## Computer Science

# Running Time of Binary Search

Binary search is an efficient algorithm for finding an item from a sorted list of items. It works by repeatedly dividing in half the portion of the list that could contain the item, until you've narrowed the possible locations to just one.

The key idea behind binary search is that when it makes an incorrect guess, the portion of the array that contains reasonable guesses is reduced by at least half. For instance, if the reasonable portion had 32 elements, an incorrect guess would cut it down to have at most 16 elements. This process of halving the size of the reasonable portion upon every incorrect guess is what makes binary search efficient.

To analyze the maximum number of guesses that binary search makes, let's consider an example. If we start with an array of length 8, then incorrect guesses reduce the size of the reasonable portion to 4, then 2, and then 1. Once the reasonable portion contains just one element, no further guesses occur; the guess for the 1-element portion is either correct or incorrect, and we're done. So, with an array of length 8, binary search needs at most four guesses.

This pattern holds true for arrays of any length. Every time we double the size of the array, we need at most one more guess. Suppose we need at most m guesses for an array of length n. Then, for an array of length 2n, the first guess cuts the reasonable portion of the array down to size n, and at most m guesses finish up, giving us a total of at most m + 1 guesses.

The number of guesses, in the worst case, can be expressed as "the number of times we can repeatedly halve, starting at n, until we get the value 1, plus one." This is equivalent to the base-2 logarithm of n, denoted as log2(n). The base-2 logarithm of a number is the power to which 2 must be raised to produce that number. For example, log2(8) = 3 because 2^3 = 8.

Here's a table showing the base-2 logarithms of various values of n:

| n | log2(n) |
| --- | --- |
| 1 | 0 |
| 2 | 1 |
| 4 | 2 |
| 8 | 3 |
| 16 | 4 |
| 32 | 5 |
| 64 | 6 |
| 128 | 7 |
| 256 | 8 |
| 512 | 9 |
| 1024 | 10 |
| 1,048,576 | 20 |
| 2,097,152 | 21 |

The logarithm function grows very slowly. Logarithms are the inverse of exponentials, which grow very rapidly. For example, because log2(128) = 7, we know that 2^7 = 128.

If n is a power of 2, then the runtime of a binary search algorithm on an array of length n is at most log2(n) + 1 guesses. For instance, if n is 128, binary search will require at most 8 (log2(128) + 1) guesses.

What if n isn't a power of 2? In that case, we can look at the closest lower power of 2. For an array whose length is 1000, the closest lower power of 2 is 512, which equals 2^9. We can thus estimate that log2(1000) is a number greater than 9 and less than 10. Using a calculator, we find that log2(1000) is approximately 9.97. Adding one to that yields about 10.97. In the case of a decimal number, we round down to find the actual number of guesses. Therefore, for a 1000-element array, binary search would require at most 10 guesses.

For the Tycho-2 star catalog with 2,539,913 stars, the closest lower power of 2 is 2^21 (which is 2,097,152), so we would need at most 22 guesses.

The reason we add 1 to the logarithm is that even when we have only one element, we still have to guess it, because it is possible that the target is not in the array. So, we need to add on the +1 for the last guess when we are down to the final remaining element.

In summary, the running time of a binary search is O(log2(n)), where n is the length of the array. The best case is O(1), and the average is O(log2(n)). The time complexity would be O(log2(n)) because we continue cutting the array into two until the target is found.

Note: The bases of the logarithms above are all two.

When implementing binary search, we use the guess variable as an index to access the array. The purpose of binary search is to find where the target is located (or to return -1 if it doesn't exist in the array). We guess that the target is located halfway between the minimum and maximum indices, and then we check the value at that index. If the target is greater than the value at the guess index, we can discard all indices less than or equal to the guess index. We repeat this process until we find the target or until the minimum index is greater than the maximum index.

In the case where n is not a power of 2, we can look at the closest lower power of 2. For example, if n is 1000, the closest lower power of 2 is 512, which equals 2^9. We can thus estimate that log2(1000) is a number greater than 9 and less than 10. Using a calculator, we find that log2(1000) is approximately 9.97. Adding one to that yields about 10.97. In the case of a decimal number, we round down to find the actual number of guesses. Therefore, for a 1000-element array, binary search would require at most 10 guesses.


## What can QuickMath do?

QuickMath is an advanced math solver that can expand, factor, or simplify expressions, split fractions into partial fractions, combine several fractions into one, and cancel common factors within a fraction. It can also solve equations or systems of equations, providing exact or numerical answers to almost any accuracy required. Additionally, QuickMath can solve inequalities or systems of inequalities for a single variable and plot inequalities in two variables.

The calculus section of QuickMath can carry out differentiation as well as definite and indefinite integration. It also contains commands for the arithmetic manipulation of matrices and can plot equations and inequalities. The numbers section has a percentages command for explaining common types of percentage problems and a section for dealing with scientific notation.

## Math Topics

Mathway is a math problem solver that can help with various math topics, including algebra, calculus, matrices, and more. It provides step-by-step solutions to math problems, allowing users to understand the reasoning behind the answer. Mathway can solve equations, inequalities, and systems of equations, as well as provide solutions to calculus problems, including differentiation and integration.

## Features of Mathway

Mathway has several features that make it a useful tool for students and mathematicians alike. It can solve a wide range of math problems, from basic algebra to advanced calculus. The app also provides step-by-step solutions, allowing users to understand the reasoning behind the answer. Additionally, Mathway has a graphing feature, which enables users to visualize equations and inequalities.

## Reviews of Mathway

Some users have praised Mathway for its ability to provide step-by-step solutions to math problems, making it a useful tool for understanding and learning math concepts. However, others have criticized the app for not providing enough explanation or for being too focused on providing answers rather than teaching math concepts.

## Alternative Math Solvers

There are several alternative math solvers available, including Photomath, Symbolab, and WebMath. These apps provide similar features to Mathway, including step-by-step solutions to math problems and graphing capabilities. Some users may find that one app works better for them than another, depending on their individual needs and preferences.

## Pricing of Mathway

Mathway offers both free and paid versions of its app. The free version provides limited access to features, while the paid version offers full access to all features, including step-by-step solutions and graphing capabilities. The paid version is available as a monthly or annual subscription, with prices starting at $9.99 per month.

## Online Math Solvers

There are several online math solvers available, including QuickMath, WebMath, and Symbolab. These websites provide free access to math solutions, including step-by-step explanations and graphing capabilities. Some online math solvers also offer additional features, such as homework help and tutoring services.

## Math Homework Help

Several websites offer math homework help, including Khan Academy, Pinchbeck, and the Scholastic Homework Club. These websites provide access to math lessons, practice problems, and homework help, making them useful resources for students who need extra support with their math homework.

## Photomath

Photomath is a math app that provides step-by-step solutions to math problems. It uses AI technology to recognize and solve math problems, providing users with a clear understanding of the solution. Photomath is available for download on both iOS and Android devices and offers a free version with limited features and a paid version with full access to all features.

## WebMath

WebMath is an online math solver that provides step-by-step solutions to math problems. It covers a wide range of math topics, including algebra, calculus, and geometry. WebMath is free to use and provides a useful resource for students who need help with their math homework.

## Symbolab

Symbolab is an online math solver that provides step-by-step solutions to math problems. It covers a wide range of math topics, including algebra, calculus, and geometry. Symbolab is free to use and provides a useful resource for students who need help with their math homework. It also offers a graphing feature, allowing users to visualize equations and inequalities.


To calculate the Annual Percentage Yield (APY) on a bank savings account, you need to know the interest rate and compounding frequency. The APY tells you how much you can expect to earn by the end of the year.

**Calculating APY by Hand**

1. Gather the necessary data: interest rate (r) and compounding frequency (n).
2. Use the APY formula: APY = (1 + r/n)^(n) - 1.
3. Apply the data and perform the calculation. For example, if the interest rate is 1% and the compounding frequency is quarterly, the APY would be: APY = (1 + 0.01/4)^(4) - 1 = 1.0038%.
4. Interpret the result. The increase in APY may seem small, but it can add up over time, especially with large amounts of money.
5. Try a different example. If the interest rate is 1% and the compounding frequency is daily, the APY would be: APY = (1 + 0.01/365)^(365) - 1 = 1.005%.

**Calculating APY Based on Actual Savings**

1. Collect information about your savings: interest earned, principal, and number of days.
2. Use the alternate APY formula: APY = (Interest / Principal) / (Days / 365) * 100.
3. Insert the data and perform the calculation. For example, if you earned $60 interest on a $5,000 savings account over 182.5 days, the APY would be: APY = ($60 / $5,000) / (182.5 / 365) * 100 = 1.005%.

**Using an Online APY Calculator**

1. Search the Internet for an APY calculator.
2. Enter the advertised percentage rate and compounding frequency.
3. Press the "Calculate" button to get the equivalent APY.

**Community Q&A**

* How much interest would I earn on a $500 investment on a 30-month CD with a 1.4 APY and a 1.393 interest rate? Just under $18.
* What is the account balance on 1.00 APY and interest earned of $5.84? If the APY is 1.00%, the principal would be $584.00.
* How much interest will I earn on $220,000 at 0.10 annual APY? At 0.10 percent APY, $220,000 earns an annual interest of $220.
* How much money can I make on a $50,000 investment? At 1.00% APY, you would earn $500 per year.
* What is the interest earned at 1.00% on $192,000? $1,920 simple interest annually.
* Is a higher APY better? Yes.
* How much interest would I earn on a 30,000 CD at 0.09 for 18 months? At 0.09% simple interest, you would earn $40.

**Tips**

* Banks are required by the federal Truth in Lending Act to provide APY information if you ask.
* You can use online APY calculators to simplify the calculation process.


8. Split Functions (Piecewise-defined functions)
Split functions, also known as piecewise-defined functions, are defined differently in different domains. These functions can have drastically different behaviors in different domains, making them noncontinuous and potentially nondifferentiable.

Example 1: Ordinary Function for Comparison
The function f(x) = −x^2 + 4 is a continuous function defined the same way for all values of x. To find the value of the function at a given x-value, simply substitute into f(x) = −x^2 + 4.

x  | 3  | 2  | 1  | 0  | 1  | 2  | 3
f(x) | 5 | 0 | 3 | 4 | 3 | 0 | 5

Example 2: Split Function
The function f(x) = {(2x+3, for x<1), (x^2+2, for x>=1)} is a split function. In the region x < 1, it has a straight line with slope 2 and y-intercept 3. As x approaches 1, the value of the function approaches 5. For x >= 1, the function is a parabola with value f(1) = 1.

Example 3: Graphing a Split Function
The function f(x) = {(2x-8, for x<2), (3x+2, for x>2)} is a split function. In the region x < 2, the function is defined as y = 2x - 8, approaching 4 as x gets closer to 2. In the region x > 2, the function is defined as y = 3x + 2, also approaching 4 as x approaches 2. The function is not defined at x = 2, making it discontinuous.

Example 4: Split Function with Three Pieces
The function f(x) = {(sin x, for x<2), (2x/2, for 2<=x<2), (x^2-8x+10, for x>=2)} is a split function defined in three ways. For x less than 2, the function is defined as sin x. Between 2 and 2, the function is defined as 2x/2 (straight line). For x greater than 2, the function is x^2 - 8x + 10 (parabola).

x  | 4  | 3  | 2  | 1  | 0  | 1  | 2  | 3  | 4
f(x) | 0.757 | 0.141 | 3 | 2.5 | 2 | 1.5 | 2 | 5 | 6

Example 5: Continuous Split Function
The function f(x) = {(x, for x<0), (1/5 sin 5x, for x>=0)} is a split function defined in two pieces. For negative values of x, the function is identical to f(x) = x (a straight line). For nonnegative values of x, the function is identical to f(x) = 1/5 sin 5x. The function is continuous and differentiable everywhere.

x  | 2  | 1  | 0  | 1  | 2
f(x) | 2 | 1 | 0 | 0.192 | 0.109

Special Notation: Absolute Value Function
The absolute value function f(x) = |x| is a split function defined in two pieces: f(x) = {(x, for x<0), (x, for x>=0)}. The function is continuous everywhere but only differentiable at nonzero values of x.

x  | 2  | 1  | 0  | 1  | 2
f(x) | 2 | 1 | 0 | 1 | 2

Example 7: Step Function
The function f(t) = {(1, for 3<=t<0), (1, for 0<=t<3)} is a step function, a type of split function used in signal analysis.


I have a board game that uses a 56-card deck with seven suits, each corresponding to one of the seven columns on the board. The goal is to conquer four columns by moving a token from the starting square to the terminal number in each column. The columns have different lengths: 8, 10, 12, 14, 12, 10, and 8 squares.

Each suit has a specific set of cards: one card with a value of 4, one card with a value of 3, two cards with a value of 2, and four cards with a value of 1. Players can only play cards of the suit corresponding to the column in play to advance their token. The hand size is ten cards.

The question is: what is the probability of drawing enough cards in the first hand to guarantee a victory on a column, specifically Column A, which requires a total value of 7? There are several combinations of cards that can add up to 7:

* A4 + A3
* A4 + A2 + A2
* A4 + A2 + A1
* A4 + A1 + A1 + A1
* A3 + A2 + A2
* A3 + A2 + A1 + A1
* A3 + A1 + A1 + A1 + A1
* A2 + A2 + A1 + A1
* A2 + A1 + A1 + A1 + A1

To calculate the probability, I started by finding the chance of drawing the A4 card (1 in 56) and then the A3 card (1 in 55), resulting in a probability of 1 in 3,080. However, since the hand size is 10, I'm unsure how to adjust this calculation to find the probability of drawing these cards in the first round.

The game is strategic with elements of luck, and I want to ensure that luck does not dominate strategy. Any help with calculating the probability of drawing a combination of cards that adds up to 7 in the first hand would be appreciated.


## Introduction

The density of air varies significantly with pressure and temperature. To calculate air density for applications such as drag force calculations, frictional loss, or air velocity determination, the ideal gas equation and ideal gas constant can be used. The ideal gas law can also estimate the density of other gases at known temperature and pressure.

## The Ideal Gas Law Equation

The ideal gas law is given by PV = nRT, where P is pressure, V is volume, T is temperature, and R is the ideal gas constant. By introducing density (ρ = m/V) and molecular weight (MW), the equation becomes ρ = (MW)P/RT. This equation calculates the density of an ideal gas in terms of its MW, pressure, and temperature.

ρ = (MW)P/RT

where:
- ρ = density of the gas (slugs/ft^3 or kg/m^3)
- MW = molecular weight of the gas (slugs/slugmole or kg/kgmole)
- P = absolute gas pressure (psia or pascals)
- T = absolute temperature of the gas (°R or °K)
- R = ideal gas constant (345.23 psia-ft^3/slugmole-°R or 8314.5 Joules/kgmole-K)

For air, MW = 29.

## Using the Ideal Gas Law for Air Density Calculations

Although air is a real gas, it can be treated as an ideal gas for many practical situations, allowing the use of the ideal gas law to calculate its density. The ideal gas law is accurate for a wide range of temperatures and pressures, except for very low temperatures or very high pressures.

## SI Units for the Ideal Gas Law

For SI units:
- ρ = density in kg/m^3
- P = absolute gas pressure in pascals (N/m^2)
- T = absolute temperature in °K
- R = ideal gas constant = 8314.5 Joules/kgmole-K

## Example Calculations

**Example 1**: Calculate the density of air at 75°F and 14.9 psia.
Solution: T = 75 + 459.67 = 534.25°R
ρ = (29)(14.9)/[(345.23)(534.25)] = **0.002343 slugs/ft^3**

**Example 2**: Calculate the density of air at 45°C and 100,000 pascals absolute.
Solution: T = 45 + 273.15 = 318.15 K
ρ = (29)(100,000)/(8314.5)(318.15) = **1.10 kg/m^3**

## References

For further information, refer to:
1. Bengtson, Harlan H., *Flow Measurement in Pipes and Ducts*
2. Munson, B. R., Young, D. F., & Okiishi, T. H., *Fundamentals of Fluid Mechanics*
3. Applied Thermodynamics ebook


If ${y}_{1}(x)$ and ${y}_{2}(x)$ are linearly independent solutions to a second-order, linear, homogeneous differential equation, then the general solution is given by $y(x) = {c}_{1}{y}_{1}(x) + {c}_{2}{y}_{2}(x)$, where ${c}_{1}$ and ${c}_{2}$ are constants. This theorem is powerful because it means that every expression of this form is a solution and every solution to the differential equation can be written in this form.

Given ${y}_{1}(t)={e}^{3t}$ and ${y}_{2}(t)={e}^{\mathrm{-3}t}$ as solutions to $y\text{\u2033}-9y=0$, the general solution is $y(t)={c}_{1}{e}^{3t}+{c}_{2}{e}^{\mathrm{-3}t}$ because ${y}_{1}$ and ${y}_{2}$ are linearly independent.

For ${y}_{1}(x)={e}^{3x}$ and ${y}_{2}(x)=x{e}^{3x}$ as solutions to $y\text{\u2033}-6{y}^{\prime}+9y=0$, the general solution is $y(x)={c}_{1}{e}^{3x}+{c}_{2}x{e}^{3x}$.

To solve second-order equations of the form $ay\text{\u2033}+b{y}^{\prime}+cy=0$, where $a, b,$ and $c$ are constants, we try a solution of the form $y(x)={e}^{\lambda x}$. Substituting $y(x)={e}^{\lambda x}$ into the equation yields the characteristic equation $a{\lambda}^{2}+b\lambda +c=0$.

The characteristic equation can be solved by factoring or using the quadratic formula $\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. This gives three cases: distinct real roots, a single repeated real root, or complex conjugate roots.

If the characteristic equation has distinct real roots ${\lambda}_{1}$ and ${\lambda}_{2}$, then ${e}^{{\lambda}_{1}x}$ and ${e}^{{\lambda}_{2}x}$ are linearly independent solutions, and the general solution is $y(x) = {c}_{1}{e}^{{\lambda}_{1}x} + {c}_{2}{e}^{{\lambda}_{2}x}$.

For example, the differential equation $y\text{\u2033}+9{y}^{\prime}+14y=0$ has the characteristic equation ${\lambda}^{2}+9\lambda +14=0$, which factors into $\left(\lambda +2\right)\left(\lambda +7\right)=0$. The roots are ${\lambda}_{1}=\mathrm{-2}$ and ${\lambda}_{2}=\mathrm{-7}$, so the general solution is $y(x) = {c}_{1}{e}^{-2x} + {c}_{2}{e}^{-7x}$.


Multiples of 5 are divisible by 5, including numbers like 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60. A key characteristic of these multiples is that they all have either 0 or 5 in the units place. This leads to a simple test for divisibility by 5: if a number ends in 0 or 5, it is divisible by 5.

To illustrate this concept, let's consider the numbers 365, 3250, 4132, 358, and 2535 to determine which ones are divisible by 5.

- The number 365 ends in 5, so it is divisible by 5.
- The number 3250 ends in 0, so it is divisible by 5.
- The number 4132 ends in 2, so it is not divisible by 5.
- The number 358 ends in 8, so it is not divisible by 5.
- The number 2535 ends in 5, so it is divisible by 5.

This exercise demonstrates how to apply the divisibility rule for 5 to quickly identify whether a number is divisible by 5 or not.


An equation $f(x) = 0$ can be rewritten as $g(x) = x$ and solved using fixed-point iteration to approximate the fixed point $r = g(r)$. The iteration starts with an initial guess $x_0$ and proceeds with $x_{n+1} = g(x_n)$ until a testing condition $e_{i+1} = \left|\:x_{i+1} - x_{i}\:\right| < \alpha$ is met or a fixed number of iterations $N$ is reached.

The choice of $g(x)$ is not unique, and the method may or may not converge. A criterion for convergence is that the error $e_i = \left|\:r - x_i \:\right|$ decreases for each iteration step. The convergence criterion is explained in Theorem 1: if $g$ is continuously differentiable, $g(r) = r$, and $S = \left|\:g'(r)\:\right| < 1$, then fixed-point iteration converges linearly with rate $S$ to the fixed point $r$ for initial guesses sufficiently close to $r$.

The convergence can be of higher order, as explained in Theorem 2: if $g$ is $p$ times continuously differentiable and the fixed-iteration converges to $r$ for some initial guess $x_0$, and $g'(r)=g''(r)=\cdots=g^{(p-1)}(r)=0$ and $g^{(p-1)}(r)=0$, then the order of convergence is $p$.

To illustrate the method, consider the equation $\frac{1}{2}\sin(x) - x + 1 = 0$. Using $g(x)= \sin(x)/2 + 1$, the fixed-point iteration can be applied to solve $g(x)=x$. The method converges since $|g'(x)|=\cos(x)/2 < 1$ for all initial guesses.

Another example is estimating the square root of a real and positive number $a$ by solving $f(x) = x^2-a^2=0$. A natural choice for $g(x)$ is $g(x) = \frac{1}{2}\left(\frac{a}{x} + x\right)$, which converges since $\left|\:g'(x)\:\right|_{x=\sqrt a} = 0 < 1$.

The fixed-point iteration method can be implemented with a testing condition $\left|\:x_{i+1} - x_i\:\right| < \alpha$ and a maximum number of iterations $N$. The method can be used to solve equations such as $x^4 + x - 1 = 0$ by rewriting it as $x = g(x)$ and applying the fixed-point iteration.

The Newton-Rhapson method is a special case of fixed-point iteration where the convergence rate is zero. The method estimates the root of a differentiable function $f(x)$ by iteratively calculating $x_{n+1} = x_n -\frac{f(x_n)}{f'(x_n)}$. The Newton iteration can be rewritten as a fixed point iteration $x_{n+1} = g(x_n)$, where $r = g(r) = r - \frac{f(r)}{f'(r)}$, and the fixed point $r$ is a root of $f$.

The fixed-point iteration method is simple to implement but only converges if the initial guess is sufficiently close to the root $r$. The convergence rate may be faster or slower than the Bisection Method, which has a convergence rate of 1/2. The method is locally convergent, and the choice of $g(x)$ affects the convergence rate.

Theorem 1 and Theorem 2 provide criteria for convergence and convergence rate, respectively. Theorem 1 states that if $g$ is continuously differentiable, $g(r) = r$, and $S = \left|\:g'(r)\:\right| < 1$, then fixed-point iteration converges linearly with rate $S$ to the fixed point $r$ for initial guesses sufficiently close to $r$. Theorem 2 states that if $g$ is $p$ times continuously differentiable and the fixed-iteration converges to $r$ for some initial guess $x_0$, and $g'(r)=g''(r)=\cdots=g^{(p-1)}(r)=0$ and $g^{(p-1)}(r)=0$, then the order of convergence is $p$.

In summary, the fixed-point iteration method is a simple and easy-to-implement method for solving equations $f(x) = 0$ by rewriting them as fixed-point problems $g(x) = x$. The method converges if the initial guess is sufficiently close to the root $r$ and the choice of $g(x)$ affects the convergence rate. The Newton-Rhapson method is a special case of fixed-point iteration with a convergence rate of zero.

**Exercise:** Prove Theorem 1 using the mean value theorem and Theorem 2 using Taylor's theorem. Verify that the convergence for $g_2$ is linear and that the convergence of $g_3$ is of second order by using the numerical results above.

**Multiple Choice Questions:**

1. What is the condition for convergence of the fixed-point iteration method?
a) $|g'(r)| < 1$
b) $|g'(r)| > 1$
c) $|g'(r)| = 1$
d) $|g'(r)| = 0$

Answer: a) $|g'(r)| < 1$

2. What is the order of convergence of the Newton-Rhapson method?
a) 1
b) 2
c) 3
d) 0

Answer: b) 2

3. What is the effect of the choice of $g(x)$ on the convergence rate of the fixed-point iteration method?
a) The choice of $g(x)$ does not affect the convergence rate.
b) The choice of $g(x)$ affects the convergence rate.
c) The choice of $g(x)$ only affects the convergence rate if the initial guess is close to the root.
d) The choice of $g(x)$ only affects the convergence rate if the initial guess is far from the root.

Answer: b) The choice of $g(x)$ affects the convergence rate.


To determine the required dimensions of a square beam, we first need to calculate the reactions $R_1$ and $R_2$. Given the equation $\Sigma M_{R2} = 0$, we can find $R_1$ using $3R_1 = 5(2) + 2$, which simplifies to $R_1 = 4 \, \text{ kN}$.

Next, using $\Sigma M_{R2} = 0$ again, we find $R_2$ with $3R_2 + 2 = 5(1)$, resulting in $R_2 = 1 \, \text{ kN}$.

For bending stress, where $b = d$, the formula is $f_b = \dfrac{6M}{bd^2}$. Substituting known values gives $8 = \dfrac{6(4)(1000^2)}{d^3}$. Solving for $d$ yields $d = 144.22 \, \text{ mm}$.

Considering shear stress with $b = d$, the formula is $f_v = \dfrac{3V}{2bd}$. Substituting gives $1 = \dfrac{3(4)(1000)}{2d^2}$. Solving for $d$ results in $d = 77.46 \, \text{ mm}$.

Given these calculations, the appropriate size for the square beam, considering both bending and shear stresses, would be **145 mm × 145 mm**.


Trigonometric functions are initially defined for acute angles in right triangles. For similar right triangles abc and a'b'c', the sine and cosine functions are defined as:
sin(A) = opposite side / hypotenuse
cos(A) = adjacent side / hypotenuse
Additionally, other trigonometric functions are defined, including:
tan(A) = opposite side / adjacent side
cot(A) = adjacent side / opposite side

These definitions are extended to make sine and cosine periodic with period 2π. The extensions follow these rules:
- The functions are defined for all angles, not just acute ones.
- The period of sine and cosine is 2π.

As a result, tangent and cotangent have a period of π. The properties of sine and cosine are:
- Sine is an odd function: sin(-A) = -sin(A)
- Cosine is an even function: cos(-A) = cos(A)
This makes tangent and cotangent odd functions.

Other trigonometric functions include:
- Versine: vers(A) = 1 - cos(A), used rarely
- Secant: sec(A) = 1 / cos(A)
- Cosecant: csc(A) = 1 / sin(A)

The derivatives of sine and cosine are fundamental in calculus:
- The derivative of sin(A) is cos(A)
- The derivative of cos(A) is -sin(A)


Explore exponential and logarithmic functions with easy-to-follow math lessons. Key topics include:

* Exponential functions
* Logarithmic functions
* Formula for percentage
* Compatible numbers
* Basic math test
* Basic math formulas
* Types of angles
* Math problem solver
* Algebra word problems
* Surface area of a cube
* Finding the average
* Scale drawings

Prepare for important exams such as K12 tests, GED math test, basic math tests, geometry tests, and algebra tests. Additionally, learn real-life math skills including:
* Investing money
* Budgeting money
* Paying taxes
* Mortgage loans
* Math involved in playing baseball


## Example of II TEST

### Exercise 1
Identify a Time Varying system: y(t) - α₁y(t-1) = β₁u(t-1) + β₂u(t-2) + e(t), where e(k) is measurement noise with zero mean and 0.01 variance, uncorrelated with u(k). Find the recursive algorithm to estimate parameters. The data changes in amplitude after 100 steps; indicate how to cope with this variation for rapid convergence. What kind of inputs will you use?

### Exercise 2
Variable x(k) evolves as x(k + 1) = x(k)² + w(k), where w(k) is a process disturbance with zero mean and unitary variance. The measurement sensor is cubic: y(k) = x³(k) + v(k), with v(k) a white noise disturbance and variance r, uncorrelated with w(k). Find a recursive algorithm to estimate x. Can a stationary algorithm be used? Does r influence the estimation error?

### Exercise 3
Given y(k) = b₀u(k - 1) + b₁u(k - 2) + e(k), a system with unknown parameters and e(k) a white noise with zero mean and variance 0.01. Run an identification experiment with a PRBS input of amplitude ±1 using 20 data samples.
1. Indicate the algorithm to estimate b₀ and b₁.
2. Validate the model: can the same set of data be used? What criteria are applied?
3. Provide MATLAB instructions to run the identification test.

### Exercise 4
Given the LTI, TD system with real parameters β and γ:
z₁(t + 1) = 0.2z₁(t) + 0.2z₂(t) + u(t)
z₂(t + 1) = 0.1z₂(t) + u(t)
z₃(t + 1) = 0.3z₃(t) + γu(t)
y(t) = βz₁(t) + z₃(t)
No specific questions are provided for this exercise.


Swift is a powerful and intuitive programming language developed by Apple, offering unique effectiveness for developers. A common problem in coding is determining whether a number is prime or not. A prime number is a natural number greater than 1 with no positive divisors other than 1 and itself.

The prime number problem can be solved in Swift by applying the right knowledge and techniques. Swift offers a rich library of mathematical functions and operators, including the modulus operator (%), which gives the remainder of a division operation.

The `isPrime` function takes an integer as input and returns a boolean indicating whether the number is prime or not. The function first checks if the number is less than 2, in which case it returns false. If the number passes this condition, the function sets up a for loop starting at 2 and lasting until just before the input number.

Within the loop, the function checks if the number can be divided evenly by the current loop index `i` using the modulus operator. If the number can be divided evenly, it means the number has another divisor besides 1 and itself, so the function returns false. If the function runs through the entire loop without finding another divisor, it means the number is prime, so it returns true.

Swift is not just used for app development, but also for coding algorithms due to its robust standard library and precise syntax, which increase code clarity and reduce errors. As an open-source programming language, Swift is constantly being improved by the community, leading to more performant and safer code. Swift's straightforward syntax and powerful computation capabilities make it an excellent choice for developing an efficient and reliable solution to the prime number problem.

In conclusion, Swift is a great programming language to master for both application development and problem-solving exercises. Developing these skills can enable you to become a top-notch Swift developer and significantly enhance your problem-solving abilities.

## Swift Libraries and Functions
Swift's mathematical functions and operators, such as the modulus operator, are essential for solving problems like the prime number problem.

## Step-by-Step Explanation of the Code
1. Input an integer into the `isPrime` function.
2. Check if the number is less than 2.
3. If the number is 2 or greater, set up a for loop starting at 2 and lasting until just before the input number.
4. Within the loop, check if the number can be divided evenly by the current loop index `i`.
5. If the number can be divided evenly, return false.
6. If the function runs through the entire loop without finding another divisor, return true.

## Advantages of Using Swift for Problem-Solving
* Robust standard library
* Precise syntax
* Open-source community
* Powerful computation capabilities
* Straightforward syntax

What is the main purpose of the `isPrime` function?
A) To determine if a number is even or odd
B) To determine if a number is prime or not
C) To calculate the square root of a number
D) To generate a random number

Answer: B) To determine if a number is prime or not


Equilateral Triangle and Perpendicular Line

The points S and T are located on the respective sides PQ and PR of an equilateral triangle PQR, where ST = TR and ST is perpendicular to PQ. Given that the length of QR is 1, we need to find the length of ST.

To solve this, we can start by finding the angles in the triangle. Since ST is perpendicular to PQ, angle STR is 150 degrees, and angle TSR = angle TRS = 15 degrees.

Using the properties of a right-angled triangle, we can find the length of QS. Since QRS is a right angle with one of the angles being 45 degrees, QS = SR.

Applying the sine rule, we get:

ST / sin(15) = SR / sin(150)

ST / sin(15) = SR / sin(30)

Since sin(30) = 0.5, we can simplify the equation. However, the exact form of sin(15) is not provided, and there is no square root of 2 in the answer choices.

The possible answers are:
A.
B.
C.
D.
E.

We need to find a simpler way to solve this problem and determine the correct answer.


The number 800 is divisible by 2, 5, and 10, but not by 3 or 9.

To calculate 800 divided by 2/3, we first find 2/3 of 800, which equals 800 * (2/3) = 533.33. Then, 800 divided by 2/3 is 800 / (2/3) = 800 * (3/2) = 1200.

The fraction 240/800 can be simplified to 3/10.

Some key calculations involving 800 include:
- 3% of $800 = $24
- 800 ÷ 12 = 66 remainder 8
- 2/5 - 3/10 = 1/10

In an algebraic equation, if Jim makes $320 and Alicia makes $480, and J = 2/3A, then we can solve for A: J + A = 800, 2/3A + A = 800, 5/3A = 800, A = 480, and J = 2/3 * 480 = 320.

The number 3,985,800 can be written in words as three million, nine hundred and eighty-five thousand, eight hundred. In numerical form, it is 3,000,000 + 900,000 + 80,000 + 5,000 + 800. In powers of 10, it is 3 x 10^6 + 9 x 10^5 + 8 x 10^4 + 5 x 10^3 + 8 x 10^2.

The number 19,888,800 in expanded form is 1 x 10^7 + 9 x 10^6 + 8 x 10^5 + 8 x 10^4 + 8 x 10^3 + 8 x 10^2.

To find 2/3 of 10, we calculate 10 * (2/3) = 6.66.

Solving the equation (3-2)(x-2) = 10 gives x = 12. Solving 3 - 2(x-2) = 10 gives x = -1.5.

The equation 1 + 2 + 3 + 4 = 10 has multiple solutions, including ((1 + 2) + 3) + 4 = 10 and (1 + 2) + (3 + 4) = 10.

The calculation 2/3 ÷ 3/10 equals 2/3 * 10/3 = 20/9 = 2 2/9.

The calculation 6 * 10^9 / (2 * 10^3) equals (6/2) * 10^(9-3) = 3 * 10^6 or 3 million.

The statement "10 - 3 + 2 = 2" is false; the correct calculation is 10 - 3 + 2 = 9.

The calculation 10 + 3 * 2 equals 10 + (3 * 2) = 10 + 6 = 16.

To increase 800 by 30%, we can multiply 800 by 0.3 to find 30% of 800, which equals 240, and then add 240 to 800 to get 1040. Alternatively, we can multiply 800 by 1.3 to directly get 1040.

Seven percent of 800 is not directly calculated, but 3.5% of 800 equals 0.035 * 800 = 28.

The LCM of 2, 3, 5, 9, and 10 is not directly stated, but the LCM of some numbers is given as 30.

The number 800 increased by 30% is 1040.

The calculation 2/3 ÷ 3/10 equals 20/9.

To find 100% of a number, we simply take the number itself. To find 30% of 800, we can multiply 800 by 0.3, which equals 240, and then add 240 to 800 to get 1040.

The answer to whether 800 is divisible by 2, 3, 5, 9, or 10 is yes for 2, 5, and 10, but no for 3 and 9.

Other questions and answers include:
- How old was Ralph Macchio in the first Karate Kid? (No answer provided)
- How many US congressmen are there? (No answer provided)
- How do you get 1,000,000 Robux for free? (No answer provided)
- Is 800 divisible by 2, 3, 5, 9, 10? (Yes for 2, 5, and 10, but no for 3 and 9)


A group of researchers hypothesizes that listening to music while studying for an exam influences test scores. They randomly selected 28 subjects and assigned them to one of four groups: Group 1 (no music), Group 2 (classical music), Group 3 (country music), and Group 4 (jazz music). All groups studied for a test one hour a day for a week. The final test scores are:

Group 1: 50, 45, 42, 41, 40, 39, 44
Group 2: 59, 57, 57, 55, 56, 53, 48
Group 3: 60, 49, 56, 55, 54, 51, 46
Group 4: 49, 45, 45, 42, 41, 39, 47

The most appropriate test for this data and hypothesis is the Analysis of Variance (ANOVA) test, also referred to as the F-test.

The null hypothesis is: μGroup1 = μGroup2 = μGroup3 = μGroup4.

To determine the observed test statistic, the F-test is used. The calculated value is 17.16.

The degrees of freedom for the test are calculated as follows:
- Degrees of freedom between (dfb) = k - 1 = 4 - 1 = 3
- Degrees of freedom within (dfw) = N - k = 28 - 4 = 24
So, the degrees of freedom for the test are 3 and 24.

If the alpha level chosen is 0.01, the researchers can conclude from the F-test results that:
- There is a significant difference among these groups, if the null hypothesis is rejected.
- There are no significant differences among the groups, if the null hypothesis is not rejected.

Given the F-test results, the correct conclusion is:
- There is a significant difference among these groups.


SWBAT represent relationships between two quantities visually by developing a connection between ratio tables and coordinate grids.

The lesson begins with a 15-minute activity where students complete the Equivalent Ratios and tables worksheet to re-activate prior learning, using tables to find equivalent ratios (SMP 7). This worksheet helps students understand the relationship between ratio tables and coordinate grids, a concept addressed in a prior lesson.

Next, students will use ratio tables to find at least 3 equivalent ratios to 4/16 and plot them on a coordinate grid. The teacher will explain that the numerator belongs on the x-axis and the denominator on the y-axis, modeling how to graph the points. Students will be asked to identify the relationship between the table and the graph (SMP 7: structure) and find two additional points on the line, justifying that these points are still equivalent ratios (SMP 3: arguing). As an extension, students will be asked to make a generalization about all points on the line (SMP 8: Patterns), with the expected response being that all points on the line are equivalent ratios if they form a straight line.

In the next 20-minute section, students will use their tables from the Equivalent ratio and tables worksheet to plot points on the coordinate grid, verifying that their work is correct as the plotted points form a line. As an extension, students will use the graph to find a value not in the table and justify their answer by adding it to the table (SMP 6).

Students will then practice independently using the In-Class worksheet (Massachusetts DOE, 2012) for 15 minutes, with the teacher walking around to check for understanding. Students can use their tablemates for assistance.

To conclude the lesson, students will create their own ratio table and graph it on the grid within 10 minutes. The teacher will ask students to identify if their graph shows a ratio and how they know, and then use the grid to find a ratio not from their table and justify their answer.


PA and PB are the tangents to the circle, so OA ⊥ PA and ∠OAP = 90°. In ΔOPA, sin ∠OPA = OA/OP = r/2r = 1/2 = sin 30°, which implies ∠OPA = 30°. Similarly, ∠OPB = 30°.

Now, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°. In ΔPAB, PA = PB because the lengths of tangents drawn from an external point to a circle are equal. This equality implies ∠PAB = ∠PBA, as equal sides have equal angles opposite to them.

The angle sum property states that ∠PAB + ∠PBA + ∠APB = 180°. Substituting ∠APB = 60° and ∠PAB = ∠PBA, we get ∠PAB + ∠PAB = 180° - 60° = 120°, which simplifies to 2∠PAB = 120°. Therefore, ∠PAB = 60°.

Since ∠PAB = ∠PBA = ∠APB = 60°, ΔPAB is an equilateral triangle.


The function $f$ preserves only the even digits of its input, defined as $f(x) = \sum_k a_{2k} 10^k$, where $x = \sum_k a_k 10^k$ with $a_k \in\{ 0,1,\dots,9\}$. For example, $f(59842.1073) = 582.03$.

The function $f$ is self-similar, as shown by the equation $f\left(10^{2k}x\right) = 10^k f(x)$, which holds for any $k \in \Bbb Z$. This self-similarity is evident in the function's graph, which exhibits the same shape when plotted over different ranges, such as $\left[0, 10^{2k}\right]\times \left[0,10^k\right]$.

To demonstrate that $f$ is Riemann integrable, recall that a bounded function $f$ is integrable over an interval $[a,b]$ if its lower and upper integrals coincide. The upper sum of $f$ in the interval $\left[0,10^{2k}\right]$ with a specific partition is equal to $\frac12\cdot 10^{3k}$.

A sequence of lower sums $\mathcal L^{(k)}_n$ can be constructed to converge to $\frac12\cdot 10^{3k}$. The lower sum $\mathcal L^{(k)}_0$ is $0$, obtained with a trivial partition. Increasing the number of sub-intervals to $100$ yields $\mathcal L^{(k)}_1 = 45 \cdot 10^{3k-2}$. Further increasing the number of sub-intervals by a factor of $100$ results in $\mathcal L^{(k)}_2 = 45\cdot 10^{3k-2} + 10^2\cdot\mathcal L^{(k-1)}_1 = 45 \cdot \left(10^{3k-2}+10^{3k-3}\right)$.

By iterating this procedure, it can be shown that for $n \geq 1$, $\mathcal L^{(k)}_n = 45\cdot 10^{3k-2} \cdot \sum_{h=0}^{n-1}10^{-h}$. The sequence $\left(\mathcal L^{(k)}_n\right)$ converges to $\frac12 \cdot 10^{3k}$, demonstrating that $f$ is Riemann integrable and that $\int_0^{10^{2k}} f(x) dx = \frac12 \cdot 10^{3k}$.

Key steps in this demonstration include:
1. Showing that the upper sum of $f$ in the interval $\left[0,10^{2k}\right]$ is equal to $\frac12\cdot 10^{3k}$.
2. Constructing a sequence of lower sums $\mathcal L^{(k)}_n$ that converges to $\frac12\cdot 10^{3k}$.
3. Demonstrating that the sequence $\left(\mathcal L^{(k)}_n\right)$ converges to $\frac12 \cdot 10^{3k}$, thus establishing the Riemann integrability of $f$ and the value of the integral $\int_0^{10^{2k}} f(x) dx$.

The self-similarity of $f$ and the construction of the sequence of lower sums are crucial in demonstrating the Riemann integrability of $f$ and evaluating the definite integral. The function's self-similarity allows for the iteration of the procedure to construct the sequence of lower sums, ultimately leading to the convergence of the sequence and the establishment of the integral's value.

In conclusion, the function $f$ is Riemann integrable, and the value of the definite integral $\int_0^{10^{2k}} f(x) dx$ is $\frac12 \cdot 10^{3k}$. This result is a direct consequence of the function's self-similarity and the construction of the sequence of lower sums, which converges to the upper sum, thus satisfying the condition for Riemann integrability.

Consider the following:
- What is the upper sum of the function $f$ in the interval $\left[0,10^{2k}\right]$ with partition $\left(0,10^{2k-1},2\cdot 10^{2k-1},\dots, 9\cdot 10^{2k-1},10^{2k}\right)$?
A) $\frac14\cdot 10^{3k}$
B) $\frac12\cdot 10^{3k}$
C) $\frac34\cdot 10^{3k}$
D) $10^{3k}$

Answer: B) $\frac12\cdot 10^{3k}$

- What is the value of $\mathcal L^{(k)}_1$?
A) $45 \cdot 10^{3k-3}$
B) $45 \cdot 10^{3k-2}$
C) $45 \cdot 10^{3k-1}$
D) $45 \cdot 10^{3k}$

Answer: B) $45 \cdot 10^{3k-2}$

- What is the value of $\int_0^{10^{2k}} f(x) dx$?
A) $\frac14\cdot 10^{3k}$
B) $\frac12\cdot 10^{3k}$
C) $\frac34\cdot 10^{3k}$
D) $10^{3k}$

Answer: B) $\frac12\cdot 10^{3k}$


Piper works at a camera store, earning an hourly rate plus a 16% commission on his sales. One week, he worked 20 hours and sold $1500 worth of cameras, earning a total of $515. To find his hourly rate, let's denote it as x. His hourly wages for 20 hours are 20x. Adding his commission, which is 16% of $1500 (0.16 * 1500 = $240), his total earnings are:

20x + 0.16(1500) = 515
20x + 240 = 515

Subtracting 240 from both sides gives:
20x = 275

Dividing both sides by 20 yields:
x = 275 / 20
x = 13.75

Therefore, Piper's hourly rate is $13.75.


If four times a number, added to 3, is divided by the number plus 7, the result is seven halves. Assuming the number is x, we can create an equation: (4x + 3) / (x + 7) = 7/2.

Multiplying both sides by x + 7 gives: 4x + 3 = (7/2)(x + 7). Using the distributive property, we get: 4x + 3 = (7/2)x + (7/2)(7).

Moving similar terms to the same side yields: 4x - (7/2)x = 49/2 - 3. Combining similar terms results in: (1/2)x = 43/2. Dividing both sides by 1/2 (or multiplying both sides by 2) gives: x = 43.

The number is 43. Let the unknown number be x, and the solution is found through this equation.


Consider two differentiable functions \(f\) and \(g\) defined in an open interval \((a,b)\), where \(b\) might be \(\infty\). Given that
\[\lim\limits_{x \to b^-} f(x) = \lim\limits_{x \to b^-} g(x) = \infty\]
and \(g^\prime(x) \neq 0\) in some interval \((c,b)\), a version of l’Hôpital’s rule states that if
\[\lim\limits_{x \to b^-} \frac{f^\prime(x)}{g^\prime(x)} = L\],
then
\[\lim\limits_{x \to b^-} \frac{f(x)}{g(x)} = L\].

A counterexample exists when \(g^\prime\) vanishes in all neighborhoods of \(b\), attributed to Otto Stolz. Let \((a,b) = (0,\infty)\) and define
\[
\begin{cases}
f(x) &= x + \cos x \sin x\\
g(x) &= e^{\sin x}(x + \cos x \sin x)
\end{cases}
\]
with derivatives
\[
\begin{cases}
f^\prime(x) &= 2 \cos^2 x\\
g^\prime(x) &= e^{\sin x} \cos x (x + \cos x \sin x + 2 \cos x)
\end{cases}.
\]
Evaluating the limits yields
\[
\lim\limits_{x \to \infty} \frac{f^\prime(x)}{g^\prime(x)} = \lim\limits_{x \to \infty} \frac{2 \cos x}{e^{\sin x} (x + \cos x \sin x + 2 \cos x)} = 0.
\]
However,
\[
\frac{f(x)}{g(x)} = \frac{1}{e^{\sin x}}
\]
does not have a limit at \(\infty\) because it oscillates between \(\frac{1}{e}\) and \(e\).


The component form of a vector is an ordered pair that describes the changes in the x- and y-values. For example, given two points (x1, y1) and (x2, y2), the ordered pair that describes the changes is (x2 - x1, y2 - y1). In a specific example, where x1 = 0, y1 = 0, x2 = 2, and y2 = 5, the ordered pair is (2-0, 5-0) or (2, 5).

Two vectors are considered equal if they have the same magnitude and direction. They are parallel if they have the same or opposite direction. When vectors are drawn on a coordinate grid, they can be described in terms of their components in the x and y directions.

Vectors can be represented in different forms, including component form and polar form. In polar form, vectors can be specified using ordered pair notation or matrix notation. The first component of the vector is r, and the second component is θ.

Key concepts in vector mathematics include vector magnitude, vector addition, vector subtraction, and vector multiplication. These concepts are crucial in solving vector problems, including those involving geometric shapes and algebraic vectors. Parallel vectors and geometric vectors with application problems are also important areas of study.

A vector calculator can be used to perform various operations, including adding, subtracting, finding the magnitude, angle, dot product, and cross product of two vectors in 2D or 3D. Detailed explanations are provided for each operation, making it easier to understand and work with vectors.


**Converting Teaspoons to Cups: A Guide**

When baking, it's common to need to convert measurement units, especially when converting teaspoons to cups. This can be frustrating, but the solution is simple math. In this guide, we'll cover how many teaspoons are in a 1/8 cup.

**What Are Teaspoons and Cups?**

A teaspoon is a small measuring unit, typically 5 milliliters (mL) or 0.2 fluid ounces (fl oz), abbreviated as "tsp". A cup is a larger unit, representing 8 fluid ounces (fl oz) or 236.6 milliliters (mL), abbreviated as "c".

**How Many Teaspoons Are In A 1/8 Cup?**

To find out, divide 8 by 1/8 (0.125), resulting in 64. Then, divide 64 by 5 (the amount of teaspoons in a milliliter), which equals 12. So, there are 12 teaspoons in a 1/8 cup.

**Why Know How Many Teaspoons Are In A 1/8 Cup?**

Knowing this conversion makes following recipes and baking projects easier. For example, if a recipe calls for one tablespoon of sugar, but you only have teaspoons, you can easily figure out that 6 teaspoons make up one tablespoon.

**Tips For Converting Teaspoons to Cups**

* One teaspoon is equal to 5 milliliters (mL).
* One cup is equal to 8 fluid ounces (fl oz) or 236.6 milliliters (mL).
* To convert teaspoons to cups, divide the number of teaspoons by 48.

**Tools and Methods For Converting Teaspoons to Cups**

Converting teaspoons to a 1/8 cup can be done by dividing or multiplying the number of teaspoons by 48. If a recipe calls for one tablespoon of sugar, but you only have teaspoons, remember that 6 teaspoons make up one tablespoon.

**Conclusion**

Now you know how many teaspoons are in a 1/8 cup. With this knowledge and helpful conversion tools, you'll be able to accurately measure ingredients for cooking or baking.

**FAQs: Teaspoons in a 1/8 Cup**

* **Is 2 tablespoons equal to 1/8 cup?**: Yes, according to the US conversion formula.
* **How many ounces is a 1/8 cup?**: 1/8 cup is equal to one ounce.
* **What can I use for 1/8 cup?**: Measure 1/8 cup with a tablespoon (2 tablespoons), weigh 1 ounce of ingredient, or use a standard shot glass (1.5 fluid ounces).
* **What is 1/8 + 1/8 cup equal to?**: 1/8 + 1/8 is equal to 2/8, or 1/4 when reduced to its simplest form.
* **How many spoonfuls is 1/8 cup?**: 1/8 cup is equal to 2 US tablespoons.


**Math Questions**

1. What is the side length of a square with an area of 64 square inches?
Answer: 8 in

2. Solve for x: 3x = 18
Answer: 6

3. What is 37 + 70 + 170?
Answer: 277

4. What is the AMS mascot?
Answer: Bears

5. What does the word "sum" refer to in math?
Answer: Addition

6. What is the value of a complementary angle?
Answer: 90 degrees (Note: The original answer was incorrect, a complementary angle is 90 degrees)

7. How many months have 28 days?
Answer: All of them

8. What is the perimeter of a square with all even sides and one side is 11 cm?
Answer: This question is incorrect as 11 is an odd number, a correct question would be: What is the perimeter of a square with one side is 10 cm?
Answer: 40 cm

9. What is the side length of a square with the volume of 8 meters cubed?
Answer: 2 meters

10. What is -155 + 245?
Answer: 90

11. What is 100^0?
Answer: 1

12. What does the word "product" refer to in math?
Answer: Multiplication

13. What is the value of a straight angle?
Answer: 180 degrees

**DOUBLE JEOPARDY**

14. Find the slope of the line from the table:
x | 0 | -1 | -2
y | 1 | 3 | 5
Answer: Slope = 2 (Note: The original answer was incorrect, the slope is 2)

15. What is the area of a square with the length of 12m and a width of 12m?
Answer: This question is incorrect as the width is given as 9m, a correct question would be: What is the area of a rectangle with the length of 12m and a width of 9m?
Answer: 108m

16. Solve for x: 2x + 5 = 29
Answer: 12

17. What is 198 + (-299)?
Answer: -101

18. What does "equivalent" mean?
Answer: Equal to

19. David's father has three sons: Snap, Crackle, and?
Answer: David

20. Solve for x: 5x = 225
Answer: 45

21. A rectangle has a length of 12 cm and height of 6 cm. If a triangle is made with a diagonal from corner to corner. What is the area of the triangle?
Answer: This question is incorrect as the area of the triangle cannot be determined with the given information, a correct question would be: What is the area of the rectangle?
Answer: 72 cm squared

22. Solve the inequality: 4 - 3x > 40
Answer: x < -12

23. What is 30 + __ = 131?
Answer: 101

24. Is pi rational or irrational?
Answer: Irrational

25. What does the word "quotient" refer to in math?
Answer: Division

26. A cowboy rode into town on Friday. He stayed for three nights and rode out on Friday. How is this possible?
Answer: His horse's name is Friday

27. What is 60% as a decimal?
Answer: 0.6

28. What does congruent mean?
Answer: Equal

29. Elijah's mom said to take a break after lunch at 12:25 p.m. His break was over at 1:10 p.m. How long was his break?
Answer: 45 minutes

30. What is -143 + 134?
Answer: -9

31. What is 2/5 as a decimal?
Answer: 0.4

32. What does Factor mean in math?
Answer: The numbers you multiply together to get the product

33. One angle of a supplementary angle is 60 degrees. What is the measure of the other angle?
Answer: 120 degrees

34. A shape is graphed in quad I then reflected across the y-axis. Which quad is it in now?
Answer: Quad II

35. The area of a rectangle is 210mm. One side of it is 10mm. What is the length of the other side of the rectangle?
Answer: 21mm

36. What is -2(-5 + 7) + 24/6?
Answer: 8

37. What is ^{-45 - (-20)}?
Answer: -25

38. What is the expression for 40 less than x?
Answer: x - 40

39. When a fair die is thrown, what is the probability of getting a number greater than 4?
Answer: 1/3

40. Are vertical angles congruent?
Answer: Yes

41. A circle has a diameter of 10 inches. What is the approximate area of the circle? (use 3.14 for pi)
Answer: A = 78.5 inches squared


**Theorem**

The intersections of the adjacent perpendicular bisectors of the sides of a parallel-hexagon form a parallelo-hexagon, i.e., a hexagon with opposite sides parallel and equal.

**Alternative Formulation**

The circumcentres G, H, I, J, K, and L of triangles ABC, BCD, CDE, DEF, and EFA of a hexagon ABCDEF with opposite sides parallel form a parallelo-hexagon.

**Challenge**

1. The result may seem easy to prove using properties of parallelograms, but a purely geometric proof has not been found.
2. The theorem can be attacked using vector methods or complex algebra, but the algebra becomes complicated. A proof by computer technology, such as Mathematica, is convenient, but a purely geometric proof is still needed.

**Hint**

A possible construction for a novel geometric proof can be found by clicking the Hint button in the sketch.

**Novel Proof**

A geometric proof using ideas from calculus, as well as a Mathematica proof, can be found in the International Journal for Mathematical Education in Science & Technology (IJMEST).

**Note**

The result is not true for arbitrary perpendiculars to the sides of a general hexagon with opposite sides parallel. The perpendiculars must intersect at the midpoints of the sides of the original hexagon, i.e., be perpendicular bisectors.

**More Proofs**

More elementary and explanatory proofs have been given by:
- Dubrovsky, V. (2021) in a post at the Romantics of Geometry group on Facebook.
- Frank, R. & Schumann, H. (2023) in the International Journal of Mathematical Education in Science and Technology.

**Another Result**

Another interesting result involving a parallel-hexagon is the Parallel-hexagon Concurrency Theorem, where the lines connecting the midpoints of the opposite sides are concurrent.

**Related Links**

- Conway's Circle Theorem as a special case of the Side Divider Theorem
- Area ratios of some polygons inscribed in quadrilaterals and triangles
- Triangle Centroids of a Hexagon form a Parallelo-Hexagon: A generalization of Varignon's Theorem
- Some Parallelo-hexagon Area Ratios
- A side trisection triangle concurrency
- A generalization of a Parallelogram Theorem to Parallelo-hexagons, Hexagons, and 2*n*-gons in general
- A 1999 British Mathematics Olympiad Problem and its dual
- Easy Hexagon Explorations

Created by Michael de Villiers, updated on several occasions between 2019 and 2023.


**Calculating Square Roots and Other Roots**

The simplest way to calculate the decimal digits of a square root is to figure out each digit in turn. First, determine how many digits to the left of the decimal point the square root will have. A number with one digit to the left of the decimal point will have a square with one or two digits to the left of the decimal point. A number with two digits to the left of the decimal point will have a square with three or four digits to the left of the decimal point.

To calculate the square root, start from the left and figure out each digit in turn. For example, to find the square root of 1234 to 3 decimal places, first determine that the square root has two digits to the left of the decimal point. Then, find the first digit by determining that 40^2 = 1600 is too big, so the first digit is 3. Next, find the second digit by determining that 36^2 = 1296 is too big, so the second digit is 5.

This method is slow and inefficient, but it can be improved with shortcuts. A more efficient method is to use an iteration, such as *x* = (1/2)(*x* + *a*/*x*), where *a* is the number for which you want to find the square root. Start with an initial value of *x*, and then calculate subsequent values using the iteration. For example, to find the square root of 1234, start with *x* = 40 and calculate subsequent values: 40, 35.425, 35.129578, and so on.

**Calculating Other Roots**

The methods for computing square roots can be generalized to handle any rational exponent. If the degree of the root is a whole number, use a similar method to find the root. For example, to find the cube root of 7809, first determine that the root is of the form 1A.BCD, and then find the subsequent digits.

If the degree of the root is a fraction, first take the *n*th root of *a* and then raise this number to the *m*th power. A faster approach is to use an iteration, such as *y* = (1/*n*) \* (*m*/*x*^(*n*-1)) \* *a*, where *x* is an initial guess. Repeat the process using *y* as the new guess, and the sequence of numbers will converge quickly to the desired root.

**The Square Root of 2**

The square root of 2 is the one and only positive real number whose square is 2. The digits of the square root of 2 are 1.41421356, but these digits do not repeat and cannot be written down completely. A mathematician would not say that "1.41421356" is an answer to the question "what is the square root of 2", but rather that the square root of 2 is the positive real number whose square is 2.

**The Square Root Symbol**

The square root symbol can be referred to as either a "radical sign" or a "surd symbol". It is believed to have originated from a corrupted form of the initial letter of the word "radix", so "radical sign" may be the most appropriate term.

**Rational Roots**

To calculate rational roots, such as 2.5, use a similar method to find the root. First, take the *n*th root of *a* and then raise this number to the *m*th power. A faster approach is to use an iteration, such as *y* = (1/*n*) \* (*m*/*x*^(*n*-1)) \* *a*, where *x* is an initial guess. Repeat the process using *y* as the new guess, and the sequence of numbers will converge quickly to the desired root.

**Methods for Computing Roots**

There are several methods for computing roots, including the iteration method and the method of finding each digit in turn. The iteration method is generally faster and more efficient, but the method of finding each digit in turn can be useful for small roots or for educational purposes.

**Conclusion**

Calculating square roots and other roots can be done using several methods, including the iteration method and the method of finding each digit in turn. The iteration method is generally faster and more efficient, but the method of finding each digit in turn can be useful for small roots or for educational purposes. The square root of 2 is a special case, and its digits do not repeat and cannot be written down completely. The square root symbol can be referred to as either a "radical sign" or a "surd symbol", and rational roots can be calculated using a similar method to find the root.


Time.deltaTime is the time since the last frame. In Lerp, the third parameter is a float between 0 and 1. If it's 0, you get the first parameter in Lerp. If it's 1, you get the second. Therefore, with a statement like `transform.position = Vector3.Lerp(transform.position, target.position, Time.deltaTime);` in Update, you get a position that's part-way between `transform.position` and `target.position`, depending on the time difference between the current frame and the last frame.

For example, a Time.deltaTime value of .01 (100fps) would return a value that's a combo of 99% `transform.position` and 1% `target.position`. As this is repeated each frame, the `transform.position`, assuming `target.position` is stationary, becomes closer and closer to `target.position` but never *quite* reaches it. (Unless the framerate drops to 1fps or less, in which case Time.deltaTime would be 1 or greater, so the return value would be 100% `target.position`.)

This is a "trick" usage of Lerp, which creates a "slow down as you approach the target" effect. It's mathematically incorrect, though, since it will produce results that vary somewhat depending on the framerate. A better approach is to use a coroutine that moves a transform from `startPos` to `endPos` over a given time, resulting in completely linear movement, without the "slowdown" effect.

As for Time.time, consider `transform.position = Vector3.Lerp(startPos, endPos, Time.time);` in Update. This moves the transform from `startPos` to `endPos` as the time advances from 0.0 to 1.0. Once the time is past one second, no further movement will occur since the third parameter in Lerp is clamped between 0 and 1.

Another script that prevents the jag effect is:
```csharp
var pointA:Vector3;
var pointB:Vector3;
var time:float = 10.0;
private var i:float = 0.0;
private var rate:float = 0.0;
function Update ()
{
    MoveObject(this.transform, pointA, pointB, time);
}
function MoveObject (thisTransform : Transform, startPos : Vector3, endPos : Vector3, time : float)
{
    rate = 1.0/time;
    if (i < 1.0)
    {
        i += Time.deltaTime * rate;
        thisTransform.position = Vector3.Lerp(startPos, endPos, i);
    }
}
```
However, using `Time.deltaTime` can lead to unexpected high values, causing errors. A simpler solution to achieve the "ease out" effect is to use `curPos = Vector3.Lerp(curPos, endPos, 1-Pow(2, -Time.deltaTime * rate));`, where `curPos` is the current position, `endPos` is the end position, and `rate` is the rate of movement.

This solution works because it takes into account the time difference between frames, resulting in a smooth movement that is independent of the framerate. For example, if `Time.deltaTime = 1`, the calculation will run 1 time, resulting in `curPos = 0.5`. If `Time.deltaTime = 0.25`, the calculation will run 4 times, resulting in the same final value of `curPos = 0.5`.

To make a movement frame-independent, you can add `Time.deltaTime` to the end of the current distance, like so: `currentDistance = (currentFrame / target_Frame) + (0.01f * Time.deltaTime);`. This ensures that the movement is smooth and consistent, regardless of the framerate.

In summary, using `Time.deltaTime` with Lerp can create a "slow down as you approach the target" effect, but it's mathematically incorrect and can lead to errors. A better approach is to use a coroutine or a simpler solution that takes into account the time difference between frames, resulting in a smooth and frame-independent movement.

Multiple choice questions:
1. What is Time.deltaTime?
a) The time since the last frame
b) The time since the game started
c) The framerate of the game
d) The time until the next frame

Answer: a) The time since the last frame

2. What is the effect of using `Time.deltaTime` with Lerp?
a) It creates a "slow down as you approach the target" effect
b) It creates a linear movement
c) It creates a random movement
d) It creates no movement

Answer: a) It creates a "slow down as you approach the target" effect

3. How can you make a movement frame-independent?
a) By using `Time.time`
b) By using `Time.deltaTime`
c) By adding `Time.deltaTime` to the end of the current distance
d) By using a coroutine

Answer: c) By adding `Time.deltaTime` to the end of the current distance


This problem illustrates James Tanton's Problem Solving Tip #6: **Eliminate Wrong Answer Choices**. The solution involves eliminating incorrect answer choices to arrive at the correct one.

Consider a problem where the answer choices are (A), (B), (C), and (D). Choice (A) can be eliminated because it would result in both children being wrong. Choice (B) is also incorrect because Juan would be wrong. Similarly, choice (C) can be eliminated as it would mean Lashana is wrong.

After eliminating choices (A), (B), and (C), the only remaining option is (D). Choice (D) could be correct if there were a specific distribution, such as 3 horses on one side and 4 on the other. This distribution makes choice (D) a plausible solution.

The problem-solving technique demonstrated here is from the book *Avoid Hard Work! … And Other Encouraging Mathematical Problem-Solving Tips** **for the Young, the Very Young, and the Young at Heart *by Maria Droujkova, James Tanton, and Yelena McManaman. For more information, visit http://naturalmath.com/avoid-hard-work/.


In classical geometry, a radius of a circle or sphere is a line segment from its center to its perimeter. The term "radius" comes from the Latin word for "ray" or "spoke of a chariot wheel." The plural of radius can be either "radii" or "radiuses," and the typical abbreviation is "R" or "r." The diameter (D) is defined as twice the radius.

If an object lacks a center, the term "radius" may refer to its circumradius, the radius of its circumscribed circle or sphere. In this case, the radius may be more than half the diameter, which is usually defined as the maximum distance between any two points of the figure. The inradius of a geometric figure is the radius of the largest circle or sphere contained in it. The inner radius of a ring, tube, or other hollow object is the radius of its cavity.

For a circle, the radius is the same as its circumradius. The inradius of a regular polygon is also called the apothem. In graph theory, the radius of a graph is the minimum distance from any vertex to any other vertex.

The radius of a circle with perimeter (circumference) C is given by the formula: r = C / (2π). The radius of a circle passing through three non-collinear points (x1, y1), (x2, y2), and (x3, y3) can be expressed as: r = abc / (4Δ), where Δ is the area of the triangle formed by the three points.

The radius of a regular polygon with sides of length "s" is given by: Rn = s / (2 sin(π/n)), where "n" is the number of sides. Values of Rn for small values of "n" are provided in a table.

In polar coordinates, the distance from the pole (origin) is called the radial coordinate or radius, and the angle is the angular coordinate, polar angle, or azimuth. The radius and azimuth are together called the polar coordinates, corresponding to a two-dimensional polar coordinate system.

The distance from the axis may be called the radial distance or radius, while the angular coordinate is sometimes referred to as the angular position or azimuth. The radius and azimuth are used in cylindrical coordinates, where the third coordinate may be called the height, altitude, longitudinal position, or axial position.


## How to Calculate Growth Rate

To calculate the average growth rate, organize your data in a table and use a growth rate equation that takes into account the number of time intervals in your data. The compound annual growth rate (CAGR) metric smoothes out lumpy growth to calculate a theoretical annual growth rate.

### Formula for Calculating CAGR

The formula for CAGR is:
CAGR = (End Value / Beginning Value)^(1 / Number of Years) - 1

### Average Annual Growth Rate (AAGR)

AAGR is the average increase of an investment over a period of time. It is a standard for measuring average returns of investments over several time periods. AAGR works the same way that a typical savings account works, with interest compounded annually.

### Calculating Average Growth Rate Over Regular Time Intervals

1. Organize your data in a table.
2. Use a growth rate equation which takes into account the number of time intervals in your data.
3. Isolate the "growth rate" variable.
4. Solve for your growth rate.

### Population Growth and Compound Interest

Population growth can be calculated based on annual increase, and it can also be compared to compound interest. Logistic growth including carrying capacity is another method of calculating population growth.

### GDP Growth Rates

To calculate GDP growth rates, use the formula: (GDP in year 2 / GDP in year 1) - 1. This will give you the growth rate of the economy from one year to the next.

### Determining Growth Rates

To determine growth rates, you can use a growth rate calculator or create your own formula. The percent growth rate calculator is used to calculate the annual percentage growth rate.

### FAQs

* What is the formula for calculating the percent growth rate?
* How do you calculate the average annual growth rate?
* What is the difference between CAGR and AAGR?

### Example

If a company has a beginning value of $100 and an end value of $150 after 2 years, the CAGR would be:
CAGR = (150 / 100)^(1 / 2) - 1 = 22.47%

This means that the company's value grew at an average annual rate of 22.47% over the 2-year period.


**Question: **Why did you decide to use the exclusive definition of a trapezoid?

**Answer: **There are two definitions of a trapezoid: inclusive and exclusive. The inclusive definition considers all parallelograms (including rectangles, squares, and rhombuses) as trapezoids, while the exclusive definition does not.

In choosing a definition, we considered the following factors:

- Most elementary textbooks use the exclusive definition, although the Common Core State Standards do not specify which definition to use. Colleges and universities often use the inclusive definition, but their mathematical goals differ.
- The exclusive definition is suitable for elementary grades, as it supports the mathematical goal of classifying 2-d shapes and understanding how changing attributes affects classification. This definition complicates the hierarchy of 4-sided 2-d shapes, requiring students to identify parallel sides and their numbers.

The existence of two competing definitions can be surprising, as mathematics is often perceived as precise with a single right answer. This discrepancy can be a fascinating investigation for fourth or fifth-grade students, who can explore the implications of using different definitions. For example, "What would happen if we use the definition of a trapezoid that says 'at least one pair of parallel sides'? How would that change your categories and the shapes within them?"

This topic sparked a lively debate in a 5th-grade math class, where students discussed whether a parallelogram, rhombus, rectangle, and square could be considered a trapezoid. The exclusive definition prompted students to think critically about the classification of 2-d shapes and the importance of definitions in mathematics. The discussion highlighted the value of exploring inclusive and exclusive definitions in geometry, allowing students to develop a deeper understanding of mathematical concepts.


**Math Concepts**

1. **Place Value**: The value of a digit's position in a number.
2. **Remainder**: The amount left over when one number is divided by another.
3. **Order of Operations**: Follows the PEMDAS rule.

**Fractions and Decimals**

1. **Fraction to Decimal**: Divide the numerator by the denominator.
2. **Decimal to Fraction**: e.g., 0.9 = 9/10.
3. **Improper Fraction**: A fraction with a numerator larger than the denominator.
4. **Mixed Number**: A whole number and a fraction combined.
5. **Comparing Fractions**: Find a common denominator or convert to decimals.
6. **Comparing Decimals**: Line up numbers and fill in missing zeros.

**Operations with Fractions and Decimals**

1. **Adding/Subtracting Fractions (like denominators)**: Add/subtract numerators, keep denominator the same.
2. **Adding/Subtracting Fractions (unlike denominators)**: Find a common denominator first.
3. **Adding/Subtracting Decimals**: Line up decimal points, then add or subtract.
4. **Dividing Decimals**: Bring the decimal point up, add zeros if needed.
5. **Multiplying Decimals**: Ignore decimal point until answer, count decimal places in factors.

**Geometry and Graphs**

1. **Ordered Pairs**: Pairs of numbers (x, y) representing a point in a coordinate plane.
2. **Edge of a Prism**: The line where two surfaces meet.
3. **Face of a Prism**: A polygon forming a surface of a prism.
4. **Vertex**: A point of a three-dimensional shape.
5. **Area of a Parallelogram**: A = bh or A = l*w.
6. **Volume of a Prism**: V = Bh or V = l*w*h.
7. **Surface Area of a Prism**: The sum of all face areas.
8. **Cube**: A prism with six square faces.

**Data Analysis**

1. **Mean**: Average.
2. **Median**: Middle number.
3. **Range**: Difference between highest and lowest scores.
4. **Double Bar Graph**: Compares two sets of categorical data.
5. **Line Graph**: Shows changes over time using line segments.

**Miscellaneous**

1. **Associative Property**: Changing number grouping doesn't change the value.
2. **Distributive Property**: a(b + c) = ab + ac.
3. **Expression**: A mathematical phrase without an equal sign.
4. **Equation**: A mathematical sentence with an equal sign.
5. **Greater Than**: >, **Less Than**: <, **Greater Than or Equal To**: ≥, **Less Than or Equal To**: ≤.
6. **Estimate**: A careful guess, not an exact number.
7. **Rounding**: Expressing a number to the nearest specified value.


MergeSorting a hand of playing cards

We've been exploring relatively simple quadratic (O(n^{2})) sorting algorithms, which are incremental and sort one element at a time. Another category of algorithms is divide-and-conquer, where the data set is partitioned into smaller portions, solved, and then integrated back together.

Mergesort is a simple example of a divide-and-conquer algorithm. It can be described as divide-conquer-combine:
* **Divide:** Split the items in half, creating a left set and a right set, each with n/2 items.
* **Conquer:** Recursively apply the divide step to each subset until reaching subsets of a single item, which are trivially sorted.
* **Combine:** Merge the sorted subsets back together to produce the full sorted set.

This process is conceptually simple, but implementing it without higher-level constructs can be more complicated. Mergesort is a divide-and-conquer algorithm that efficiently sorts data by recursively dividing and merging subsets.


# How many 5 digit numbers can you make starting with 5 without repeating the numbers 0 to 5?

## Introduction to 5-Digit Numbers

To calculate the number of 5-digit numbers starting with 5 without repeating the numbers 0 to 5, we first need to understand the total possible 5-digit numbers and then subtract the ones that do not meet our criteria.

## Total Possible 5-Digit Numbers

In the base 10 system, for each digit, there are 10 possibilities (0-9). However, for a 5-digit number, the ten thousands place cannot be 0, or it would be a 4-digit number. Thus, the ten thousands place can be filled in 9 ways (1-9), and the rest of the places can each be filled in 10 ways. But since we're considering all possible 5-digit numbers without any restrictions first, we calculate this as 9 * 10 * 10 * 10 * 10 = 90,000.

## 5-Digit Numbers Without Repeated Digits

To find how many 5-digit numbers have no repeated digits, we consider the first digit can be any of 9 digits (1-9, since 0 cannot lead), the second digit can then be any of the remaining 9 digits (including 0, but excluding the first digit chosen), the third digit can be any of the remaining 8 digits, and so on. Thus, the total number of 5-digit numbers with no repeated digits is 9 * 9 * 8 * 7 * 6 = 27,216.

## Numbers Including Specific Digits

For numbers that include both 5 and 8, we consider the positions of these two digits and the possibilities for the remaining digits. However, the exact calculation for this specific scenario isn't directly provided in the initial analysis.

## Repeated Digits in 5-Digit Numbers

To find how many 5-digit numbers have one or more repeated digits, we subtract the number of 5-digit numbers with no repeated digits from the total possible 5-digit numbers. The total possible 5-digit numbers, considering the first digit cannot be 0, is 9 * 10 * 10 * 10 * 10 = 90,000. The numbers with no repeated digits are 9 * 9 * 8 * 7 * 6 = 27,216. Thus, the numbers with at least one repeated digit are 90,000 - 27,216 = 62,784.

## Specific Scenarios

- **Starting with 5 without repeating 0 to 5**: Since the first digit is fixed as 5, and we cannot repeat 0 to 5, we consider the remaining digits. For the second digit, there are 5 choices (6, 7, 8, 9, since 0, 1, 2, 3, 4, and 5 are excluded), for the third digit, there are 4 choices (excluding the second digit chosen and 0 to 5), and so on. However, the precise calculation for this scenario requires considering all possible combinations under these constraints.

## Base 10 System

In the base 10 system, each digit has 10 possibilities (0-9). For a 5-digit number, if all digits could freely be any number 0-9, there would be 10^5 = 100,000 possible numbers. However, the first digit cannot be 0, reducing our total to 9 * 10 * 10 * 10 * 10 = 90,000 possible 5-digit numbers.

## Conclusion

Understanding the constraints and applying them to the total possible combinations gives us the number of 5-digit numbers under specific conditions. For a 5-digit number starting with 5 and not repeating the numbers 0 to 5, we need to calculate based on the available choices for each subsequent digit, considering the exclusion of 0 to 5 and the fixed position of 5.

**Multiple Choice Questions:**

1. How many 5-digit numbers can be made without repeating any digit?
- A. 90,000
- B. 27,216
- C. 62,784
- D. 100,000

Answer: B. 27,216

2. What is the total number of possible 5-digit numbers in the base 10 system, considering the first digit cannot be 0?
- A. 100,000
- B. 90,000
- C. 10,000
- D. 1,000

Answer: B. 90,000


Given the polynomial f(x) = x^{2} – 5x + k with zeros α and β, where α – β = 1, we need to find the value of k. The coefficients of the polynomial are a = 1, b = -5, and c = k.

We know that α + β = -b/a and αβ = c/a. Given α – β = 1, we can express α and β in terms of their sum and difference.

Let α + β = 5 (i) and α – β = 1 (ii). Solving these equations, we get 2α = 6 and 2β = 4, which gives α = 3 and β = 2.

Now, using αβ = c/a, we have (3)(2) = k/1, which simplifies to k = 6. Therefore, the value of k is 6.


## Description

The problems examine the use of LDA, QDA, logistic regression, and other classification methods.

**Problem 1 (30%)**

1. Split the Default dataset from ISLR into a training set (70%) and a test set (30%). Compare the classification error of LDA, QDA, and logistic regression when predicting default as a function of selected features. Determine which method works best.
2. Report the confusion table for each classification method, labeling the predicted class and true class. Analyze the results.
3. Examine if LDA assumptions are satisfied when predicting default as a function of balance only. Use qqnorm and qqline to check for normality and compare standard deviations of class distributions. Also, assess if QDA assumptions are satisfied.
4. Discuss whether to use LDA over QDA when LDA assumptions are violated (e.g., different conditional standard deviations).

**Problem 2 (30%)**

1. Using the MNIST dataset, fit classification models to predict the digit 1 (vs all others). Compare classification errors for:
	* Logistic regression
	* K-NN with two reasonable k choices
	* LDA
2. Explore feature transformations (e.g., combinations) and re-run the methods.
3. Determine which method works best. Split the data into a training set and a test set (a subsample of 10,000 datapoints is acceptable).

**Problem O2 (35% - optional for extra credit)**

Solve Exercises 1.11 and 1.13 in [Bishop, C. M. (2006). Pattern Recognition and Machine Learning]. This problem can substitute for Problem 2, with the better score considered.

**Problem 3 (20%)**

Prove that the logistic function for predicting class probabilities is equivalent to assuming a linear model for the log-odds. Use algebraic manipulation to show the equivalence between the two expressions.

**Problem 4 (20%)**

1. Compare the expected performance of LDA and QDA on an arbitrary training set.
2. If the Bayes decision boundary is linear, compare the expected performance of LDA and QDA on the training and test sets.
3. Analyze how the sample size affects the prediction accuracy of QDA relative to LDA.
4. True or False: Even with a linear Bayesian decision boundary, QDA will likely achieve a superior test error rate due to its flexibility. Justify the answer.


**Gravitational Field**

A gravitational field is the space around a mass where it can exert a gravitational force on other masses. This field is a result of the interaction between two masses, which can be explained by the concept of action at a distance. The gravitational field intensity, denoted by E, is the force experienced by a unit mass placed at a point in the field. It is a vector quantity with a unit of N kg^-1.

Consider a body of mass M at point Q and another body of mass m at point P, a distance r from Q. The mass M develops a field E at P, exerting a force F = mE. The gravitational force of attraction between the masses m and M is F = GMm/r^2, where G is the gravitational constant. The gravitational field intensity at P is E = F/m = GM/r^2.

**Gravitational Potential Difference**

The gravitational potential difference between two points is the amount of work done in moving a unit mass from one point to another against the gravitational force of attraction. It is defined as dv = -E dr, where E is the gravitational field intensity and dr is the distance between the two points. The negative sign indicates that work is done against the gravitational field.

**Gravitational Potential**

The gravitational potential at a point is the amount of work done in moving a unit mass from the point to infinity against the gravitational field. It is a scalar quantity with a unit of N m kg^-1. The expression for gravitational potential at a point is v = -GM/r, where G is the gravitational constant, M is the mass, and r is the distance from the center of the mass.

**Gravitational Potential Energy**

The gravitational potential energy of a mass m at a distance r from another mass M is the amount of work done in moving the mass m from a distance r to infinity. It is defined as U = GMm/r. The gravitational potential energy is zero at infinity and decreases as the distance decreases, due to the attractive nature of the gravitational force.

**Gravitational Potential Energy near the Surface of the Earth**

Near the surface of the Earth, the gravitational potential energy of a mass m at a height h above the surface is U = mgh, where g is the acceleration due to gravity. This is an approximation of the more general expression U = GMmh / R(R+h), where M is the mass of the Earth, R is its radius, and h is the height above the surface.


Using Waves to Determine Primes, Composites, Number's Factors, and their Distributions

Introduction:

This paper presents a method utilizing periodic functions to determine primality, count factors, list factors, calculate prime distribution, and find the Nth prime. The method is described in a step-by-step manner, with equations and graphs illustrating each key step to facilitate visualization of the reasoning and results. The paper concludes with a discussion of afterthoughts and ongoing questions regarding the technique.

The method aims to provide a straightforward approach to:
- Check for primality
- Count factors
- List factors
- Calculate prime distribution
- Determine the Nth prime

The paper is supported by visual aids, including graphs, to help illustrate the reasoning and results of each step. Feedback is sought, particularly regarding the topics addressed in the afterthoughts section.


# Volume of Triangular Prism

A triangular prism is a prism with two congruent parallel triangles as its bases and rectangular lateral faces. The formula for the volume of a triangular prism is given by Volume V = A × h, where A is the area of the base triangle and h is the height of the prism.

The area of the base triangle, A, can be calculated using the following formulas:
- A = $\frac{1}{2}$bh, where b is the base of the triangle and h is the height of the triangle.
- A = $\sqrt{s(s-a)(s-b)(s-c)}$, where a, b, and c are the sides of the triangle and s = (a+b+c)/2.
- A = $a^2 \sqrt{3}/ 4$, where a is the side of an equilateral triangle.

To find the volume of a triangular prism, follow these steps:
**Step 1:** Calculate the area of the base triangle using one of the above formulas.
**Step 2:** Multiply the area of the base triangle by the height of the prism to get the volume.

Example solutions:
- For a triangular prism with a base triangle of area calculated as $\frac{1}{2}$ × 14 × 8 and a height of 10, the volume V = $\frac{1}{2}$ × 14 × 8 × 10 = 560 cubic feet.
- For a triangular prism with a base triangle of area calculated as $\frac{1}{2}$ × 14 × 8 and a height of 6, the volume V = $\frac{1}{2}$ × 14 × 8 × 6 = 336 cubic feet.


Curriculum: Grade 5, Module 4, Topic C - Multiplication of a Whole Number by a Fraction

To calculate the fraction of a whole number, use a model to represent the given information. Select the number of parts and calculate the value of one part. Then, determine how many parts are represented by the fraction.

Key skills include:
A. Determining fractions of a whole number based on a model
B. Applying the model to solve word problems
C. Advanced determination of fractions using a model
E. Finding a whole number given a fraction of that number using a model
F. Multiplying a whole number by a fraction
G. Relating fractions to multiplication and solving problems
H. Determining a fraction of a measurement by converting units and multiplying
I. Identifying common factors of two whole numbers
J. Dividing whole numbers by a common factor
K. Reducing fractions by dividing by a common factor
L. Reducing factors before multiplying a whole number by a fraction (basic)
M. Reducing factors before multiplying a whole number by a fraction (advanced)
N. Multiplying a whole number by a fraction, with reduction before or after multiplication.


Vertical force balance and horizontal force balance yield two key equations. Assuming small angles, we can derive simplified expressions. From these equations, we find that the angle of attack is related to the ratio of the aircraft's weight to its maximum lift, and the angle of controlled ascent is related to the thrust developed by the engine.

The relationship between the angle of attack and the weight-to-lift ratio is given by Equation (9.113), while the relationship between the angle of controlled ascent and the thrust is given by Equation (9.114). For an unpowered aircraft, such as a glider, the angle of controlled descent can be minimized by making the wing aspect-ratio as large as possible. This is why gliders and albatrosses have long thin wings.

For a powered aircraft, the critical thrust-to-weight ratio required to maintain level flight is given by Equation (9.116). This ratio is minimized by minimizing the glide angle, which is why long-haul aircraft tend to have long thin wings. However, if the angle of attack exceeds a critical value, boundary layer separation occurs, resulting in increased drag and a phenomenon known as a stall.

To avoid a stall, the airspeed must be kept above the critical value, known as the stall speed. This is a crucial consideration in aerodynamics, as it affects the safety and efficiency of flight. By understanding the relationships between angle of attack, thrust, and airspeed, aircraft designers can optimize their designs to minimize fuel consumption and maximize performance.

Key equations include:
- Expression (9.113): relates the angle of attack to the weight-to-lift ratio
- Expression (9.114): relates the angle of controlled ascent to the thrust
- Equation (9.115): describes the angle of controlled descent for an unpowered aircraft
- Equation (9.116): gives the critical thrust-to-weight ratio for level flight
- Equation (9.117): describes the stall phenomenon and its relation to airspeed.

The main concepts to remember are:
- The importance of angle of attack and its relation to weight and lift
- The role of thrust in controlled ascent and level flight
- The impact of wing aspect-ratio on glide angle and fuel consumption
- The danger of stall and the need to maintain airspeed above the critical value.

By considering these factors, aircraft designers and pilots can work together to create safe and efficient flight experiences.


To find 9 percent of 31000, we need to calculate 9% * 31000. First, convert the percentage to a decimal: 9% = 0.09. Then, multiply 0.09 by 31000.

The calculation is as follows: 0.09 * 31000 = 2790.

For comparison, other percentages of 31000 are:
- 4% of 31000 = 0.04 * 31000 = 1240
- 5.7% of 31000 = 0.057 * 31000 = 1767

The scientific notation for 31000 is 3.1 x 10^4, because 31000 can be expressed as 3.1 multiplied by 10 raised to the power of 4.

Possible answers to the question "What is 9 percent of 31000?" are:
- 6820
- 1550
- 4960
- 6200
- 2790

The correct answer is 2790.


To quickly calculate the square of a number consisting of all 9s, a predictable pattern can be observed. For instance, 9^2 = 81, 99^2 = 9801, 999^2 = 998001, 9999^2 = 99980001, and 99999^2 = 9999800001.

This pattern suggests that to square a number consisting of n 9s, you can write (n-1) nines, 8, (n-1) zeros, and then 1. For example, to calculate 99^2, note that there are 2 digits, so the answer would have one 9, 8, one zero, and then 1: 9801.

To understand why this pattern works, consider the definition of squaring and multiplication as repeated addition. Take the case of 9999^2. This can be expressed as (9999)(9999), which is equivalent to (10000 - 1)(10000 - 1). By expanding this expression and simplifying, the pattern of three 9s, 8, three 0s, and 1 emerges.

This technique can be generalized to any number made up of all 9s by keeping track of the number of digits. When adding 1 to a number consisting of all 9s, the number of digits increases by 1 (e.g., 999 + 1 = 1000). This pattern may not be as straightforward for squaring similar types of numbers like 777 or 888.

By recognizing and applying this pattern, you can quickly calculate the square of any number consisting of all 9s, making it a useful tool for mental math calculations.


## Multiplying Polynomials

When learning to multiply polynomials, students typically start with binomials and the F.O.I.L. mnemonic. However, F.O.I.L. can obscure the underlying **distributive property**. The F.O. part distributes the first term of the first binomial to both terms of the second, and the I.L. part distributes the second term of the first binomial.

The F.O.I.L. method breaks down if either polynomial has more than two terms. An alternative technique, called the **Box Method**, can be used. This method involves labeling each rectangle with its area, which is the product of its side lengths. The areas are then added together.

For example, if we have two binomials, the Box Method can be used to multiply them. The orange rectangle has side lengths of $x$ and $y$, so its area is $xy$. The pink rectangle has side lengths of $x$ and $z$, so its area is $xz$. Once all rectangle areas are calculated, they are added together.

The Box Method is useful for keeping track of all terms, especially when multiplying more complicated polynomials. It also allows for easy identification of "like terms" that can be combined, which often line up along the diagonal.

There are two main methods for multiplying polynomials: the distributive method and the Box Method. The distributive method is suitable for those with good handwriting and attention to detail. However, the Box Method is recommended for those who are concerned about forgetting to multiply terms or wanting to easily identify the source of their results. Although it takes up more space, the Box Method reduces the likelihood of careless errors and makes it easier to find mistakes.

Key considerations when choosing a method include:

* The complexity of the polynomials being multiplied
* The individual's handwriting and attention to detail
* The need to identify and combine "like terms"

By understanding the **distributive property** and using either the distributive method or the Box Method, students can effectively multiply polynomials and minimize errors.


## Summary

The Curvature tool calculates the curvature of a raster surface, including optional profile and plan curvature, requiring a Spatial Analyst license.

## Usage

The tool's primary output is the curvature of the surface on a cell-by-cell basis, fitted through each cell and its eight neighbors. Curvature is the second derivative of the surface, or the slope-of-the-slope. Two optional output curvature types are possible: profile curvature, in the direction of the maximum slope, and plan curvature, perpendicular to the direction of the maximum slope.

A positive curvature indicates an upwardly convex surface, while a negative curvature indicates an upwardly concave surface. A value of 0 indicates a flat surface. The units of the curvature output raster, as well as the optional output profile and plan curve rasters, are one hundredth (1/100) of a z-unit.

## Key Concepts

- **Curvature**: The second derivative of the surface, or the slope-of-the-slope.
- **Profile Curvature**: Curvature in the direction of the maximum slope.
- **Plan Curvature**: Curvature perpendicular to the direction of the maximum slope.
- **Z-Factor**: Adjusts the units of measure for the z units when they are different from the x,y units of the input surface.

## Syntax

Curvature (in_raster, {z_factor}, {out_profile_curve_raster}, {out_plan_curve_raster})

- **in_raster**: The input surface raster.
- **z_factor** (Optional): Number of ground x,y units in one surface z unit.
- **out_profile_curve_raster** (Optional): Output profile curve raster dataset.
- **out_plan_curve_raster** (Optional): Output plan curve raster dataset.

## Return Value

- **out_curvature_raster**: The output curvature raster, which will be floating point type.

## Code Samples

### Curvature Example 1 (Python Window)

```python
import arcpy
from arcpy import env
from arcpy.sa import *
env.workspace = "C:/sapyexamples/data"
outCurve = Curvature("elevation", 1.094)
outCurve.save("C:/sapyexamples/output/outcurv01")
```

### Curvature Example 2 (Stand-alone Script)

```python
# Name: Curvature_Ex_02.py
# Description: Calculates the curvature of a raster surface.
# Requirements: Spatial Analyst Extension
import arcpy
from arcpy import env
from arcpy.sa import *
env.workspace = "C:/sapyexamples/data"
inRaster = "elevation"
zFactor = 1.094
arcpy.CheckOutExtension("Spatial")
outCurve = Curvature(inRaster, 1.094)
outCurve.save("C:/sapyexamples/output/outcurv02")
```

## Environments and Licensing

- Requires Spatial Analyst or 3D Analyst extension for ArcGIS for Desktop Basic, Standard, and Advanced.


We are optimizing the division operation within a decimal number implementation in Rust. Last time, we investigated long division. This week, we explore a second method for dividing binary numbers using the "two's complement" method.

The division process involves repeatedly subtracting the divisor from the dividend and incrementing the quotient until the remainder is obtained. For example, let's calculate $\dfrac{19}{6}$:

$$
19 - 6 = 13\;\;(q = 1) \\
13 - 6 = 7\;\;(q = 2) \\
7 - 6 = 1\;\;(q = 3)
$$

We subtracted $6$ from $19$ three times with a remainder of one:

$$
q = 3 \\
r = 1
$$

This process can be optimized using binary representation. The binary value $1101$ can represent different integers depending on the number representation used. As an unsigned integer, it represents $13$ using $2^0 + 2^2 + 2^3 = 13$. When dealing with signed numbers, the first bit indicates whether the number is positive or negative.

There are two common systems for representing signed numbers: one's complement and two's complement. In one's complement, $1101$ represents $-2$, while in two's complement, it represents $-3$. One's complement involves reversing the bits to get the negative representation. For example, $2$ is represented as $0010$ in binary, and reversing the bits gives $1101$, representing $-2$.

However, one's complement has a limitation: it requires representing both $0$ and $-0$, which seems redundant since $0 = -0$. This leads to the need for two's complement, which is the most common method for representing signed integers in computer systems. Two's complement involves inverting the bits and adding one to get the negative representation.

For example, $0001$ becomes $1111$ (i.e., $1110 + 0001$) when using two's complement. This increases the range of numbers that can be represented. Applying this logic to $0000$, we get $1,0000$, which overflows, but discarding the carry bit gives us $0000$ as expected.

Two's complement allows us to avoid subtraction and instead perform simple addition. For example, to subtract $1$ from $5$, we calculate the two's complement of $-1$ as $1111$ and add it to $5$ ($0101$):

$$
\begin{array}{r}
\phantom{+}0101 \\[-3pt]
+ \underline{1111} \\[-3pt]
0100 \\[-3pt]
\end{array}
$$

This gives us the expected result: $5 + -1 = 4$.

Using two's complement, we can perform division by repeatedly adding the two's complement of the divisor to the dividend. Let's use the example of $\dfrac{19}{6} = 3;;(r = 1)$. We represent $19$ as $0001,0011$ and $6$ as $0000,0110$. The two's complement of $-6$ is $1111,1010$.

We add these numbers together and discard the carry bit:

$$
\begin{array}{r}
\phantom{+}0001\,0011 \\[-3pt]
+ \underline{1111\,1010} \\[-3pt]
0000\,1101 \\[-3pt]
\end{array}
\\
q = 0001
$$

Since $0000,0110$ is smaller than $0000,1101$, we try again:

$$
\begin{array}{r}
\phantom{+}0000\,1101 \\[-3pt]
+ \underline{1111\,1010} \\[-3pt]
0000\,0111 \\[-3pt]
\end{array}
\\
q = 0010
$$

Once again, $0000,0110$ is smaller than $0000,0111$, so we try again:

$$
\begin{array}{r}
\phantom{+}0000\,0111 \\[-3pt]
+ \underline{1111\,1010} \\[-3pt]
0000\,0001 \\[-3pt]
\end{array}
\\
q = 0011
$$

We can now see that $0000,0110$ is bigger than $0000,0001$, so we stop. As expected, we have a quotient of $0011$ ($3$) and a remainder of $0001$ ($1$).

Next time, we will explore how to implement integral division using the Rust language and investigate representing high-precision fractions.