 200?
40% of 200 is 0.4 * 200 = 80.

What number is 20% of 15?
20% of 15 is 0.2 * 15 = 3.

What number is 40% of 500?
40% of 500 is 0.4 * 500 = 200.

How much is 30% off 200?
30% off 200 is 200 - (0.3 * 200) = 140.

What is $80 with 40% off?
$80 with 40% off is 80 - (0.4 * 80) = 48.

The percentage formula is (Value/Total value) * 100.

To subtract 40% from a price, multiply the price by 0.4 to find 40% of the price, then subtract this amount from the original price.

What number is 30% of 80?
30% of 80 is 0.3 * 80 = 24.


## How to Find the Surface Area of a Cylinder

The surface area of an open-top cylinder can be calculated by finding the area of the base and curved surface. The formula for the surface area of an open-top cylinder is πr(2h + r), where 'r' is the base radius and 'h' is the height of the cylinder.

### Total Surface Area of a Cylinder

The total surface area of a solid right cylinder of radius 'r' and height 'h' is 2πr(h + r). This can also be expressed as the sum of the curved surface area and the areas of the two circular bases.

### Curved Surface Area of a Cylinder

The curved surface area of a cylinder (CSA) is defined as the area of the curved surface of any given cylinder having base radius 'r' and height 'h'. The formula for the curved surface area is 2πrh.

### Surface Area of a Right Circular Cylinder

The total surface area of a right circular cylinder can be calculated using the formula: Total Surface Area (TSA) = 2πrh + 2πr² = 2πr(h+r) square units.

### Effect of Doubling Radius and Height on Surface Area and Volume

When the radius and height of a cylinder are doubled, the surface area increases by a factor of 4, and the volume increases by a factor of 8.

### Formula for Volume of a Cylinder

The formula for the volume of a cylinder is V = πr²h, where 'r' is the radius and 'h' is the height of the cylinder.

### Slant Height of a Cylinder

The slant height of a cylinder can be calculated using the formula a² + b² = c², where 'a' is the altitude, 'b' is the distance from the center of the base to the point where the slant height segment starts, and 'c' is the slant height.

### Calculating Surface Area

Surface area is the sum of the areas of all faces (or surfaces) on a 3D shape. For a cylinder, the surface area can be calculated by adding the areas of the two circular bases and the curved surface area.

### Finding Base Radius of a Cylinder

To find the base radius of a cylinder, you need to know the surface area and height of the cylinder. The formula for the surface area of a cylinder can be rearranged to solve for the radius.

### Cavalieri's Principle

Cavalieri's Principle states that if two solids have the same height and equal cross-sectional areas, then they have the same volume. This principle applies to cylinders, whether they are right or oblique.

### Multiple Choice Questions

1. What is the formula for the surface area of an open-top cylinder?
a) πr(2h + r)
b) 2πr(h + r)
c) πr²h
d) 2πrh + 2πr²

Answer: a) πr(2h + r)

2. What happens to the surface area of a cylinder when the radius is doubled?
a) It increases by a factor of 2
b) It increases by a factor of 4
c) It decreases by a factor of 2
d) It remains the same

Answer: b) It increases by a factor of 4

3. What is the formula for the curved surface area of a cylinder?
a) 2πrh
b) 2πr²
c) πr²h
d) 2πr(h + r)

Answer: a) 2πrh


A free response question for AP Physics B & C aspirants involving magnetic fields was posted on February 11, 2008. The question consists of four parts:

(a) An electron enters a non-uniform static magnetic field and comes out following an irregular path. Would its final speed be equal to the initial speed if it suffered no collisions during its motion in the magnetic field? Justify your answer.

The magnetic force on a moving charge is always perpendicular to the magnetic field, so no work is done by the field. The kinetic energy and speed of the charged particle remain unchanged.

(b) In a cloud chamber photograph, two circular tracks emerging from a common point can be seen. The tracks curve in opposite directions in a plane normal to the magnetic field. However, in a liquid hydrogen bubble chamber, two spiral tracks are seen. Explain why.

In a liquid, charged particles lose energy by collision to a much greater extent than in a gas, causing greater ionization. The velocity of the charged particle decreases, and the radius of the circular path decreases according to the expression r = mv/qB. Instead of a circular path, a spiral path is followed.

(c) Explain why a charged particle projected into a uniform magnetic field at an arbitrary angle θ follows a helical path. Derive an expression for the pitch of the helix.

When a charged particle is projected with velocity v into a magnetic field at angle θ, the velocity component v cosθ moves the particle forward, while the velocity component v sinθ makes the particle move along a circle of radius r = (mv sinθ)/qB. The charged particle moves along a helical path. The pitch of the helix is obtained by multiplying the forward velocity v cosθ with the period T of the circular component of motion, given by T = 2πm/qB.

(d) A positively charged particle of mass m and charge q travels with constant velocity v in the positive X-direction and enters a uniform magnetic field B directed along the negative Z-direction. Derive an expression for the minimum velocity required for the particle to cross the magnetic field.

The charged particle will move along a semicircular path just before crossing the magnetic field. The radius of the semicircle is x2 – x1. The limiting velocity v is given by mv^2/r = qvB, where r = x2 – x1. Therefore, v = qB(x2 – x1)/m.


# Bayesian Model Choice for the Poisson Model

In the Bayesian Case Studies course, we explored Bayesian model choice and basic Monte Carlo methods using data on the number of oral questions asked by French deputies. We aimed to determine whether male and female deputies behave differently in this matter.

## Modeling

We modeled the number of questions asked by a deputy using a Poisson random variable and denoted their gender (1 for men, 2 for women). Two models were considered:

- **Model 1**: A single parameter is used to model the behavior of men and women.
- **Model 2**: Separate parameters are used to model the behavior of men and women.

## Data Analysis

First, we downloaded the data and set some variables:
```r
deputes = read.csv2('http://www.ceremade.dauphine.fr/~ryder/CaseStudies/deputes.csv')
attach(deputes)
summary(questions_orales)
hist(questions_orales, breaks = (0:6 - 0.5))
n = length(questions_orales)
nh = sum(sexe == "H")
nf = sum(sexe == "F")
qtot = sum(questions_orales)
qh = sum(questions_orales[sexe == "H"])
qf = sum(questions_orales[sexe == "F"])
```

## Prior and Posterior Distributions

Since the Gamma prior is a conjugate prior for the Poisson model, we obtained a Gamma distribution as our posterior. Using a Gamma prior with parameters `a` and `b`, we compared the prior and posteriors for the two models.

## Bayes Factor

To compare the two models, we computed a Bayes factor. The Bayes factor is available in closed form, but an initial computation did not work due to large values. However, we were able to compute the Bayes factor analytically by using the log-scale instead:
```r
BF_analytical2 = function(a, b, y1, n1, y2, n2) {
  m1 = a * log(b) - lgamma(a) + lgamma(a + y1 + y2) - (a + y1 + y2) * log(b + n1 + n2)
  m2 = 2 * a * log(b) - 2 * lgamma(a) + lgamma(a + y1) - (a + y1) * log(b + n1) + lgamma(a + y2) - (a + y2) * log(b + n2)
  return(exp(m2 - m1))
}
BF_analytical2(2, 2, qh, nh, qf, nf)
```
Using Jeffrey's scale of evidence, this corresponds to "substantial" evidence in favor of Model 1.

## Monte Carlo Methods

We also approximated the Bayes factor using Monte Carlo methods, including vanilla Monte Carlo, the harmonic mean method, and importance sampling.

### Vanilla Monte Carlo
```r
lkd.model1 = function(y, n, lambda) {
  return(exp(-n * lambda + y * log(lambda)))
}
lkd.model2 = function(y1, n1, y2, n2, lambda1, lambda2) {
  return(exp(-n1 * lambda1 + y1 * log(lambda1) - n2 * lambda2 + y2 * log(lambda2)))
}
BF_MC = function(a, b, y1, n1, y2, n2, M) {
  lambda1 = rgamma(M, a, b)
  m1 = cumsum(lkd.model1(y1 + y2, n1 + n2, lambda1)) / (1:M)
  lambda2.1 = rgamma(M, a, b)
  lambda2.2 = rgamma(M, a, b)
  m2 = cumsum(lkd.model2(y1, n1, y2, n2, lambda2.1, lambda2.2)) / (1:M)
  return(m2 / m1)
}
M = 100000
BF_MC(2, 2, qh, nh, qf, nf, M)[M]
```

### Harmonic Mean Method
```r
BF_HM = function(a, b, y1, n1, y2, n2, M) {
  lambda1 = rgamma(M, a + y1 + y2, b + n1 + n2)
  m1 = (1:M) / cumsum(1 / lkd.model1(y1 + y2, n1 + n2, lambda1))
  lambda2.1 = rgamma(M, a + y1, b + n1)
  lambda2.2 = rgamma(M, a + y2, b + n2)
  m2 = (1:M) / cumsum(1 / lkd.model2(y1, n1, y2, n2, lambda2.1, lambda2.2))
  return(m2 / m1)
}
BF_HM(2, 2, qh, nh, qf, nf, M)[M]
```

### Importance Sampling
```r
BF_IS = function(a, b, y1, n1, y2, n2, M) {
  mean1 = (a + y1 + y2) / (b + n1 + n2)
  mean2.1 = (a + y1) / (b + n1)
  mean2.2 = (a + y2) / (b + n2)
  sigma1 = sqrt((a + y1 + y2) / (b + n1 + n2)^2)
  sigma2.1 = sqrt((a + y1) / (b + n1)^2)
  sigma2.2 = sqrt((a + y2) / (b + n2)^2)
  lambda1 = rnorm(M, mean1, sigma1)
  m1 = cumsum(lkd.model1(y1 + y2, n1 + n2, lambda1) * dgamma(lambda1, a, b) / dnorm(lambda1, mean1, sigma1)) / (1:M)
  lambda2.1 = rnorm(M, mean2.1, sigma2.1)
  lambda2.2 = rnorm(M, mean2.2, sigma2.2)
  m2 = cumsum(lkd.model2(y1, n1, y2, n2, lambda2.1, lambda2.2) * dgamma(lambda2.1, a, b) * dgamma(lambda2.2, a, b) / (dnorm(lambda2.1, mean2.1, sigma2.1) * dnorm(lambda2.2, mean2.2, sigma2.2))) / (1:M)
  return(m2 / m1)
}
BF_IS(2, 2, qh, nh, qf, nf, M)[M]
```
Importance sampling converged and provided a more precise estimate than vanilla Monte Carlo.

With this computation of the Bayes factor, we chose Model 1, indicating that a single parameter can be used to model the behavior of men and women.


The number 246 is even, as it is divisible by 2: 246/2 = 123.

The factors of 246 are all the numbers between -246 and 246 that divide 246 without leaving any remainder. These factors include:
-246, -123, -82, -41, -6, -3, -2, -1, 1, 2, 3, 6, 41, 82, 123, and 246.

Multiples of 246 are all integers divisible by 246, with a remainder of zero. There are infinite multiples of 246, including:
0 (since 0 × 246 = 0) and
246 (since 246 is divisible by 246, with 246 / 246 = 1 and a remainder of zero).

The number 246 is not a prime number. To determine primality, various algorithms can be used, such as:
- Trying all divisors below the number (up to its square root, approximately 15.684 for 246)
- The Eratosthenes screen
- The Atkin screen
- Probabilistic tests
- The cyclotomic test

Previous numbers: 244, 245
Previous prime number: 241
Next prime number: 251


To solve the equation 3x^2 + 5x + 2 = 0, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a, where a = 3, b = 5, and c = 2.

First, we calculate the discriminant: b^2 - 4ac = 5^2 - 4*3*2 = 25 - 24 = 1.

Since the discriminant is positive, the equation has two distinct solutions. Plugging the values into the quadratic formula, we get: x = (-5 ± √1) / (2*3) = (-5 ± 1) / 6.

Simplifying, we get two possible solutions: x = (-5 + 1) / 6 = -4/6 = -2/3 or x = (-5 - 1) / 6 = -6/6 = -1.

Therefore, the solutions to the equation 3x^2 + 5x + 2 = 0 are x = -2/3 and x = -1.

In general, when solving quadratic equations of the form ax^2 + bx + c = 0, we can use the quadratic formula to find the solutions. The quadratic formula is: x = (-b ± √(b^2 - 4ac)) / 2a.

We can also factor the quadratic expression, if possible, to find the solutions. For example, the equation x^2 + 4x + 4 = 0 can be factored as (x + 2)(x + 2) = 0, which gives us the solution x = -2.

It's also important to note that some quadratic equations may have no real solutions. For example, the equation x^2 + 2x + 5 = 0 has no real solutions, since the discriminant is negative (b^2 - 4ac = 2^2 - 4*1*5 = 4 - 20 = -16).

In summary, to solve quadratic equations, we can use the quadratic formula or factor the quadratic expression, if possible. We should also be aware of the possibility of no real solutions.


We are given that $f(f(x)) = x$ for all $x$. For $x < 2$, $f(x) = ax + b$. Substituting into the first equation yields $f(ax + b) = x$. Since $f(ax + b) = 8 - 3(ax + b)$ when $ax + b \geq 2$, it must be that $ax + b \geq 2$, implying $x \geq -\frac{2}{a}$.

For $-\frac{2}{a} \leq x < 2$, $f(x) = ax + b$. Substituting into the first equation gives $(ax + b) + b = x$, simplifying to $ax + 2b = x$. Since $ax + b \geq 2$, $x$ must equal $2$. However, $f(2) = 8 - 3(2) = 2$, making this case impossible.

The only possible value of $x$ is $x = -\frac{2}{a}$. Here, $f(x) = f\left(-\frac{2}{a}\right) = 8 - 3\left(-\frac{2}{a}\right) = \frac{10}{a}$. Substituting into the first equation yields $\frac{10}{a} + b = -\frac{2}{a}$, simplifying to $a + b = -10$. Therefore, $a + b = -10$.


A classic one-line program for the Commodore 64 microcomputer is:
10 PRINT CHR$(205.5+RND(1)); GOTO 10.
This program, featured in the Commodore 64 User’s Guide (1982) under “Random Graphics,” generates a random maze that builds up on-screen.

A 309-page book by Montfort et al. (MIT Press, 2013) explores this program from various angles. The book, available as a free PDF download, was written by ten authors from digital media, art, literature, and computer science backgrounds.

Translations of the program exist in multiple languages, but a MATLAB version was lacking. Here is a MATLAB equivalent:
while 1, fprintf('%s\n',char(rand(1,80)+9585.5)); pause(.2), end.
The `pause` command slows down the printing and may require adjustment based on the computer.

Other interesting MATLAB one-liners include:
- Mike Croucher’s example: x=[-2:.001:2]; plot(x,(sqrt(cos(x)).*cos(200*x)+sqrt(abs(x))-0.7).*(4-x.*x).^0.01)
- A Mandelbrot set computation: [x,y]=ndgrid(-1.5:5e-3:1.5); c=x+1i*y; z=c; for k=1:50, z=z.^2+c; end, contourf(x,y,abs(z)<1e6)

Readers are invited to share additional MATLAB one-liners in the comments.


Cars in a Garage Task is a challenge to find all the ways to park cars in garages, familiar to students of all ages. This task has a From The Classroom section, which includes students' journal records and links to videos explaining their reasoning.

The task is the tip of a learning iceberg, with more to explore beyond the initial challenge. Possible investigations include using simple calculators to carry out multiplications in sequence, such as 2x1, 3x2x1, 4x3x2x1, and so on. This work can be shared around the class, highlighting the limits of technology when the calculator runs out of screen digit space. At which number of cars does this happen?

Older students can continue using a scientific calculator to explore further. What is the last multiplication string shown in decimal notation, and can students explain the new notation that appears? This investigation is also included in Maths At Home, with fresh context, purpose, and additional resources.

The task can be modified to become a whole-class investigation, modeling how a mathematician works. The whole-class lesson for this task is detailed in Maths300 Lesson 128, Cars in a Garage, which extends the investigation with software. At least three model cars per pair of students are needed, and Poly Plug can be used as a red, yellow, and blue car.

For more ideas and discussion, visit Maths300 Lesson 128, Cars in a Garage, which includes software, a class set of cutout cars and garages, and a visit to Cars in a Garage on Poly Plug & Tasks. The Cars in a Garage task is an integral part of Maths With Attitude, a set of hands-on learning kits available from Years 3-10, which structures the use of tasks and whole-class investigations into a week-by-week planner.


A two-sample independent *t*-test is used to determine if the mean of a normally distributed numerical outcome variable differs across two independent groups. For instance, it can be used to compare the mean GPA between freshman and senior college students.

**Hypotheses:**

* *H*_{o}: The population mean of one group equals the population mean of the other group, or μ_{1} = μ_{2}
* *H*_{A}: The population mean of one group does not equal the population mean of the other group, or μ_{1} ≠ μ_{2}
* Alternative directional hypothesis:
	+ *H*_{o}: The population mean of one group equals the population mean of the other group, or μ_{1} = μ_{2}
	+ *H*_{a}: The population mean of one group is greater than the population mean of the other group, or μ_{1} > μ_{2}

The test statistic is calculated by taking the difference in the two sample means and dividing by the estimated standard error, which can be either pooled or unpooled. The estimated standard error measures the variation in both groups.

**Relevant Equations:**

* Degrees of freedom: The smaller of *n*_{1} – 1 and *n*_{2} – 1, where *n* is the sample size for each group.

**Assumptions:**

1. Random samples
2. Independent observations
3. The population of each group is normally distributed
4. The population variances are equal

If the third assumption is not met, the Mann-Whitney *U*-Test can be used. If the fourth assumption is not met, Welch's Test is used with the unpooled estimated standard error.

**Examples:**

1. **Hand Calculation:** A study compared the mean number of hours slept for male and female students. The result was *t*(*df*=40)=1.73, which was more extreme than the critical value of 1.68, indicating that female students sleep more than male students on average.
2. **Excel 2016:** A tutorial used the Data Analysis ToolPak to determine if males and females differ in their level of happiness, assuming unequal variances (Welch's Test). The result was *t*(*df*=28)=2.29, *p*<0.05, indicating that female students are happier than male students on average.
3. **RStudio:** A study compared the mean number of hours slept for male and female students, assuming unequal variances (Welch's Test). The result was *t*(*df*=27)=-0.42, *p*>0.05, indicating no evidence of a difference in sleep amounts between male and female students on Tuesday nights.


A set of 8 non-linear equations is given, with 8 knowns and 8 unknowns. The system can be solved, but using the Matlab function `solve` results in an empty `sym 0 by 1` matrix.

The 8 knowns are defined as real symbols: `Lx`, `Ly`, `Mx`, `My`, `Nx`, `Ny`, `Ox`, `Oy`. The 8 unknowns are also defined as real symbols: `k11`, `k12`, `k13`, `k14`, `k21`, `k22`, `k23`, `k24`. A constant `cte` is defined as a real symbol.

The system of equations is defined as:
```matlab
equations = [Lx == k21 * k11 - k12 * k23,...
             Mx == - ( k22 * k12 ) / cte,...
             Ny == k21 * k13 + k23 * k14,...
             Ly == ( k21 * k12 ) / cte,...
             My == k11 * k22 + k12 * k24,...
             Nx == ( k21 * k14 ) / cte,...
             Ox == k24 * k14 - k22 * k13,...
             Oy == ( k22 * k14 ) / cte];
```
To solve the equations, the following approach can be used:
```matlab
sol1_6 = solve(equations(1:6), [k11 k12 k13 k21 k22 k23])
eqn2 = expand(subs(equations(7:end), sol1_6))
```
This approach solves the first 6 equations for `k11`, `k12`, `k13`, `k21`, `k22`, and `k23`, and then substitutes the solutions into the remaining 2 equations.

Alternatively, constraint equations can be used to give values to all of the `k` variables:
```matlab
constraint_equations = [k11 * k14 + k12 * k13 == 1; k12 * k24 + k22 * k23 == 1]
sol1_6 = solve(equations(1:6), [k11 k12 k13 k21 k22 k23])
sol4 = solve(constraint_equations, [k14 k24])
```
However, it is noted that the system of equations may not have a solution, even though it has the same number of unknowns as equations. The system is non-linear, and the parameter `cte` is not known as a numerical constant. Therefore, numerical tools like `vpasolve` cannot be applied, and the problem may not have a general solution in an algebraic form.


On Airplane Bandwidth and Latency
=================================
Having recently used in-flight Internet access, this discussion compares the bandwidth and latency of typical Internet connections to those associated with transporting data on an airplane via hard drives.

Let's compare a high-speed DSL connection to taking a moderately large hard drive on a plane for data transfer between San Francisco and Boston:
- Bandwidth:
  - DSL: 15 Mbit/s
  - Airplane: approximately 150 Mbit/s (calculated from 500 GB / 7 hours, converted to Mbit/s)
- Latency:
  - DSL: approximately 100 ms
  - Airplane: over 7 hours

For a larger comparison, consider OC-768 vs a Boeing 747-400F plane filled with 2 TB hard drives:
- Bandwidth:
  - OC-768: 38 Gbit/s
  - 747-400F: approximately 93 Tbit/s (calculated from 250,000 lb / 7 hours, with 2 TB per 1.7 lb, converted to Tbit/s)
- Latency:
  - OC-768: less than 100 ms
  - 747-400F: over 7 hours

In summary, hard drives on an airplane offer higher bandwidth but suffer from significantly higher latency compared to typical Internet connections. The choice between these options depends on the specific application and its requirements.


In teaching, several key concepts often go unlearned by JC students. To address this, a few essential points are highlighted below.

The MF26 formula booklet contains crucial formulas, including double angle, integration, and differentiation formulas. It is essential to review these formulas to understand how to apply them, rather than simply memorizing them.

Key mathematical concepts include understanding that 0 is neither positive nor negative, which is critical when encountering terms like nonnegative or nonpositive. Additionally, 0 is both an integer and a real number.

When solving limits, specific formulas yield straightforward results:  gives 1, and  gives 0. However, questions about the relative size of 0 and infinity can be tricky. For instance, comparing different types of 0 and infinity requires careful consideration.

In Complex Numbers, a common misconception is writing . This highlights a poor understanding of the underlying definitions.

Distinguishing between variables and constants is vital. For example, when performing integration with substitution, it is alarming that some students mistakenly consider 'u' to be a constant. The expression  can be rewritten as , and students should be comfortable identifying this.

Understanding these concepts is crucial for a strong foundation in mathematics. Reviewing basic probability and other fundamental topics can help solidify knowledge and address common misconceptions.


# Calculate Spring Constant

To determine the spring constant of an elastic cord, a student hangs the cord from a rod, attaches various weights, and measures the entire length of the cord for each weight. The data is recorded in a table.

However, the provided text does not contain the table or sufficient information to calculate the spring constant. Instead, it presents several unrelated problems.

## Related Problems

1. A 20 kg student goes down a slide with a 3 N frictional force opposing his movement, starting from 0 m/s. Find his velocity at the bottom.

2. A 60 kg student is launched from a spring-loaded canon with a spring constant of 200 N/kg. The spring is compressed by 3 m. Find the student's velocity when leaving the canon and after the spring decompresses to 2 m.

3. A rubber ball of mass $m$ is dropped from a cliff, subject to air drag $bv^2$, where $b$ is a constant drag coefficient and $v$ is the instantaneous speed. The drag coefficient $b$ depends on the cross-sectional area and air density but not on the ball's mass.

## Questions about the Rubber Ball

A. Draw and label the forces on the ball before it reaches terminal speed.

B. State whether the magnitude of the acceleration increases, decreases, or remains the same as the ball approaches terminal speed.

C. Write a differential equation for the instantaneous speed $v$ in terms of time $t$, given quantities, and fundamental constants.

D. Determine the terminal speed $v_t$ in terms of given quantities and fundamental constants.

E. Determine the energy dissipated by the drag force during the fall from height $h$ to terminal speed, in terms of given quantities and fundamental constants.

## Multiple Choice Question

A small sphere moves at a constant speed in a vertical circle. Which of the following quantities is changing?

A. i and ii only (kinetic energy and potential energy)
B. i and iii only (kinetic energy and momentum)
C. i and ii only (kinetic energy and potential energy)
D. ii only (potential energy)
E. iii only (momentum)
F. ii and iii only (potential energy and momentum)

Correct answer: A. i and ii only (kinetic energy and potential energy)


## A Quick Introduction to Numpy Argmax

The Numpy argmax function returns the index of the maximum value of a Numpy array. It's similar to the Numpy maximum function, but instead of returning the maximum value, it returns the index of the maximum value.

### Numpy Arrays Store Data

A Numpy array is a data structure that stores data in a grid format, typically storing numeric data. Each location in a Numpy array has a position, and in Python, this position is called the "index." Numpy array indexes start at 0, and index values can be used to identify or retrieve specific elements of an array.

### Numpy Argmax Identifies the Maximum Value and Returns the Associated Index

When using Numpy argmax, the function identifies the maximum value in the array and returns the index associated with the maximum value. The np.argmax function can be used along particular axes to retrieve the index of the maximum value for a particular array axis.

## The Syntax of np.argmax

The syntax of np.argmax is simple, with only three parameters: `a`, `axis`, and `out`. The `out` parameter is rarely used and won't be discussed here.

### The Parameters of np.argmax

* `a` (required): specifies the input array to operate on. This parameter can accept Numpy arrays or "array-like" objects, such as Python lists.
* `axis`: controls the axis along which to use argmax. If not used, argmax will flatten out the array and retrieve the index of the maxima of the flattened array.

## Examples of How to Use Numpy Argmax

### Example 1: Use argmax on a 1-Dimensional Array

Create a 1-dimensional Numpy array: `my_1d_array = np.array([1,2,3,100,5])`. Apply np.argmax: `np.argmax(a = my_1d_array)`. Output: `3`. The maximum value (100) is at index position 3.

### Example 2: Apply argmax to a 2-Dimensional Array

Create a 2-dimensional Numpy array: `my_2d_array = np.array([[100,2,3],[4,5,600]])`. Apply np.argmax: `np.argmax(my_2d_array)`. Output: `5`. The array is flattened, and the maximum value (600) is at index position 5.

### Example 3: Use argmax Along Axis 0

Apply np.argmax to a 2D array with `axis = 0`: `np.argmax(my_2d_array, axis = 0)`. Output: `array([0, 1, 1])`. Argmax identifies the maximum along axis 0 (downward) and returns the row index.

### Example 4: Use argmax Along Axis 1

Apply np.argmax to a 2D array with `axis = 1`: `np.argmax(my_2d_array, axis = 1)`. Output: `array([0, 2])`. Argmax identifies the maximum along axis 1 (horizontally) and returns the column index.

To understand Numpy argmax, it's essential to understand Numpy indexes and axes. If you're still confused, consider reading more about indexes and axes or joining a course to learn more about Numpy.


To solve a second-order ordinary differential equation (ODE) using MATLAB's ode45 routine, it is essential to understand the required inputs. The given ODE is:

2y'' + 3y' + 5y = 7 exp(-x), y(0) = 11, dy/dx(0) = 13

To use ode45, the equation must be converted into state variable form. By letting y' = z, the equation becomes:

y' = z, with initial condition y(0) = 11
2z' + 3z + 5y = 7 exp(-x), with initial condition z(0) = 13

Simplifying the second equation yields:

z' = (7 exp(-x) - 3z - 5y) / 2

As required by MATLAB, let y = y(1) and z = y(2). The equations then become:

dy(1)/dx = y(2), with initial condition y(1) at x = 0 being 11
dy(2)/dx = (7 exp(-x) - 3y(2) - 5y(1)) / 2, with initial condition y(2) at x = 0 being 13

These equations are implemented in a MATLAB function called odestate.m:

dy = zeros(2,1);
dy(1) = y(2);
dy(2) = (7*exp(-x) - 3*y(2) - 5*y(1)) / 2;

To solve the ODE, the inputs to ode45 are:

1. The function odestate
2. The output interval [0 17]
3. The initial conditions [11 13]

The outputs are:

1. X: an array of x values between 0 and 17
2. Y: a matrix with two columns, where the first column is y(x) and the second column is dy/dx(x)

The MATLAB code to solve the ODE is:

[X,Y] = ode45(@odestate, [0 17], [11 13]);


# When to use the geometric mean?

The geometric mean is a better choice than the arithmetic mean in certain situations.

- Consider a scenario where wealth increases by 30% one year and 20% the previous year. To calculate the average growth, let's denote the initial wealth as *w*. The current wealth is 1.3 * 1.2 * *w*. If *t* represents the average growth factor over two years, then the current wealth can be expressed as *t* * *t* * *w*. Setting *t* = 1.25 is incorrect. Instead, choosing *t* = sqrt(1.3 * 1.2) provides the correct solution.
- Another case where the geometric mean is suitable is when averaging non-comparable numbers, such as the time to build a data cube versus the average query time. If *a* and *b* are two numbers with *a* being much smaller than *b*, then the arithmetic mean (2* *a* + *b*)/2 is approximately equal to (*a* + *b*)/2. This approach is flawed because it gives the same average performance despite one component being significantly worse. In this case, computing sqrt(*a* * *b*) is more sensible.


# Enhancing Perlin Noise in Creative Coding

This article provides tips to make creative coding works with Perlin noise more interesting. It is intended for individuals who have experience with the `noise()` function in Processing or p5.js. If you are new to Perlin noise, refer to the "Come on! Feel the noise!" series for an introduction.

## Ordinary Creative Coding with Perlin Noise

To start, let's explore an ordinary creative coding example using Perlin noise. It is recommended to use HSB color mode, which allows for easy control of hue, saturation, and brightness values.

```javascript
colorMode(HSB, 360, 100, 100, 100);
let pHue = initHue + noise(n02) * hueStep;
let pSat = noise(10, n01, n02) * 100;
let pBri = noise(20, n01, n02) * 100;
fill(pHue % 360, pSat, pBri, 100);
```

A basic example using Perlin noise to generate a painting can be seen in the following code:

```javascript
const w = 570;
const h = 360;
const wDiv = 285;
const hDiv = 180;
const nStep = 0.005;

function setup() {
  createCanvas(w, h);
  colorMode(HSB, 360, 100, 100, 100);
  noLoop();
}

function draw() {
  let cellW = w / wDiv;
  let cellH = h / hDiv;
  let baseHue = random(360);
  background(0, 0, 0, 100);
  noStroke();
  for (let x = 0; x < w; x += cellW) {
    let nX = x * nStep;
    for (let y = 0; y < h; y += cellH) {
      let nY = y * nStep;
      let nValH = noise(10, nX, nY);
      let nValS = noise(20, nX, nY);
      let nValB = noise(30, nX, nY);
      let nHue = (baseHue + nValH * 240) % 360;
      let nSat = 30 + 60 * nValS;
      let nBri = 20 + 80 * nValB;
      fill(nHue, nSat, nBri, 100);
      rect(x, y, cellW, cellH);
    }
  }
}

function mouseClicked() {
  let dt = new Date();
  noiseSeed(dt.getTime());
  redraw();
}
```

The smoothness of the noise can be controlled by adjusting the `nStep` value. For example, using `const nStep = 0.002;` or `const nStep = 0.008;` will produce different effects.

## Adding Ingenuity to Parameters

To make the Perlin noise more interesting, we can introduce a new parameter `nP` and calculate it using the `noise()` function. This creates a nested noise effect.

```javascript
let nP = noise(nX, nY);
let nValH = noise(10 + nP, nX, nY);
let nValS = noise(30 + nP, nX, nY);
let nValB = noise(40 + nP, nX, nY);
```

Multiplying `nP` by a constant can amplify the effect:

```javascript
let nP = noise(nX, nY) * 5;
```

## Devising Parameter Calculations

We can create interesting paintings by changing the parameter calculations of the Perlin noise. For example:

```javascript
let nP = noise(nX + nY) * 5;
let nP = noise(nX * nY) * 5;
let nP = noise(nX * nX + nY * nY) * 5;
```

Using trigonometric functions as parameters can also produce unique effects:

```javascript
let nP = (cos(nX * 2) + sin(nY)) * 5;
let nP = cos(nX * sin(nY * 2.0)) * 3;
```

## Forming Shapes with Distance Functions

We can use distance functions to create shapes and patterns:

```javascript
let nP = sin(dist(x, y, w * 0.5, h * 0.5) / w * HALF_PI) * 5;
let d = dist(x, y, w * 0.5, h * 0.5);
let dx = cos(PI * d / w) * w;
let dy = sin(PI * d / h) * h;
let nP = dist(x, y, dx, dy) * nStep * 5;
```

More complex shapes can be created by combining distance functions with trigonometric functions:

```javascript
let dx = x;
let dy = (sin(x / w * PI * 3) * 0.2 * x / w + 0.5) * h;
let nP = dist(x, y, dx, dy) * nStep * 3;
let dx = constrain(x, 0, w * 0.7);
let dy = (sin(x / w * PI * 5) * 0.2 * (1 - x / w) + 0.5) * h;
let nP = dist(x, y, dx, dy) * nStep * 3;
```

By experimenting with different parameter calculations and functions, you can create unique and interesting creative coding works with Perlin noise.


First, objects in computer-generated movies are modeled as wire skeletons made up of simple polygons, such as triangles. These objects are then realistically shaded to mimic real light using a mathematical method called ray tracing. Ray tracing involves tracing a ray from the viewpoint to a facet and determining if it reflects off and intersects a light source.

The first step in creating a computer-generated movie is to create the characters and the world they live in. Each object is modeled as a surface made up of connected polygons, usually triangles, with the corners of each triangle stored as a point in three-dimensional space in the computer memory.

Once the scene is set and lit, mathematics brings the images to life by enabling the characters to move. One of the most basic ways an object can move is to rotate around a given axis and through a given angle. This movement is made simple by using complex numbers, which involve the number i, where i is the imaginary unit.

Complex numbers have a geometric interpretation, where they are associated with points in the two-dimensional plane. For example, the number 1+i corresponds to the point (1,1). Multiplication by complex numbers has a geometric description, which is rotation. However, the wire frames of characters are stored as a collection of points in three-dimensional space.

The mathematician Sir William Rowan Hamilton developed a way to represent three-dimensional rotations using quaternions. Quaternions are the most efficient way to rotate objects in three dimensions and are essential for creating realistic computer-generated images. Hamilton's discovery of quaternions was a significant breakthrough, although not everyone was immediately convinced of their value.

Despite the importance of mathematics in creating computer-generated movies, it is often overlooked. Many movies, such as those featuring complex animations and special effects, would not be possible without the geometry of ray tracing or quaternions spinning objects in space. The next time you watch a computer-generated movie, consider the crucial role that mathematics plays in creating the visuals.

Mathematics is a hidden star of the show, and its contributions should not be missed. The variety of creative talent that goes into making a successful movie is vast, including writers, directors, actors, costume designers, and prop builders. However, mathematics is an essential component of the movie-making process, and its impact should be recognized.

In conclusion, mathematics is a vital part of creating computer-generated movies. From ray tracing to quaternions, mathematical concepts are used to bring images to life and create realistic visuals. The importance of mathematics in movie-making should not be underestimated, and its contributions should be acknowledged and appreciated.


# Collection

Majka had a collection of calendars. She gave 15 calendars to Hanke, 6 calendars to Julke, and 10 calendars to Petke. After giving away these calendars, she had 77 calendars left. How many calendars did Majka have in her collection initially?

To solve this problem, we need to find the initial number of calendars Majka had. We know that she gave away a total of 15 + 6 + 10 = 31 calendars. Since she had 77 calendars left, the initial number of calendars can be calculated as 77 (remaining calendars) + 31 (calendars given away) = 108 calendars.

## Next similar math problems:

- Math test

A class has 26 students, and the teacher gave tests to 13 students in 6.5 minutes. How many students still haven't taken the test? If the teacher needs to give tests to 43 students, how long will it take?

To find the number of students who still haven't taken the test, we subtract the number of students who took the test (13) from the total number of students (26), which gives us 26 - 13 = 13 students.

As for the time it takes to give tests to 43 students, we first need to find the time it takes to give a test to one student. Since the teacher gave tests to 13 students in 6.5 minutes, the time per student is 6.5 minutes / 13 students = 0.5 minutes per student. For 43 students, the total time would be 0.5 minutes/student * 43 students = 21.5 minutes.

- Collection of stamps

Jano, Rado, and Fero have a collection of stamps in a ratio of 5:6:9. If two of them had 429 stamps together, how many stamps did their shared collection have?

Let's denote the number of stamps Jano, Rado, and Fero have as 5x, 6x, and 9x, respectively. Since two of them had 429 stamps together, we can set up the equation 5x + 6x = 429 (assuming Jano and Rado are the two). Solving for x, we get 11x = 429, so x = 39.

Now, we can find the total number of stamps in their shared collection by adding the stamps of all three: 5x + 6x + 9x = 20x. Substituting x = 39, we get 20 * 39 = 780 stamps.

- Speed of Slovakian trains

Rudolf took the train from 'Ostratice' to 'Horné Ozorovce'. The train timetable shows the train Os 5409 with the following schedule: Chynorany 15:17, Ostratice 15:23, Rybany 15:27, Dolné Naštice 15:31, Bánovce nad Bebravou 15:35.

This problem doesn't provide enough information to calculate the speed of the train, as it only gives the schedule without distances between the stations.

- Phone numbers

How many 7-digit telephone numbers can be compiled from the digits 0, 1, 2, ..., 8, 9, with no digit repeated?

Since the first digit of a phone number cannot be 0, we have 9 choices for the first digit (1-9). For the second digit, we have 9 choices (since one digit is already used), for the third digit we have 8 choices, and so on. Therefore, the total number of 7-digit phone numbers without repeated digits is 9 * 9 * 8 * 7 * 6 * 5 * 4 = 544,320.

- Chickens and rabbits

In a yard, there were chickens and rabbits with a total of 27 heads and 86 legs. How many chickens and rabbits were in the yard?

Let's denote the number of chickens as C and the number of rabbits as R. We know that chickens have 2 legs and rabbits have 4 legs. We can set up two equations based on the given information:

1. C + R = 27 (total heads)
2. 2C + 4R = 86 (total legs)

We can solve this system of equations by multiplying the first equation by 2 and subtracting it from the second equation:

2C + 2R = 54
2C + 4R = 86

Subtracting the first equation from the second, we get:
2R = 32
R = 16

Now, substituting R = 16 into the first equation, we get:
C + 16 = 27
C = 11

So, there were 11 chickens and 16 rabbits in the yard.

- Farmers

Farmers loaded a truck with fruit and vegetables: 10 boxes of 5 kg pears, 8 boxes of 6 kg plums, 7 boxes of 9 kg carrots, and 10 bags of 50 kg potatoes. How many kilograms of fruit and vegetables were loaded in total?

To find the total weight, we calculate the weight of each type of fruit and vegetable and add them together:
- Pears: 10 boxes * 5 kg/box = 50 kg
- Plums: 8 boxes * 6 kg/box = 48 kg
- Carrots: 7 boxes * 9 kg/box = 63 kg
- Potatoes: 10 bags * 50 kg/bag = 500 kg

Total weight = 50 kg + 48 kg + 63 kg + 500 kg = 661 kg

- Princess

Princess had some beads, but she lost 321 beads. Prince George found all but 83 beads on the first day and 68 beads on the second day. On the third day, he found the rest, but the princess had lost 76 more beads. How many beads did the princess have after the third day?

Initially, the princess had some beads, but we don't know the exact number. Let's denote the initial number of beads as x.

After losing 321 beads, she had x - 321 beads.

Prince George found all but 83 beads on the first day, which means he found 321 - 83 = 238 beads. So, after the first day, the princess had x - 321 + 238 = x - 83 beads.

On the second day, Prince George found 68 beads, making the total x - 83 + 68 = x - 15 beads.

On the third day, he found the rest of the lost beads (83 - 68 = 15 beads), but the princess lost 76 more beads. So, after the third day, the princess had x - 15 + 15 - 76 = x - 76 beads.

Since we don't know the initial number of beads (x), we cannot determine the exact number of beads the princess had after the third day.

- Pencils

Emil had 23 pencils, 18 of which were broken. Emily trimmed 9 pencils, broke 2, trimmed 4, and went to school. When she returned, she had broken 5 pencils. How many pencils did she trim at school?

Initially, Emil had 23 pencils, and 18 were broken, leaving 23 - 18 = 5 unbroken pencils.

Emily trimmed 9 pencils (which doesn't change the number of broken pencils), broke 2 (making 18 + 2 = 20 broken pencils), and trimmed 4 more (still not changing the number of broken pencils).

Before going to school, she had 20 broken pencils and 3 unbroken pencils (since 5 - 2 = 3).

When she returned, she had broken 5 pencils, which means she broke 2 more while at school (since she already had 20 broken pencils and now has 5 more broken, but we know 3 of those 5 were broken before she left, so only 2 additional pencils were broken at school).

However, the question asks how many pencils she trimmed at school, not how many she broke. Since the number of broken pencils increased by 2 at school (from 20 to 22), and we know she broke 5 in total after returning, but 3 of those were before she left for school, the information about breaking pencils at school doesn't directly answer the question about trimming.

Given the information provided, we can't directly calculate how many pencils she trimmed at school because the actions of trimming and breaking are mixed in the narrative, and the specific question about trimming at school isn't directly answered by the provided numbers.

- Beautiful alley

In an alley, 4 rows of 10 trees and 5 rows of 8 trees blossomed. Six of the trees have already blossomed. How many trees still bloom in the alley?

First, calculate the total number of trees:
- 4 rows * 10 trees/row = 40 trees
- 5 rows * 8 trees/row = 40 trees
Total trees = 40 + 40 = 80 trees

Since 6 trees have already blossomed, the number of trees that still bloom is 80 - 6 = 74 trees.

- Four prisms

The prism has dimensions a = 2.5 cm, b = 100 mm, and c = 12 cm. What is its volume?

First, convert all dimensions to the same unit (centimeters):
- a = 2.5 cm
- b = 100 mm = 10 cm
- c = 12 cm

The volume of a prism is calculated as V = a * b * c. So, the volume is:
V = 2.5 cm * 10 cm * 12 cm = 300 cm^3

- Theater

The theater has 19 seats in each row. Tickets for the first 10 rows cost 30 USD, and for the next rows, they cost 15 USD. The theater was completely sold out, and the revenue was 12255 USD. How many rows are in the theater?

Let's denote the number of rows as R. The first 10 rows are sold at 30 USD/seat, and the remaining rows are sold at 15 USD/seat.

Revenue from the first 10 rows = 10 rows * 19 seats/row * 30 USD/seat = 5700 USD

Revenue from the remaining rows = (R - 10) rows * 19 seats/row * 15 USD/seat

Total revenue = 5700 USD + (R - 10) * 19 * 15 USD

Given that the total revenue is 12255 USD, we can set up the equation:
5700 + (R - 10) * 19 * 15 = 12255

Simplifying the equation:
5700 + 285(R - 10) = 12255

Subtract 5700 from both sides:
285(R - 10) = 6555

Divide both sides by 285:
R - 10 = 23

Add 10 to both sides:
R = 33

So, there are 33 rows in the theater.

- Number train

The numbers 1, 2, 3, 4, 5, 6, 7, 8, and 9 traveled by train. The train had three cars, and each car carried just three numbers. No. 1 rode in the first carriage, and in the last carriage were all odd numbers. The conductor calculated the sum of the numbers in the first carriage. How many pupils still haven't taken the test is not relevant to this problem, as it seems to be a mix-up with another question.

Given the odd numbers are in the last carriage, and No. 1 is in the first carriage, the odd numbers (3, 5, 7, 9) must be in the last carriage. Since each carriage can only hold three numbers, and we know the first carriage has No. 1, we need to figure out the combinations that satisfy these conditions.

However, the problem statement seems incomplete or mixed with another question, as it doesn't provide enough information to calculate the sum of numbers in the first carriage or determine which numbers are in each carriage beyond the odd numbers being in the last carriage and No. 1 being in the first.

- Railways - golden parachutes

This text appears to be a commentary or opinion piece rather than a math problem.

- Eggs

Mother bought 39 domestic eggs from neighbors and got back 4.54 EUR from 10 EUR. How many euros will she get back from a 20 EUR banknote if she buys 94 eggs from a neighbor?

First, find out how much 1 egg costs:
She paid 10 EUR - 4.54 EUR = 5.46 EUR for 39 eggs.
So, the cost per egg = 5.46 EUR / 39 eggs = 0.14 EUR per egg.

For 94 eggs, the cost = 94 eggs * 0.14 EUR/egg = 13.16 EUR.

If she pays with a 20 EUR banknote, she will get back:
20 EUR - 13.16 EUR = 6.84 EUR.

- Wire

From a 53-meter long wire, 12.1 meters and then 13.1 meters were cut off. How many meters of wire are left?

Initially, the wire was 53 meters long.
After cutting 12.1 meters, 53 - 12.1 = 40.9 meters were left.
After cutting another 13.1 meters, 40.9 - 13.1 = 27.8 meters of wire are left.

- Piggy

A man bought a pig for 60 EUR, sold it for 75 EUR, bought it back for 90 EUR, and sold it again for 100 EUR. How many euros did he earn?

First sale: 75 EUR - 60 EUR = 15 EUR profit.
Second sale: 100 EUR - 90 EUR = 10 EUR profit.

Total profit = 15 EUR + 10 EUR = 25 EUR.

- Cycling

A cyclist drove for 4 hours at an average speed of 44 kilometers per hour and then for another 4 hours at a speed of 36 kilometers per hour. How many total kilometers did the cyclist drive?

Distance = speed * time.
For the first part: 44 km/h * 4 h = 176 km.
For the second part: 36 km/h * 4 h = 144 km.

Total distance = 176 km + 144 km = 320 km.


Finding the Greatest Common Factor (GCF) of Two Different Numbers

To find the GCF, it's essential to understand the concept and its relation to factoring trees. The GCF of two numbers is the largest whole number that divides evenly into both of them.

Consider the numbers 18 and 24. Using factoring trees, we can express them as:
18 = 2 x 3 x 3
24 = 2 x 2 x 2 x 3

Every factor of 18 is a combination of up to one 2 and two 3s, resulting in factors such as:
2, 3, 2 x 3 = 6, 3 x 3 = 9

For a number to be the GCF of 18 and 24, it must be a factor of 18, composed of some combination of 2s and 3s (with at most one 2 and at most two 3s). Additionally, the GCF must be a factor of 24, meaning it's composed of 2s and 3s, with no more than three 2s and no more than one 3.

Combining these conditions, the GCF must have only 2s and 3s as its factors, with no more than one 2 and no more than one 3. The possible choices for the GCF are:
2 (one 2, no 3s)
3 (no 2s, one 3)
2 x 3 = 6 (one 2, one 3)

The largest of these options, 6, is the greatest common factor.

To further understand the pattern, try applying this method to several other pairs of numbers.

Key points to remember:
- The GCF is the largest whole number that divides evenly into both numbers.
- Use factoring trees to express the numbers as products of prime factors.
- The GCF must be a factor of both numbers, composed of the common prime factors.
- The GCF is the product of the common prime factors, raised to the smallest power they appear in either factorization.


**Factorial Notation:**
Let $n$ be a positive integer. Then, factorial $n$, denoted $n!$ is defined as: $n! = n(n – 1)(n – 2) … 3.2.1$.
Examples:
- $0! = 1$.
- $4! = (4 x 3 x 2 x 1) = 24$.
- $5! = (5 x 4 x 3 x 2 x 1) = 120$.

**Permutations:**
The different arrangements of a given number of things by taking some or all at a time, are called permutations.
Examples:
- All permutations made with the letters $a$, $b$, $c$ by taking two at a time are $ab$, $ba$, $ac$, $ca$, $bc$, $cb$.
- All permutations made with the letters $a$, $b$, $c$ taking all at a time are $abc$, $acb$, $bac$, $bca$, $cab$, $cba$.

**Number of Permutations:**
The number of all permutations of $n$ things, taken $r$ at a time, is given by:
$^{n}P_{r}= n(n– 1)(n– 2) … (n–r+ 1) = \frac{n!}{(n–r)!}$.
Examples:
- $^{6}P_{2}= (6 x 5) = 30$.
- $^{7}P_{3}= (7 x 6 x 5) = 210$.
Corollary: The number of all permutations of $n$ things, taken all at a time = $n!$.

**An Important Result:**
If there are $n$ subjects of which $p_{1}$ are alike of one kind; $p_{2}$ are alike of another kind; $p_{3}$ are alike of third kind and so on and $p_{r}$ are alike of $r^{th}$ kind, such that $(p_{1}+p_{2}+ …+p_{r}) = n$. Then, the number of permutations of these $n$ objects is $= \frac{n!}{p_{1}!p_{2}!…..p_{r}!}$.

**Combinations:**
Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.
Examples:
- Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.
Note: AB and BA represent the same selection.
- All the combinations formed by taking two out of three letters $a$, $b$, $c$ are $ab$, $bc$, $ca$.
- The only combination that can be formed of three letters $a$, $b$, $c$ taken all at a time is $abc$.
- Various groups of 2 out of four persons A, B, C, D are: $AB, AC, AD, BC, BD, CD$.
Note that $ab$ and $ba$ are two different permutations but they represent the same combination.

**Number of Combinations:**
The number of all combinations of $n$ things, taken $r$ at a time is:
$^{n}C_{r}= \frac{n!}{r!(n–r)!}$.
Note: $^{n}C_{n}= 1$ and $^{n}C_{0}= 1$. $^{n}C_{r}=^{n}C_{(n – r)}$.
Examples:
i. $^{11}C_{4}=\frac{(11 x 10 x 9 x 8)}{(4 x 3 x 2 x 1)} = 330$.
ii. $^{16}C_{13}=^{16}C_{(16 – 13)}=^{16}C_{3}=\frac{16 x 15 x 14}{3 x 2 x 1} = 560$.


Probability binning is an algorithm used for partitioning data in the Chi Square test. The Probability Binning (PB) algorithm divides the distribution into bins of the same height, with each bin containing the same number of events. This approach differs from the standard binning algorithm, which selects bins of equal width. By using bins with equal event counts, the PB algorithm minimizes the maximum expected variance for the bins, allowing for a more accurate estimate of the probability of difference between test and control histograms.

The PB algorithm calculates a Chi Squared value based on differences in event counts between corresponding bins in test and control samples. The resulting X2 value ranges from 0 to 2 standard deviations, indicating no difference (0) or maximum difference (2) between the samples. However, the Chi Square value does not enable ranking of multiple samples by their differences. To address this limitation, the T(X) metric was developed to facilitate sample ranking.

When applying the Chi Squared test with multiple dimensions, the number of bins used in Probability Binning should be decreased due to the "curse of dimensionality." As the number of dimensions increases, the number of regions created (e.g., squares, cubes, hyperplanes) grows exponentially, while the number of data points remains constant. This leads to a point where most bins are empty, and differences become trivial (e.g., 1 event vs. none). To mitigate this issue, the number of bins is reduced, helping to maintain meaningful comparisons between samples.


# Areas Of A Series Paper Sizes

The area of each A paper size is defined by the A paper size standards, with A0 having an area of 1 square meter. However, due to rounding to the nearest millimeter, A0 is 841mm x 1189mm, resulting in an area of 0.999949 square meters. Theoretically, the area of a sheet of size An can be calculated as 1/2^n m^2.

## Table of Theoretical and Actual Areas of A Paper Sizes

| Size | Theoretical Area (sq m) | Actual Area (sq m) | Difference (sq m) | Difference (%) |
| --- | --- | --- | --- | --- |
| 4A0 | 4 | 3.999796 | 0.000204 | 0.0051% |
| 2A0 | 2 | 1.999898 | 0.000102 | 0.0051% |
| A0 | 1 | 0.999949 | 0.000051 | 0.0051% |
| A1 | 0.5 | 0.499554 | 0.000446 | 0.0892% |
| A2 | 0.25 | 0.24948 | 0.00052 | 0.208% |
| A3 | 0.125 | 0.12474 | 0.00026 | 0.208% |
| A4 | 0.0625 | 0.06237 | 0.00013 | 0.208% |
| A5 | 0.03125 | 0.03108 | 0.00017 | 0.544% |
| A6 | 0.015625 | 0.01554 | 0.000085 | 0.544% |
| A7 | 0.0078125 | 0.00777 | 0.0000425 | 0.544% |
| A8 | 0.00390625 | 0.003848 | 0.00005825 | 1.4912% |
| A9 | 0.001953125 | 0.001924 | 0.000029125 | 1.4912% |
| A10 | 0.0009765625 | 0.000962 | 0.0000145625 | 1.4912% |

The discrepancy between theoretical and actual areas increases at smaller sizes, with A8, A9, and A10 having a difference of approximately 1.5%. In contrast, larger sizes have significantly lower differences.

## A Series Paper Sizes Defined

The A series paper sizes are defined in ISO 216 by the following requirements:
- The length-to-width ratio is 1.4142.
- A0 has an area of 1 square meter.
- Each subsequent size A(n) is defined as A(n-1) cut in half parallel to its shorter sides.
- The standard length and width of each size are rounded to the nearest millimeter, as the root 2 aspect ratio does not always result in whole numbers.


# MATH 125- Final Exam Guide - Comprehensive Notes for the exam

## Document Summary

The equations of lines and planes can be summarized as:
- For a line: x = p1 + td1, y = p2 + td2
- For a plane: x = p1 + td1, y = p2 + td2, z = p3 + td3

Key points to note:
- When solving problems, all steps should be shown, including row reduction of matrices.
- There are often multiple methods to solve a problem; one common approach is provided here.

Example Problem: Compute the determinant of a matrix Q.
To solve this, one approach is to expand along the first column.

Additional concepts to review:
- Reduced row echelon form of an augmented matrix for a homogeneous system Ax = 0, which should equal [b|0].

Remember, it's crucial to demonstrate all work when solving problems on the exam, as some steps may be skipped in provided solutions for brevity.


$9−4x_{2}x$ का समाकलन ज्ञात कीजिए।
$=∫2x dx$

यह प्रश्न समन्वय ज्यामिति विषय से संबंधित है, जो गणित की एक शाखा है। यह प्रश्न कक्षा 12 के स्तर का है।

इस प्रश्न का उत्तर वीडियो समाधान के रूप में दिया जा सकता है, जिसकी औसत अवधि 4 मिनट है। इस प्रश्न को 110 उपवोट मिले हैं।

अन्य प्रश्न जो छात्रों ने पूछे हैं:
**प्रश्न 1** (५,९४१ दृश्य)
**प्रश्न 2** (५,२८३ दृश्य)
**प्रश्न 3** (५,६०५ दृश्य)
**प्रश्न 4** (५,१६६ दृश्य)

यदि आपको इस प्रश्न या इसके उत्तर के बारे में कोई संदेह है, तो आप हमारे गणित ट्यूटर से ऑनलाइन जुड़ सकते हैं और इस प्रश्न का चरण-दर-चरण समाधान प्राप्त कर सकते हैं।


# Covering Transformations

## Definition
A covering transformation of a covering space $(\tilde{X}, p)$ of a topological space $X$ is a homeomorphism $f : \tilde{X} \to \tilde{X}$ such that the diagram commutes. The set of all covering transformations is denoted by $A(\tilde{X}, p)$ or $A(\tilde{X})$.

## Group Structure
The set $A(\tilde{X})$ forms a group under function composition. To see this, let $f, g \in A(\tilde{X})$. Then $f \circ g$ and $g \circ f$ are covering transformations since $(f \circ g) \circ p = f \circ (g \circ p) = f \circ p = p$ and $(g \circ f) \circ p = g \circ (f \circ p) = g \circ p = p$. The operation is associative, the identity function $\mathrm{id}_{\tilde{X}}$ is the identity element, and for each $f \in A(\tilde{X})$, the inverse $f^{-1}$ is a homeomorphism such that $f^{-1} \circ p = p$, so $f^{-1} \in A(\tilde{X})$.

## Fixed Points
If $f \in A(\tilde{X})$ and $f(x_0) = x_0$ for some $x_0 \in X$, then $f = \mathrm{id}_{\tilde{X}}$. This means no covering transformation apart from the identity has fixed points.

## Unique Lifts
For every $x \in X$ and $\tilde{x_1}, \tilde{x_2} \in p^{-1}(x)$, $p_*(\pi_1(\tilde{X}, \tilde{x_1})) = p_*(\pi_1(\tilde{X}, \tilde{x_2}))$ if and only if there exists a unique covering transformation $f$ such that $f(\tilde{x_1}) = \tilde{x_2}$. This is because if $p_*(\pi_1(\tilde{X}, \tilde{x_1})) = p_*(\pi_1(\tilde{X}, \tilde{x_2}))$, then there exist unique lifts $f : \tilde{X} \to \tilde{X}$ and $g : \tilde{X} \to \tilde{X}$ such that $f(\tilde{x_1}) = \tilde{x_2}$ and $g(\tilde{x_2}) = \tilde{x_1}$. By the fixed point property, $f \circ g = \mathrm{id}_{\tilde{X}} = g \circ f$, so $f \in A(\tilde{X})$. Conversely, if there exists $f \in A(\tilde{X})$ such that $f(\tilde{x_1}) = \tilde{x_2}$, then $f$ is a lift, and since $f$ is a homeomorphism, $f^{-1}$ is also a lift, implying $p_*(\pi_1(\tilde{X}, \tilde{x_1})) = p_*(\pi_1(\tilde{X}, \tilde{x_2}))$.


## Calculating Cumulative Sum in SAS

A **Cumulative Sum**, also known as the **Running Total**, is a sum of items that changes each time a new item is added. To calculate the cumulative sum in SAS, a new column is created that contains the cumulative sum.

### Calculate the Cumulative Sum

Given a dataset with daily revenue, the goal is to create a new column with the total revenue until that day. The first value of the cumulative sum is equal to the value in the first row. The `_N_` variable is used to check if the current row is the first row. After the first row, the value of the revenue column in the current row is added to the value of the cumulative revenue column of the previous row. The `RETAIN` statement is used to let SAS remember the value of the previous row.

```sas
data work.revenue;
  input date :date9. revenue;
  format date date9.;
  datalines;
  01jan2020 100
  02jan2020 120
  03jan2020 80
  04jan2020 50
;
run;

data work.cum_sum;
  set work.revenue;
  retain cumulative_revenue;
  if _n_ = 1 then cumulative_revenue = revenue;
  else cumulative_revenue = cumulative_revenue + revenue;
run;
```

### Calculate the Cumulative Sum per Group

To calculate the cumulative sum per group, a dataset with revenue data and groups is used. The goal is to calculate the cumulative revenue for each group.

#### Step 1: Order the Dataset

The dataset must be ordered by group and date using the `PROC SORT` procedure.

```sas
data work.revenue_by_group;
  input group $ date :date9. revenue;
  format date date9.;
  datalines;
  A 01jan2020 100
  B 01jan2020 50
  A 02jan2020 120
  B 02jan2020 60
  A 03jan2020 80
  B 03jan2020 60
  A 04jan2020 50
  B 04jan2020 100
;
run;

proc sort data=work.revenue_by_group out=work.revenue_by_group_srt;
  by group date;
run;
```

#### Step 2: Calculate the Cumulative Sum by Group

The cumulative sum is calculated using the `RETAIN` statement and `IF` statement. The `FIRST.variable` is used to determine the first row of each group.

```sas
data work.cum_sum_by_group;
  set work.revenue_by_group_srt;
  by group;
  retain cumulative_revenue;
  if first.group then cumulative_revenue = revenue;
  else cumulative_revenue = cumulative_revenue + revenue;
run;
```


# Fractions and Decimals – The Answers Revealed!

AQA's 2017 Maths GCSE question on fractions and decimals has been analyzed, revealing common misconceptions among students. The question is one of at least four multiple-choice, diagnostic questions on each GCSE Maths exam, both Higher and Foundation.

The most common incorrect answer, A, was predicted by teachers, but the reason behind it was surprising. Students' explanations showed that the biggest misconception was a misunderstanding of the question itself. Many students read the question as "which number lies between 3/4 and 0.5?" instead of the actual question.

Examples of students' explanations include:
* "4/6 simplifies to 2/3, which as a decimal is 0.66 recurring, and 3/4 as a decimal is 0.75, and 0.5, so 0.66 recurring is in the middle"
* "because a is 0.6 and it is smaller than 3/4, which is 0.75, and more than 0.5"
* "because 3/4 equals 0.75, 4/6 is equal to 0.6, and that is below 3/4 but more than 0.5"

Other incorrect answers revealed mistakes with place value and adding fractions, such as:
* B: "0.34 + 0.5 = 0.84"
* C: "because 3/4 is 0.75 and 1.205 is above 0.5 and 0.75"
* C: "0.50 add 0.75 = 1.205"

The key lesson from this analysis is that students will be challenged to combine skills together in the new GCSE. To succeed, students must be able to convert fractions to decimals, display competent knowledge of place value, and avoid making classic mistakes. They must stay focused until the end of the question and not rely on a single skill.


Dependency Pair Analysis involves analyzing the given rules to identify key dependency pairs. The rules provided are:

F(s(x),y) -> F(f(x,y),y)
F(s(x),y) -> F(x,y)
f(0,y) -> 0
f(s(x),y) -> f(f(x,y),y)

A Forward Instantiation Transformation can be applied to the rule F(s(x),y) -> F(x,y), resulting in no new dependency pairs. However, this transformation leads to one new DP problem.

The dependency pairs can be analyzed using Polynomial Ordering, where the ordering is defined as:
POL(s(x)) = 1 + x
POL(F(x), x) = x

Using this ordering, the dependency pair F(s(x),y) -> F(x,y) can be strictly oriented. There are no usable rules for innermost with respect to the implicit AFS that need to be oriented.

The analysis results in one new DP problem, and the dependency graph does not yield any new DP problems. The rules and transformations are as follows:

R
↳DPs
→DP Problem 1
↳FwdInst
→DP Problem 2
↳Polo
...
→DP Problem 3

The given rules and transformations are:
f(0,y) -> 0
f(s(x),y) -> f(f(x,y),y)
innermost

The duration of the analysis is 0:00 minutes.


**Geometry Fundamentals**

* A **point** is a location with no dimension, represented by a dot.
* A **line** is a one-dimensional figure that extends infinitely in both directions, named by two points on it.
* A **plane** is a two-dimensional figure that extends infinitely in both directions, named by three non-collinear points.

**Points and Lines**

* **Collinear points** are points that lie on the same line.
* **Coplanar points** are points that lie on the same plane.
* A **line segment** is a line with two endpoints.
* A **ray** is a line with one endpoint that extends infinitely in one direction.
* **Opposite rays** are rays that share a common endpoint and lie on the same line.

**Intersections and Segments**

* An **intersection** is a point or set of points shared by two or more lines, planes, or other geometric figures.
* Two planes always intersect at a **line**.
* **Segment addition** is the process of combining two segments to form a larger segment, e.g., AB + BC = AC if B is between A and C.

**Congruence and Midpoints**

* **Congruent** means equal in size or shape, but not necessarily in measurements.
* A **midpoint** is a point that divides a line segment into two congruent segments.
* A **segment bisector** is a line, ray, or plane that intersects a segment at its midpoint.

**Formulas and Distance**

* The **midpoint formula** is (average of x-coordinates, average of y-coordinates).
* The **distance formula** is d = √((x2 - x1)^2 + (y2 - y1)^2) or d^2 = (x2 - x1)^2 + (y2 - y1)^2.

**Angles**

* An **angle** is formed by two rays that share a common endpoint.
* The **vertex** of an angle is the point where the two rays intersect.
* There are five types of angles:
	+ **Acute angle**: 0° < m < 90°
	+ **Right angle**: m = 90°
	+ **Obtuse angle**: 90° < m < 180°
	+ **Straight angle**: m = 180°
	+ **Reflexive angle**: 180° < m < 360°

**Angle Relationships**

* The **Angle Addition Postulate** states that the sum of two adjacent angles is equal to the measure of the larger angle.
* An **angle bisector** is a ray that divides an angle into two congruent angles.
* **Complementary angles** add up to 90°, while **supplementary angles** add up to 180°.
* A **linear pair** is a pair of supplementary angles that form a straight line.
* **Vertical angles** are always congruent and formed by two intersecting lines.

**Polygons**

* A **polygon** is a closed figure formed by three or more line segments, with each side intersecting exactly two other sides.
* The **vertex** of a polygon is the endpoint of a side, and the vertices determine the name of the polygon.
* A **convex polygon** has sides that stick out, while a **concave polygon** has sides that curve inward.
* A **regular polygon** has congruent sides and angles, and is always convex.

**Pythagorean Theorem**

* The **Pythagorean Theorem** states that a^2 + b^2 = c^2, where a and b are the lengths of the legs of a right triangle, and c is the length of the hypotenuse.
* The **converse of the Pythagorean Theorem** states that if the sum of the squares of the lengths of two sides of a triangle is equal to the square of the length of the third side, then the triangle is a right triangle.


# Work and Kinetic Energy – Problems and Solutions

## 1. Net Work Done on a Car
A 5000-kg car accelerates from rest to 20 m/s. Determine the net work done on the car.

Given:
- Mass (m) = 5000 kg
- Initial speed (v₀) = 0 m/s
- Final speed (vₜ) = 20 m/s

Solution:
The work-kinetic energy principle states Wₙₑₜ = ΔEK = ½m(vₜ² - v₀²).
Wₙₑₜ = ½(5000)(20² - 0²) = 1000000 J.

## 2. Net Work Done on an Object
A 10-kg object accelerates from 5 m/s to 10 m/s. Determine the net work done on the object.

Given:
- Mass (m) = 10 kg
- Initial speed (v₀) = 5 m/s
- Final speed (vₜ) = 10 m/s

Solution:
Wₙₑₜ = ΔEK = ½m(vₜ² - v₀²) = ½(10)(10² - 5²) = 375 J.

## 3. Work Done on a Decelerating Car
A 2000-kg car decelerates from 10 m/s to 5 m/s. What is the work done on the car?

Given:
- Mass (m) = 2000 kg
- Initial speed (v₀) = 10 m/s
- Final speed (vₜ) = 5 m/s

Solution:
Wₙₑₜ = ΔEK = ½m(vₜ² - v₀²) = ½(2000)(5² - 10²) = -75000 J.

## 4. Work Done by a Constant Force
A 60-N constant force is exerted on a 10-kg object for 12 seconds. The initial velocity is 6 m/s.

Given:
- Force (F) = 60 N
- Time (t) = 12 seconds
- Mass (m) = 10 kg
- Initial velocity (v₀) = 6 m/s

Solution:
First, find the acceleration: a = F/m = 60/10 = 6 m/s².
Then, find the final velocity: vₜ = v₀ + at = 6 + 6*12 = 78 m/s.
The distance traveled: s = v₀t + ½at² = 6*12 + ½*6*12² = 504 m.
Work done: W = Fs = 60*504 = 30240 J.

## 5. Larger Work Done by an Object
Which object does larger work?

Solution:
Calculate work for each object using W = ½m(vₜ² - v₀²) and compare.

## 6. Work Done on a Decelerating Car
A 4000-kg car travels at 25 m/s and decelerates to 15 m/s. What is the work done on the car?

Given:
- Mass (m) = 4000 kg
- Initial speed (v₀) = 25 m/s
- Final speed (vₜ) = 15 m/s

Solution:
Wₙₑₜ = ΔEK = ½m(vₜ² - v₀²) = ½(4000)(15² - 25²) = -800000 J.

## 7. Kinetic Energy at a Given Height
A 0.1-kg ball is thrown horizontally at 6 m/s from a height of 5 meters. Find the kinetic energy at a height of 2 meters.

Given:
- Mass (m) = 0.1 kg
- Change in height (h) = 3 meters
- Acceleration due to gravity (g) = 10 m/s²

Solution:
The initial mechanical energy (PE) = mgh = 0.1*10*3 = 3 J.
This equals the final kinetic energy (KE) = 3 J.

## 8. Work Done by a Car
A 1000-kg car accelerates from rest to 5 m/s. What is the work done by the car?

Given:
- Mass (m) = 1000 kg
- Initial speed (v₀) = 0 m/s
- Final speed (vₜ) = 5 m/s

Solution:
Wₙₑₜ = ΔEK = ½m(vₜ² - v₀²) = ½(1000)(5² - 0²) = 12500 J.

## 9. Work Done by Weight
A 0.5-kg ball is thrown vertically upward with an initial velocity of 10 m/s. Find the work done by the weight force when the ball reaches its maximum height.

Given:
- Mass (m) = 0.5 kg
- Initial velocity (v₀) = 10 m/s
- Final velocity (vₜ) = 0 m/s (at maximum height)
- Acceleration due to gravity (g) = 10 m/s²

Solution:
Wₙₑₜ = ΔEK = ½m(vₜ² - v₀²) = ½(0.5)(0² - 10²) = -25 J.

## 10. Work Done on an Object in Free Fall
A 1-kg object falls from a height of 2.5 meters. Find the work done on the object.

Given:
- Mass (m) = 1 kg
- Initial velocity (v₀) = 0 m/s
- Height (h) = 2.5 meters
- Acceleration due to gravity (g) = 10 m/s²

Solution:
First, find the final velocity using vₜ² = 2gh = 2*10*2.5 = 50, so vₜ = √50.
Then, Wₙₑₜ = ΔEK = ½m(vₜ² - v₀²) = ½(1)(50 - 0) = 25 J.

## 11. Net Work on an Object
A 2-kg object travels at 20 m/s and accelerates to 40 m/s over a distance of 400 meters. Find the net work.

Given:
- Mass (m) = 2 kg
- Initial speed (v₀) = 20 m/s
- Final speed (vₜ) = 40 m/s

Solution:
Wₙₑₜ = ΔEK = ½m(vₜ² - v₀²) = ½(2)(40² - 20²) = 1200 J.

## 12. Final Velocity of an Object
A 2-kg object is acted upon by a force doing 21 J of work. If it starts from rest, find its final velocity.

Given:
- Mass (m) = 2 kg
- Initial velocity (v₀) = 0 m/s
- Work (W) = 21 J

Solution:
W = ΔEK = ½m(vₜ² - v₀²) = 21.
21 = ½(2)(vₜ² - 0²) => vₜ² = 21, so vₜ = √21.

## 13. Final Speed of an Object
An 8-N force acts on a 16-kg object initially at rest for 4 seconds. Find the final speed.

Given:
- Force (F) = 8 N
- Mass (m) = 16 kg
- Initial speed (v₀) = 0 m/s
- Time (t) = 4 seconds

Solution:
First, find acceleration: a = F/m = 8/16 = 0.5 m/s².
Then, find final velocity: vₜ = v₀ + at = 0 + 0.5*4 = 2 m/s.

## 14. Work Required to Double Speed
An object of mass 1 kg has its speed doubled from 1 m/s to 2 m/s. Find the work required.

Given:
- Mass (m) = 1 kg
- Initial speed (v₀) = 1 m/s
- Final speed (vₜ) = 2 m/s

Solution:
Initial KE = ½m(v₀²) = ½(1)(1²) = 0.5 J.
Final KE = ½m(vₜ²) = ½(1)(2²) = 2 J.
Work = ΔKE = Final KE - Initial KE = 2 - 0.5 = 1.5 J.

## 15. Work Required to Accelerate a Car
A 1500-kg car accelerates from 10 m/s to 20 m/s. Find the work required.

Given:
- Mass (m) = 1500 kg
- Initial speed (v₀) = 10 m/s
- Final speed (vₜ) = 20 m/s

Solution:
W = ΔEK = ½m(vₜ² - v₀²) = ½(1500)(20² - 10²) = 225000 J.

## 16. Total Work on an Object
A 2-kg object accelerates from 20 m/s to 40 m/s over 400 meters. Find the total work.

Given:
- Mass (m) = 2 kg
- Initial speed (v₀) = 20 m/s
- Final speed (vₜ) = 40 m/s

Solution:
Wₙₑₜ = ΔEK = ½m(vₜ² - v₀²) = ½(2)(40² - 20²) = 1200 J.

## 17. Kinetic Energy of a Bullet
A 10-gram bullet moves at 100 m/s. Find its kinetic energy.

Given:
- Mass (m) = 10 grams = 0.01 kg
- Speed (v) = 100 m/s

Solution:
KE = ½mv² = ½(0.01)(100²) = 50 J.


# Surface of Pyramid

In a regular quadrilateral pyramid, the height of the sidewall equals the length of the edge of the base. The content of the sidewall is 32 cm². To find the surface of the pyramid, we need to calculate the area of the base and the area of the four sidewalls.

## Key Concepts

- Regular quadrilateral pyramid: A pyramid with a square base and four triangular sidewalls.
- Base edge (a): The length of one side of the square base.
- Height of the sidewall (w): The height of one of the triangular sidewalls.
- Surface area of the pyramid: The sum of the areas of the base and the four sidewalls.

## Given Information

- Content of the sidewall = 32 cm²
- Height of the sidewall = length of the edge of the base

## Related Math Problems

1. **Quadrilateral Pyramid**: Given a regular quadrilateral pyramid with a base edge a = 10 cm and a height v = 7 cm, calculate:
   - Base content
   - Casing content
   - Pyramid surface
   - Volume of the pyramid

2. **Quadrilateral Pyramid**: Find the height and surface of a regular quadrilateral pyramid with a base edge a = 8 cm and a wall height w = 10 cm.

3. **Quadrangular Pyramid**: Given a regular quadrangular pyramid with a square base, body height = 30 cm, and volume V = 1000 cm³, calculate its side a and its surface area.

4. **Triangular Prism**: The base of the perpendicular triangular prism is a right triangle with a leg length of 5 cm. The content area of the largest sidewall of its surface is 130 cm², and the height of the body is 10 cm. Calculate its volume.

5. **The Roof**: The roof of the house has the shape of a regular quadrilateral pyramid, 5 m high, and the edge of the base is 7 m. How many tiles with a content of 540 cm² are needed?

6. **Calculate**: Calculate the surface of a regular quadrilateral prism whose base edge is 2.4 dm and the height of the prism is 38 cm.

7. **Regular Quadrilateral Pyramid**: What is the volume of a regular quadrilateral pyramid if its surface is 576 cm² and the base edge is 16 cm?

8. **Quadrilateral Pyramid**: A regular quadrilateral pyramid has a volume of 24 dm³ and a base edge a = 4 dm. Calculate:
   - Height of the pyramid
   - Sidewall height
   - Surface of the pyramid

9. **Quadrilateral Pyramid**: In a regular quadrilateral pyramid, the side edge is e = 7 dm, and the diagonal of the base is 50 cm. Calculate the pyramid shell area.

10. **Special Body**: Above each wall of a cube with an edge a = 30 cm, a regular quadrilateral pyramid with a height of 15 cm is constructed. Find the volume of the resulting body.

11. **Hexagonal Pyramid**: Calculate the volume and the surface of a regular hexagonal pyramid with a base edge length of 3 cm and a height of 5 cm.

12. **Tetrahedral Pyramid**: A regular tetrahedral pyramid is given. Base edge length a = 6.5 cm, side edge s = 7.5 cm. Calculate the volume and the area of its face (side area).

13. **Truncated Pyramid**: The concrete pedestal in the shape of a regular quadrilateral truncated pyramid has a height of 12 cm, the pedestal edges have lengths of 2.4 and 1.6 dm. Calculate the surface of the base.

14. **Regular Quadrilateral Pyramid**: What is the volume of a regular quadrilateral pyramid if the edge of the base is 8 cm long and the height of the side wall is 5 cm?

15. **Surface and Volume of Cuboid**: The content area of the square base of the cuboid is Sp = 36 cm², and its height is 80 mm. Determine its surface area and volume.

16. **Quadrilateral Pyramid**: The regular quadrilateral pyramid has a base edge a = 1.56 dm and a height h = 2.05 dm. Calculate:
    - The deviation angle of the sidewall plane from the base plane
    - Deviation angle of the side edge from the plane of the base

17. **Truncated Pyramid**: The truncated regular quadrilateral pyramid has a volume of 74 cm³, a height v = 6 cm, and an area of the lower base 15 cm² greater than the upper base's content. Calculate the area of the upper base.


## Section 23.1 Field Automorphisms

The set of all automorphisms of a field \(F\) is a group under composition of functions. An automorphism of a field is a bijective homomorphism from the field to itself.

### Proposition 23.1

The set of all automorphisms of a field \(F\) is a group under composition of functions.

### Proposition 23.2

Let \(E\) be a field extension of \(F\). The set of all automorphisms of \(E\) that fix \(F\) elementwise is a group.

### Definition of Galois Group

Let \(E\) be a field extension of \(F\). The Galois group of \(E\) over \(F\), denoted by \(G(E/F)\), is the group of automorphisms of \(E\) that fix \(F\) elementwise.

### Example 23.3

Complex conjugation is an automorphism of the complex numbers.

### Example 23.4

Consider the fields \({\mathbb Q} \subset {\mathbb Q}(\sqrt{5}\, ) \subset {\mathbb Q}( \sqrt{3}, \sqrt{5}\, )\). There exist automorphisms of \({\mathbb Q}(\sqrt{3}, \sqrt{5}\, )\) that leave \({\mathbb Q}( \sqrt{5}\, )\) and \({\mathbb Q}( \sqrt{3}\, )\) fixed.

### Proposition 23.5

Let \(E\) be a field extension of \(F\) and \(f(x)\) be a polynomial in \(F[x]\). Any automorphism in \(G(E/F)\) defines a permutation of the roots of \(f(x)\) that lie in \(E\).

### Proposition 23.6

If \(\alpha\) and \(\beta\) are conjugate over \(F\), there exists an isomorphism \(\sigma : F( \alpha ) \rightarrow F( \beta )\) such that \(\sigma\) is the identity when restricted to \(F\).

### Theorem 23.7

Let \(f(x)\) be a polynomial in \(F[x]\) and suppose that \(E\) is the splitting field for \(f(x)\) over \(F\). If \(f(x)\) has no repeated roots, then \(|G(E/F)| = [E:F]\).

### Corollary 23.9

Let \(F\) be a finite field with a finite extension \(E\) such that \([E:F]=k\). Then \(G(E/F)\) is cyclic of order \(k\).

### Example 23.10

The Galois group of \({\mathbb Q}( \sqrt{3}, \sqrt{5}\, )\) over \({\mathbb Q}\) is isomorphic to \({\mathbb Z}_2 \times {\mathbb Z}_2\).

### Example 23.11

The Galois group of \({\mathbb Q}(\omega)\) over \({\mathbb Q}\) is isomorphic to \({\mathbb Z}_4\), where \(\omega\) is a primitive fifth root of unity.

### Subsection Separable Extensions

A polynomial \(f(x) \in F[x]\) of degree \(n\) is separable if it has \(n\) distinct roots in its splitting field \(E\). An extension \(E\) of \(F\) is a separable extension of \(F\) if every element in \(E\) is the root of a separable polynomial in \(F[x]\).

### Proposition 23.12

Let \(f(x)\) be an irreducible polynomial over \(F\). If the characteristic of \(F\) is \(0\), then \(f(x)\) is separable. If the characteristic of \(F\) is \(p\) and \(f(x) \neq g(x^p)\) for some \(g(x)\) in \(F[x]\), then \(f(x)\) is also separable.

### Theorem 23.13: Primitive Element Theorem

Let \(E\) be a finite separable extension of a field \(F\). Then there exists an \(\alpha \in E\) such that \(E=F( \alpha )\).


## Finding the Energy Eigenstates of an N-Well System

To find the energy eigenstates of an n-well system, we start with the n x n Hamiltonian:

$$H \; \dot{=} \left[\begin{array}{ccccc} \alpha & \beta & 0 & 0 & \dots\\ \beta & \alpha & \beta & 0 & \\ 0 & \beta & \alpha & \beta & \ddots \\ 0 & 0 & \beta & \alpha & \ddots \\ \vdots & & \ddots & \ddots & \ddots \\ \end{array}\right] \; \; . $$

Our goal is to find the possible eigenstates of the electron in the potential landscape and the energy of each eigenstate. We use the energy eigenvalue equation:

$$H\vert \psi \rangle = E \vert \psi \rangle \; \; , $$

where $\vert \psi \rangle$ is a superposition of all possible ground states:

$$\vert \psi \rangle \; = \; \sum_{i}^{n} \, c_{i}\, \vert \, i \, \rangle \; \dot{=} \; \left[\begin{array}{c} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array}\right] \; \; . $$

Let's consider an unknown state with $c_{i}=1$ for all entries as a potential eigenvector:

$$H \vert \psi \rangle = \left[\begin{array}{ccccc} \alpha & \beta & 0 & 0 & \dots\\ \beta & \alpha & \beta & 0 & \\ 0 & \beta & \alpha & \beta & \ddots \\ 0 & 0 & \beta & \alpha & \ddots \\ \vdots & & \ddots & \ddots & \ddots \\ \end{array}\right] \left[\begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ \vdots \\ \end{array}\right] = E\left[\begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ \vdots \\ \end{array}\right] \; \; .$$

Performing the matrix algebra, we find that at the ends of the vector, $E=\alpha + \beta$, and for every other entry, $E=\alpha + 2\beta$.

To find the eigenenergies, we examine the general expression for the energy eigenvalue equation:

$$ \left[\begin{array}{ccccccc} \alpha c_{1}& \beta c_{2} & 0 & 0 & 0 & 0 & \dots\\ \beta c_{1}& \alpha c_{2}& \beta c_{3}& 0 & 0 & 0 & \\ 0 & \beta c_{2} & \alpha c_{3} & \beta c_{4} & 0 & 0 & \ddots \\ & & & \vdots & & & \\ 0 & 0 & 0 & \beta c_{p-1} & \alpha c_{p} & \beta c_{p+1} & \dots \\ & & & \vdots & & & \\ \end{array}\right] = \left[ \begin{array}{c} E c_{1} \\ E c_{2} \\ E c_{3} \\ \vdots \\ E c_{p} \\ \vdots \\ \end{array} \right] \; \; . $$

For any arbitrary $p$th coefficient (except for the ends), we have:

$$ E\, c_{p} \; = \; \beta\, c_{p-1} + \alpha\, c_{p} + \beta\, c_{p+1} \; \; . $$

Assuming the coefficients are described by an envelope function, we guess:

$$c_{p} \; = \; Ae^{ikpa} \; \; ,$$

where $k$ is the wave vector of the envelope function and $p$ is the position of the basis state. Inserting this into the energy expression yields:

$$E\, Ae^{ikpa} \; = \; \beta\, Ae^{ik(p-1)a} + \alpha\, Ae^{ikpa} + \beta\, Ae^{ik(p+1)a} \; \; , $$

$$ E \; = \alpha + \beta\left(e^{ika}+e^{-ika}\right) \; \; , $$

$$E \; = \; \alpha + 2\beta \cos{ka} \; \; . $$


## Introduction to Eureka Math Grade 3
The Eureka Math Book is a valuable resource for Grade 3 students to establish a strong foundation in mathematics. Developed by professional mathematics educators, the book provides concise solutions for easy understanding. To achieve high scores, students should solve all questions and exercises in the Eureka Grade 3 Book.

## Engage NY Eureka Math 3rd Grade Module 3 Lesson 8 Answer Key
This answer key supports teachers in educating students and helps students raise their scores. Learning activities are an effective way to teach elementary school kids basic mathematical concepts like addition, subtraction, multiplication, and division.

### Eureka Math - Grade 3 Module 3 Lesson 8 Pattern Sheet Answer Key
The pattern sheet focuses on multiplication, providing answers to various multiplication questions.

## Eureka Math Grade 3 Module 3 Lesson 8 Problem Set Answer Key
This section contains a set of problems to help students practice their mathematical skills.

1. Solve the following equations:
   - (12 – 4) + 6 = 14
   - 12 – (4 + 6) = 2
   - 5 = 15 – (7 + 3)
   - 11 = (15 – 7) + 3
   - 30 = (3 + 2) × 6
   - 15 = 3 + (2 × 6)
   - 4 × (7 – 2) = 20
   - (4 × 7) – 2 = 26
   - 10 = (12 ÷ 2) + 4
   - 2 = 12 ÷ (2 + 4)
   - 9 + (15 ÷ 3) = 14
   - (9 + 15) ÷ 3 = 8
   - 60 ÷ (10 – 4) = 10
   - (60 ÷ 10) – 4 = 2
   - 37 = 35 + (10 ÷ 5)
   - 9 = (35 + 10) ÷ 5

2. Use parentheses to make the equations true.

3. Explain how placing parentheses in the equation 24 ÷ 4 + 2 can make both answers true.

4. Solve the equation 12 + 15 ÷ 3 by placing parentheses.

5. Find two possible answers to the expression 7 + 3 × 2 by placing parentheses.

### Eureka Math Grade 3 Module 3 Lesson 8 Exit Ticket Answer Key
This section assesses students' understanding of the lesson.

1. Use parentheses to make the equations true:
   - 24 = 32 – 14 + 6
   - 12 = 32 – 14 + 6
   - 2 + 8 × 7 = 70
   - 2 + 8 × 7 = 58

2. Show how the position of parentheses in the equation 24 ÷ 6 + 2 can make both answers true.

### Eureka Math Grade 3 Module 3 Lesson 8 Homework Answer Key
This section provides additional practice for students.

1. Solve the following equations:
   - 9 – (6 + 3) = 0
   - (9 – 6) + 3 = 6
   - 8 = 14 – (4 + 2)
   - 12 = (14 – 4) + 2
   - 42 = (4 + 3) × 6
   - 22 = 4 + (3 × 6)
   - (18 ÷ 3) + 6 = 12
   - 18 ÷ (3 + 6) = 2

2. Use parentheses to make the equations true:
   - 14 – 8 + 2 = 4
   - 14 – 8 + 2 = 8
   - 2 + 4 × 7 = 30
   - 2 + 4 × 7 = 42
   - 12 = 18 ÷ 3 × 2
   - 3 = 18 ÷ 3 × 2
   - 5 = 50 ÷ 5 × 2
   - 20 = 50 ÷ 5 × 2

3. Determine if the equations are true or false:
   - (15 – 3) ÷ 2 = 6
   - (10 – 7) × 6 = 18
   - (35 – 7) ÷ 4 = 8
   - 28 = 4 × (20 – 13)
   - 35 = (22 – 8) ÷ 5

4. Explain why (3 × 6) ÷ 2 and 18 ÷ 2 are equal.

5. Place parentheses in the equation 4 × 7 – 3 to solve by finding the difference between 28 and 3.

6. Determine if Johnny's statement that 2 × 6 ÷ 3 equals 4 regardless of parentheses placement is true.


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## Mathematics Concepts

Students can transfer between fractions, decimals, and percents, and recognize equivalent forms of rational numbers. They should be able to solve problems with rational numbers using all four operations, write ratios and rates, and use a rate table to solve problems. Students must also be able to write and solve proportions, use multiple representations to display data, and identify variables in a problem situation.

## Problem-Solving Skills

Students develop and refine problem-solving skills by creating a problem-solving plan and implementing it systematically. They persevere through challenging problems, recall prior knowledge, and apply it to new situations. Students make connections between previous learning and real-world problems, communicate their approaches with precision, and articulate why their strategies and solutions are reasonable.

## Lesson Plans

In a two-day lesson, students work with a partner to create and implement a problem-solving plan based on mathematical concepts of rates, ratios, and proportionality. They analyze the relationship between different-sized gummy bears to solve problems involving size and price. Students use ratios, write and solve proportions, create rate tables, and use multiple representations to organize and communicate data.

## Coordinate Plane

Students draw a figure on the coordinate plane that matches a written description. They understand that ordered pairs name locations on the coordinate plane, with the first coordinate telling how many units to go left or right of the origin along the x-axis, and the second coordinate telling how many units to go up or down from the origin along the y-axis.

## Reflections on the Coordinate Plane

Students reflect a figure across one of the axes on the coordinate plane and name the vertices of the reflection. They identify a convention for naming the coordinates of the reflected figure, understanding that when point (m,n) is reflected across the y-axis, the reflected point is (−m,n), across the x-axis, the reflected point is (m,−n), and across the origin, the reflected point is (−m,−n).

## Clarifying Statements

Clarifying statements provide additional guidance for educators to communicate the intent of the standard, supporting the development of curricular resources and assessments aligned to the 2021 math standards. These statements can be in the form of succinct sentences or paragraphs, attending to student experiences, examples, boundaries, and connections to math practices.

## Educational Resources

Various educational resources are available, including videos from Cyberchase, where Digit must make a straight line between two points and follow the path created. Another video segment shows Jackie and Matt looking at the same object along two different lines. An activity using ArcGIS Online allows students to work with measurement and shapes using live web maps and images of the Earth at various scales.

## Key Concepts and Goals

Key concepts include understanding rational numbers, problem-solving skills, and the coordinate plane. Goals and learning objectives include creating and implementing a problem-solving plan, organizing and interpreting data, analyzing relationships between variables, and articulating strategies and solutions. Students must also be able to draw figures on the coordinate plane, reflect figures across axes, and name vertices of reflections.