**Algebra II Unit #3: Polynomial Functions 5.1 Notes**

**Vocabulary:**

* A polynomial is a monomial (1-term function) or sum of monomials.
* The degree of a polynomial is the greatest exponent among its terms.
* The standard form of a polynomial arranges the terms by degree in descending numerical order.

**Classifying Polynomials:**

Classify each polynomial by degree and number of terms:

1. 4x + x + 2
2. 3 + 3x
3. 4x^5
4. x^2 + 3x - 4x^3
5. (3b^5)
6. 3m + 5m^2
7. 6(2x - 1)
8. b(b - 3)^2
9. a^3(a^2 + a + 1)
10. x(x + 5) - 5(x + 5)
11. p(p - 5) + 6
12. (3c^2)^2

**Open-Ended:**

Create your own polynomial with the following characteristics:
a) quartic trinomial
b) cubic binomial
c) quintic monomial

**Graphing Calculator Exploration:**

Sketch a graph of each function and identify the left and right end behavior:

a) f(x) = x^1, f(x) = -x^1
b) f(x) = x^2, f(x) = -x^2
c) f(x) = x^3, f(x) = -x^3
d) f(x) = x^4, f(x) = -x^4
e) f(x) = x^5, f(x) = -x^5
f) f(x) = x^6, f(x) = -x^6
g) f(x) = x^7, f(x) = -x^7
h) f(x) = x^8, f(x) = -x^8

**Writing:**

Compare the graphs above. What patterns do you notice? Which functions had matching end-behaviors? Which functions had opposite end-behaviors? What does the negative determine about the end behavior of a function?

**Practice:**

Use your calculator to sketch a graph of each polynomial function and determine the end behavior:

1. f(x) = -4x^2
2. f(x) = -5x^3 + 9
3. f(x) = -4x^4
4. f(x) = -8x^5 - 1
5. f(x) = 3x^4 + 6x^3 - x^2 + 12
6. f(x) = 50 - 3x^3 + 5x^2
7. f(x) = x^7 + x^2 + 2
8. f(x) = 12x^4 - x + 3x^7 - 1
9. f(x) = 5 + 2x + 7x^2 - 5x^3
10. f(x) = 20 - 5x^6 + 3x - 11x^3

**Vocabulary:**

* The lead coefficient is the coefficient on the term of highest degree.
* Circle the lead term on each function above.

**Key Concept:**

The degree of a polynomial affects the shape of the graph, end behavior, and maximum number of turning points.

**Turning Points:**

* Places where the graph changes directions (increasing vs. decreasing)
* Also known as local maximums and/or local minimums or relative max/min
* For a polynomial with degree n, there can be at most n - 1 turning points

**Practice:**

Sketch a graph of each function. Describe the shape including end behavior, turning points, and increasing/decreasing intervals:

1. f(x) = 2x^3 - 5x^2 + 1
2. f(x) = 3x^3 - x^2 - 3
3. f(x) = -9x^3 - 2x^2 + 5 + 3
4. f(x) = -0.5x^4 + 3x + 3x^2

**Writing:**

Why is it easiest to determine the degree, end-behavior, and maximum number of turning points from standard form? Use an example to support your answer.


The `fractions` module in Python is a useful tool for addressing the shortcomings of floating-point arithmetic in binary, particularly when working with financial data or requiring infinite precision for calculations.

### Key Concepts

*   **Decimal vs Fractional Notation**: Decimal notation is more intuitive and resembles a percentage, while fractional notation is more convenient for performing symbolic algebra by hand.
*   **Rational Numbers**: Rational numbers can be expressed as a quotient of two integers, with a non-zero denominator. They can have terminating or recurring decimal expansions.
*   **Irrational Numbers**: Irrational numbers have non-terminating and non-repeating decimal expansions, making them difficult to represent exactly in computer memory.

### Creating Fractions

*   **From Integers**: Fractions can be created from integers using the `Fraction` class, which accepts zero, one, or two arguments of various types.
*   **From Floating-Point Numbers**: Fractions can be created from floating-point numbers, but this may result in lossy conversions due to the representation error that affects `float` numbers.
*   **From Decimal Numbers**: Fractions can be created from decimal numbers using the `Fraction` class, which can help retain maximum precision.
*   **From Strings**: Fractions can be created from strings in decimal or fractional notation, with optional plus or minus signs and exponents.

### Inspecting Fractions

*   **Numerator and Denominator**: The `numerator` and `denominator` attributes of a fraction can be accessed to inspect its components.
*   **Immutable**: Fractions are immutable, meaning their internal state cannot be changed after creation.

### Converting Fractions

*   **To Floating-Point Numbers**: Fractions can be converted to floating-point numbers using the `float()` function, but this may result in lossy conversions.
*   **To Decimal Numbers**: Fractions can be converted to decimal numbers by dividing the numerator by the denominator, wrapped in a `Decimal` object.
*   **To Strings**: Fractions can be converted to strings using the `str()` or `repr()` functions, with the latter outputting a piece of Python code.

### Performing Rational Number Arithmetic

*   **Addition**: Fractions can be added using the `+` operator, which finds the common denominator and adds the numerators.
*   **Subtraction**: Fractions can be subtracted using the `-` operator, which finds the common denominator and subtracts the numerators.
*   **Multiplication**: Fractions can be multiplied using the `*` operator, which multiplies the numerators and denominators.
*   **Division**: Fractions can be divided using the `/` operator, which inverts the divisor and multiplies.

### Rounding Fractions

*   **Truncation**: Fractions can be truncated using the `int()` function or the `math.trunc()` function, which removes the fractional part.
*   **Rounding Half to Even**: Fractions can be rounded using the `round()` function, which rounds to the nearest whole number with a bias towards even numbers.

### Comparing Fractions

*   **Comparison Operators**: Fractions can be compared using comparison operators (`<`, `>`, `==`, etc.), which work as expected.

### Choosing Between Fraction, Decimal, and Float

*   **Binary Floating-Point**: `float` is suitable for most situations, with unparalleled speed due to its implementation in hardware.
*   **Decimal Floating-Point and Fixed-Point**: `Decimal` is suitable for financial calculations, with arbitrary yet finite precision.
*   **Infinite Precision Rational Number**: `Fraction` is suitable for representing rational numbers exactly, with infinite precision bounded only by available memory.

### Practical Applications

*   **Approximating Irrational Numbers**: Fractions can be used to approximate irrational numbers, such as pi, using continued fractions.
*   **Getting a Display's Aspect Ratio**: Fractions can be used to calculate the aspect ratio of a display, given its width and height.
*   **Calculating the Exposure Value of a Photo**: Fractions can be used to calculate the exposure value of a photo, given the aperture, shutter speed, and ISO speed.
*   **Solving the Change-Making Problem**: Fractions can be used to solve the change-making problem, given a set of coins and a target amount.
*   **Producing and Expanding Continued Fractions**: Fractions can be used to produce and expand continued fractions, which can be used to approximate irrational numbers.


To find the solution, first, we need to determine the component of the velocity in which it is acting. Then, apply the formula for the frequency according to the Doppler’s effect, solve the equation obtained, and conclude with the correct solution.

**Step-by-Step Answer:**

1. Draw a figure according to the given conditions, which forms a right-angled triangle. The hypotenuse, $h$, is calculated as $h = \sqrt {{3^2} + {4^2}} = \sqrt {25} = 5m$. In this triangle, $\cos \theta = \dfrac{b}{h} = \dfrac{3}{5}$.

2. Find the horizontal component of the source of the sound, which is traveling along a road. The horizontal component of the velocity of the source, ${v_s}$, is given by ${v_s} = v\cos \theta = \dfrac{{100}}{3} \times \dfrac{3}{5} = 20m{s^{ - 1}}$.

3. The formula for frequency in the case of Doppler’s effect is $f' = f\left( {\dfrac{{v \pm {v_o}}}{{v \pm {v_s}}}} \right)$. Since the observer is not moving, the velocity of the observer, ${v_o}$, is zero. Substituting the values into the equation yields $f' = 640\left( {\dfrac{{340 - 0}}{{340 - 20}}} \right) = 640 \times \dfrac{{340}}{{320}} = 680Hz$.

Therefore, the required value of the frequency is $680Hz$, which corresponds to option (B).

**Doppler’s Effect:** It is the effect observed when either the source of a wave moves or the observer moves relative to each other, resulting in an increase or decrease in the frequency of the wave.


In our discussion of numbers, functions, and graphs, a function is defined as a rule relating an input set of numbers to an output set. We will focus on specific functions arising from the study of the relationship between lines and angles, known as trigonometry. Specifically, we will examine the sine, cosine, and tangent functions.

To define these functions, we need to establish some fundamental terms. A triangle is a three-sided figure formed by the intersection of three nonparallel straight lines. A right triangle is a triangle with one right angle (90 degrees). The relationships among the sides and angles of a right triangle form the basis for the sine and cosine functions.

In a right triangle, the side opposite the right angle is called the hypotenuse. Let's label one of the acute angles "A" and the other "B". The side opposite angle "A" is labeled "a", and the side opposite angle "B" is labeled "b".

With these elements, we can define the functions:
- The sine of angle "A", sin(A), is the fraction "a" divided by the hypotenuse: sin(A) = a/hypotenuse.
- The cosine of angle "A", cos(A), is the fraction "b" divided by the hypotenuse: cos(A) = b/hypotenuse.
- The tangent of angle "A", tan(A), is the fraction "a" divided by "b": tan(A) = a/b.

These functions can also be thought of in terms of their positions relative to angle "A":
- Sine is opposite over hypotenuse.
- Cosine is adjacent over hypotenuse.
- Tangent is opposite over adjacent.

As an exercise, define the sine, cosine, and tangent of angle "B" using the same principles.


# Difference between revisions of "2005 AIME II Problems/Problem 15"

The problem involves finding the smallest possible value of $a$ for which the line $y = ax$ contains the center of a circle that is externally tangent to circle $A$ and internally tangent to circle $B$.

Given that $a = \frac{m}{n}$, where $m$ and $n$ are relatively prime integers, we need to find $m + n$.

To approach this problem, we start by rewriting the given equations of circles $A$ and $B$ as $(x - 3)^2 + y^2 = 100$ and $(x + 9)^2 + y^2 = 225$, respectively.

Let circle $C$ have center $(x, y)$ and radius $r$. Now, if two circles with radii $r_1$ and $r_2$ are externally tangent, then the distance between their centers is $r_1 + r_2$, and if they are internally tangent, it is $|r_1 - r_2|$.

So, we have the following equations:
$\sqrt{(x - 3)^2 + y^2} = r + 10$
$\sqrt{(x + 9)^2 + y^2} = |r - 15|$

Solving for $r$ in both equations and setting them equal yields
$\sqrt{(x - 3)^2 + y^2} - 10 = \sqrt{(x + 9)^2 + y^2} - 15$
or
$\sqrt{(x - 3)^2 + y^2} - 10 = 15 - \sqrt{(x + 9)^2 + y^2}$

Squaring both sides, canceling common terms, and rearranging yields
$1 = \frac{x^2}{100} + \frac{(y - 12)^2}{75}$

Squaring again and canceling yields
$1 = \frac{x^2}{100} + \frac{(y - 12)^2}{75}$

So, the locus of points that can be the center of the circle with the desired properties is an ellipse.

Since the center lies on the line $y = ax$, we substitute for $y$:
$1 = \frac{x^2}{100} + \frac{(ax - 12)^2}{75}$

Expanding yields
$1 = \frac{x^2}{100} + \frac{a^2x^2 - 24ax + 144}{75}$

We want the value of $a$ that makes the line tangent to the ellipse, which will mean that for that choice of $a$ there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is $0$.

Solving yields $a = \frac{12}{5}$, so the answer is $12 + 5 = 17$.

However, the given solution states that $a = \frac{12}{5}$ is incorrect and that the correct value of $a$ is $\frac{12}{5} = \frac{m}{n}$ where $m + n = 169$.

The correct solution is as follows:
Solving for $a$ yields $a^2 - \frac{24}{5}a - \frac{64}{3} = 0$.
Using the quadratic formula, $a = \frac{24 \pm \sqrt{576 + 768}}{10}$.
Simplifying, $a = \frac{24 \pm \sqrt{1344}}{10}$.
Further simplification yields $a = \frac{24 \pm 4\sqrt{84}}{10}$, which simplifies to $a = \frac{12 \pm 2\sqrt{84}}{5}$.
Since $a$ must be positive, we take the positive root, $a = \frac{12 + 2\sqrt{84}}{5}$.
Simplifying, $a = \frac{12 + 4\sqrt{21}}{5}$.
The value of $a$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime integers.
We need to find $m + n$.
The given solution states that $m + n = 169$.

The final answer is indeed $m + n = 169$.


#### Date of Award

1996

#### Document Type

Dissertation

#### Degree Name

Doctor of Philosophy (PhD)

#### Department

Mathematics

#### First Advisor

Robert F. Lax

#### Abstract

Let $A={\bf F}\sb{q}\lbrack x\sb1,\ ...,x\sb{s}\rbrack$ be a polynomial ring and $I$ an ideal of $A$. The set of ${\bf F}\sb{q}$-rational points of $V(I)$ is denoted by $P\sb1,\ ...,P\sb{n}$. A map $\varphi:A/I\to{\bf F}\sbsp{q}{n}$ is defined as $\varphi(\tilde f)=(f(P\sb1),\ ...,f(P\sb{n}))$, where $f$ is a preimage of $\tilde f$ under the canonical map from $A$ to $A/I$.

A basis $\{\tilde f\sb{i}\vert i\in {\bf N}\}$ of $A/I$ as an ${\bf F}\sb{q}$-vector space is used to define the affine variety code $C$ and its dual $C\sp\perp$. The code $C$ is given by $C=\varphi(\langle\tilde f\sb1,\ ...,\tilde f\sb{m}\rangle)$, and $C\sp\perp$ is the orthogonal complement of $C$ with respect to the usual inner product in ${\bf F}\sbsp{q}{n}$.

This dissertation shows that any linear code can be expressed as an affine variety code. When a code $C$ is represented as an affine variety code, problems of decoding and finding the minimum distance of $C$ can be expressed as questions about polynomial ideals. Using the theory of Grobner bases and computer programs that calculate Grobner bases, it is possible to decode and find the minimum distance of any linear code.

#### Recommended Citation

Fitzgerald, Jeanne, "Applications of Grobner Bases to Linear Codes." (1996). *LSU Historical Dissertations and Theses*. 6248.

https://repository.lsu.edu/gradschool_disstheses/6248

#### ISBN

9780591133516

#### Pages

80

#### DOI

10.31390/gradschool_disstheses.6248


Isomorphic Triangulation of Simple Polygons

Isomorphism is a map that preserves sets and relations among elements.

Spiderweb connectivity is a method for connecting a set of vertices, each with a weight representing the distance to a center point. For example, given 9 vertices with weights <3, 1, 4, 3, 3, 4, 2, 4, 3>, extra points (Steiner points) are inserted to ensure the minimum path from each vertex to the center point matches the weight.

The connectivity is layered, with Steiner points connecting to neighboring vertices in each layer. If a neighboring vertex is not present (due to a smaller weight), the point connects to its nearest neighbor. The path from each vertex to the center point is connected through Steiner points, resulting in a spiderweb-like connection. This method can be applied to any set of vertices and weights, yielding triangles and quadrilaterals, which can be triangulated.

A Steiner point is an extra point added to a vertex set that was not part of the original set, also known as a Steiner vertex. The spiderweb connectivity can be visualized using a figure, such as Figure 1, which illustrates the connectivity for the sequence <3, 1, 4, 3, 3, 4, 2, 4, 3>, starting at the red vertex.


### NBA: Does Taking More Shots Lead to Lower Accuracy?

A hypothesis suggests that good players are asked to take worse shots, such as desperation field goal attempts, which can lower their field goal percentage (FG%). To test this, a regression analysis was conducted using 2008-09 NBA data. The analysis predicted a team/position's effective field goal percentage (eFG%) using the following factors:

1. Position
2. The rest of the team's FG% (averaged by position)
3. The percentage of the team's field goal attempts (FGA) the player took

The resulting equation is:

FG% = .181 + .022 (if SG) + .016 (if SF) + .025 (if PF) + .043 (if C) + .643 * (eFG% of the rest of the team) - .114 * (percentage of team's FGA taken)

The results support the hypothesis, indicating that taking more shots leads to a lower overall percentage. For example, if a player takes 20% of their team's attempts and shoots .500, reducing their attempts to 19% would result in a .5114 shooting percentage. Conversely, increasing their attempts to 21% would result in a .4886 shooting percentage.

This translates to a significant difference in field goals made. A .500 percentage over 20 shots is 10 field goals, while a .5114 percentage over 19 shots is 9.727 field goals. The extra shot would only result in 0.283 field goals, suggesting that the player shoots .283 on those extra shots.

The analysis also reveals a high correlation between a player's position and their teammates' shooting percentage. For every percentage point the teammates shoot better than average, the individual player will shoot 0.643 points better than average.

However, the coefficient for the percentage of team's FGA taken (-.114) was only significant at the 7.2% level, with a standard error of 63 points. This suggests that more data is needed to confirm the "shoots more, therefore shoots less accurately" effect.

The comments section raises several points, including:

* The interpretation of the .283 shooting percentage on extra shots, with some arguing that it's not a meaningful measure.
* The potential for non-linear relationships between shot attempts and shooting percentage, with different marginal slopes at different points in the range.
* The impact of defensive strategies and player roles on shot selection and shooting percentage.
* The need for more data to confirm the findings and explore these issues further.

Some potential avenues for further research include:

* Analyzing shots taken in specific situations, such as the last 3 seconds of the shot clock.
* Examining the impact of defensive strategies on shot selection and shooting percentage.
* Investigating the relationship between shot attempts and shooting percentage at the player level, including the potential for non-linear relationships and varying skill curves by age, player role, and position.


## Breadth-First Search, Depth-First Search, Shortest Path, Arbitrage, Strongly Connected Components and Maximum Flow

## Requirements

- Internet connection
- Basic knowledge of data structures

## Description

This course covers advanced algorithms, focusing on graph traversal, shortest path problems, spanning trees, and maximum flow problems, with applications in web crawling, stock market arbitrage, and more.

**Section 1 – Graph Theory Basics**

- Introduction to graphs: G(V,E), adjacency matrix, and adjacency list representations

**Section 2 – Graph Traversal (Breadth-First Search)**

- Breadth-first search (BFS) fundamentals
- Using BFS for web crawling in search engines

**Section 3 – Graph Traversal (Depth-First Search)**

- Depth-first search (DFS) basics
- Implementing DFS using recursion
- Applications: topological ordering, cycle detection, and maze solving

**Section 4 – Topological Ordering**

- Topological ordering (topological sort) and directed acyclic graphs (DAGs)
- DAG shortest and longest paths
- Critical path methods and project management

**Section 5 – Cycle Detection**

- Cycles in graphs: forward and backward edges
- Cycle detection algorithms, including Tarjan's algorithm with DFS

**Section 6 – Dijkstra’s Shortest Path Algorithm**

- Shortest path in a G(V,E) graph
- Dijkstra's shortest path algorithm

**Section 7 – Bellman-Ford Shortest Path Algorithm**

- Bellman-Ford algorithm
- Handling negative cycles and finding arbitrage opportunities on the FOREX

**Section 8 – Spanning Trees (Kruskal and Prim’s Algorithms)**

- Spanning trees and union-find data structures
- Kruskal's and Prim's algorithms

**Section 9 – Strongly Connected Components (SCCs)**

- Strongly connected components and Kosaraju's algorithm
- Tarjan's algorithm for SCCs

**Section 10 – Maximum Flow Problem**

- Maximum flow problem and reducing hard problems to maximum flow
- Ford-Fulkerson algorithm and bipartite matching problem

**Section 11 – Additional Topics**

- Travelling Salesman Problem (TSP) and Hamiltonian cycles
- Eulerian paths and cycles, Hierholzer algorithm, and the Chinese Postman Problem
- Algorithms analysis: running time measurement, big O, Ω, and θ notation, complexity classes, and polynomial and non-deterministic polynomial algorithms

The course requires approximately **11 hours** to complete. To reinforce understanding, it is recommended to implement the algorithms multiple times. The source code for the entire course is available in the last lecture.

This course is suitable for anyone interested in advanced algorithmic topics, from software engineers to scientific researchers.


The inverse CDF for the normal distribution, also known as the "quantile" function, does not have a closed-form solution. The statistics module from CPython relies on the rational approximation described in Wichura's Algorithm AS 241. This algorithm is also used by the authors of the Racket math module, with additional asymptotic expansions and a Newton iteration to improve accuracy.

The algorithm's accuracy can be verified using test cases:
$$ \begin{equation} \begin{aligned} z_{0.25} &= -0.6744897501960817,\cr z_{0.001} &= -3.090232306167814,\cr z_{10^{-20}} &= -9.262340089798408\cr \end{aligned} \end{equation} $$

In Racket, the results are:
```
> (require math/distributions)
> (for ([q (in-list '(0.25 0.001 10.0e-20))])
(displayln (flnormal-inv-cdf 0.0 1.0 q #f #f)))
-0.6744897501960817
-3.090232306167813
-9.013271153126675
```

However, the last value does not match the expected result. Similarly, in R, `qnorm(10e-20)` returns the same incorrect result. In contrast, Mathematica returns the expected result:
```
In[10]:= N[InverseCDF[NormalDistribution[], 10^-20], 16]
Out[10]= -9.262340089798408
```

When working with low p-values or suspecting floating-point arithmetic issues, it is essential to double-check results. In R, the `lower.tail` argument in the `p*` and `q*` family functions can help correct for catastrophic cancellation, even in non-symmetric distributions.


Fisher's exact test is a statistical significance test used to analyze contingency tables, despite being a topic of controversy for over 80 years. This test is applied to evaluate the statistical significance of the difference between two bisulfite sequences from control and treatment groups at each CG site.

The test determines if there are nonrandom associations between two categorical variables, X and Y. Variable X represents two groups: control and treatment, while variable Y denotes the methylation status, indicating the number of times a cytosine site is found methylated (mCG) and non-methylated (CG).

This information is summarized in a 2x2 table:

|          | Y = mCG | Y = CG | R_i |
|----------|---------|--------|-----|
| Control  | a11     | a12    | a11+a12 |
| Treatment| a21     | a22    | a21+a22 |
| C_i      | a11+a21 | a12+a22| N      |

where N = ∑iR_i = ∑jC_j, and R_i and C_j are the row and column sums, respectively.

The conditional probability of getting the actual matrix, given the particular row and column sums, is calculated using the formula:

P_cutoff = (R1!R2! / N!) * (C1!C2! / ∏i,j a_ij!)

This formula is used to determine the statistical significance of the difference between the control and treatment groups. Fisher's exact test is a useful tool for analyzing contingency tables and determining nonrandom associations between categorical variables.


**Fundamentals of Calculus: Differential Equations**

Calculus evolved to support differential equations, which describe the rate of change of quantities. A key fact about differential equations is that they don't have constants to describe their graphical structure. Differentiation eliminates constants, allowing us to frame differential equations that elaborate on the actual rate of change.

For example, given $z = ax^2 + bx + c$, we can find $\frac{dz}{dx} = 2ax + b$ and $\frac{d^2z}{dx^2} = 2a$. The expression $\frac{d^3z}{dx^3} = 0$ best describes a differential equation without constants.

A normal function, such as $z = x^2 + 2x + 3$, represents a particular behavior. A family of equations, $z = ax^2 + bx + c$, expresses a family of such equations without a specified shape or orientation. The rate of change, or differential, remains ambiguous unless the constants are specified.

Consider $z = x^2 + 2x + c$, a second-degree equation with one constant. We can eliminate this constant by writing $\frac{dz}{dx} = 2x + 2$, which serves as our differential equation.

Let's analyze $z = ax^2 + 2x + 3$. Differentiating $z$ with respect to $x$ twice, we get $\frac{d^2z}{dx^2} = 2a$. Substituting this into the original expression, we obtain $z = \frac{d^2z}{2dx^2}x^2 + 2x + 3$ as our differential equation. This means $z$ is a function of $x$ for which the rate of change of the rate of change is unknown, but the rate of change and orientation are known.

A real-world example is an object moving along a single axis with varying velocity. By plotting the change in position with respect to time, we can write $z = f(t)$. If the mass of the object is $m$, the force it provides along the Z-axis can be represented by the differential equation $F = m\frac{d^2z}{dt^2}$.

Another example is $z = \frac{1}{6} \times \frac{d^2z}{dx^2}x^2$. This function can be represented as $z = ax^3$. A more challenging example is $z = x^n$, where $n$ is an unknown constant. Differentiating this $n$ times with respect to $x$ gives a differential equation, but does not eliminate the constant $n$.

**Example 1: Express $z = x^n$ as a differential equation.**

**Solution:** We can eliminate the unknown constant $n$ by differentiating and obtaining different functions. Applying this, we get $\frac{dz}{dx} = nx^{n-1}$, which can be written as $z = x^{\frac{x}{z}\times \frac{dz}{dx}}$.

**Alternate Solution:** Taking the logarithm of both sides, we get $log z = n log x$. Differentiating, we obtain $\frac{dz}{dx}\times \frac{1}{z} = \frac{n}{x}$. Rearranging, we get $log z = x\frac{dz}{dx}\frac{1}{z}log x$, eliminating the constant $n$.


## 12. Electricity

### Connecting Resistors to a Circuit

The current through a conductor depends on its resistance and the potential difference across its ends. There are two methods of joining resistors together: series and parallel.

**(i) Series:** When two or more resistances are joined end to end, the same current flows through each of them. The total equivalent resistance in series is the sum of individual resistances.

Mathematically, this is represented as:
V = V₁ + V₂ + V₃
V = IR
V₁ = IR₁, V₂ = IR₂, V₃ = IR₃
V₁ + V₂ + V₃ = I(R₁ + R₂ + R₃)
R = R₁ + R₂ + R₃

**(ii) Parallel:** When two or more resistances are connected across two points, each provides a separate path for current. The total equivalent resistance in parallel is less than any individual resistance.

When resistors R₁, R₂, and R₃ are connected in parallel, the current flowing through each resistor is different, but the potential difference across each resistor is the same.

### Effect on Current

- **Series:** When two resistors R₁ and R₂ of 3 Ω and 5 Ω are connected in series to a 6 V battery, the total resistance is R = R₁ + R₂ = 3 Ω + 5 Ω = 8 Ω, and the current is I = V/R = 6/8 = 0.75 A.
- **Parallel:** When the same devices are connected in parallel, the resulting total resistance (R) and current (I) can be calculated and compared in a table:

| Connection | Resistance | Current |
| --- | --- | --- |
| Series | 8 Ω | 0.75 A |
| Parallel | 1.875 Ω | 3.2 A |

### Battery and Ohm's Law

- A battery is a connection of two or more cells.
- Ohm's Law states that the electric current flowing through a metallic wire is directly proportional to the potential difference V across its ends, provided its temperature remains the same: V = IR.

### Heating Effect of Electric Current

The heating effect of electric current occurs when some of the energy from the source is expended in heat, raising the temperature of a gadget. In a purely resistive circuit, the source energy is dissipated entirely in the form of heat.

### Joule's Law of Heating

Joule's Law states that the heat produced in a resistor is directly proportional to the square of the current, directly proportional to resistance for a given current, and directly proportional to the time for which the current flows through the resistor: H = I²Rt.

This law is established based on the principles that the heat produced is:
(i) Directly proportional to the square of the current (I) for a given resistance.
(ii) Directly proportional to resistance (R) for a given current.
(iii) Directly proportional to the time (t) for which the current flows through the resistor.


Circle Theorems
Key concepts about angles and circles are explored in this section.

**Inscribed Angle**
An inscribed angle is formed by points on a circle's circumference. It has an apex point (B) and end points (A and C). Two important theorems related to inscribed angles are:

1. **Angle at the Center Theorem**: An inscribed angle (a°) is half of the central angle (2a°).
2. **Angles Subtended by Same Arc Theorem**: The inscribed angle (a°) remains the same, regardless of its position on the circumference, as long as the endpoints are fixed.

**Example 1**: Find the size of Angle POQ, where O is the circle's center.
Given that Angle PRQ = 62°, we can apply the Angle at the Center Theorem: Angle POQ = 2 × Angle PRQ = 2 × 62° = 124°.

**Example 2**: Find the size of Angle CBX.
Since Angle ADB = 32°, and Angle ADB = Angle ACB = Angle XCB, we can determine that Angle XCB = 32°. Using the fact that angles in a triangle add up to 180°, and given that Angle BXC = 85°, we can find the remaining angle.

**Angle in a Semicircle**
An angle inscribed in a semicircle is always a right angle (90°). This is because the inscribed angle is half of the central angle (180°), according to the Angle at the Center Theorem. Alternatively, rotating the shape by 180° creates a rectangle, which has internal right angles.

**Example 3**: Find the size of Angle BAC.
Using the Angle in the Semicircle Theorem, we know that Angle ACB = 90°. Applying the fact that angles in a triangle add up to 180°, we can find Angle BAC.

**Cyclic Quadrilateral**
A cyclic quadrilateral has all its vertices on a circle's circumference. The opposite angles of a cyclic quadrilateral add up to 180°.

**Example 4**: Find the size of Angle WXY.
Since opposite angles of a cyclic quadrilateral add up to 180°, we can determine the size of Angle WXY.

**Tangent Angle**
A tangent is a line that touches a circle at one point, forming a right angle with the circle's radius.


To find the area of a triangle with sides of lengths 11, 14, and 18 using Heron's formula, we use the formula:

Area = √(s(s-a)(s-b)(s-c))

where s is the semi-perimeter, calculated as s = (a + b + c) / 2, and a, b, and c are the lengths of the sides.

Given the sides of the triangle as a = 11, b = 14, and c = 18, we calculate the semi-perimeter:

s = (11 + 14 + 18) / 2

Then, we substitute the values into Heron's formula to find the area.


## Comparing Two Means with Log-Transformed Data

When comparing the means of two groups with skewed data, log-transformation can be used to normalize the data. However, this approach has its limitations. In this section, we will explore the process of comparing two means using log-transformed data and discuss the pros and cons of this method.

### Simulating Data

To demonstrate this concept, we will simulate data from a randomized controlled trial where the outcome is the number of days to discharge from the hospital. We will generate 100 random values for the control and treated groups from log-normal distributions with identical locations but different shapes.

### Log-Transforming Data

After generating the data, we will log-transform the days to discharge. This transformation helps to normalize the data, making it more suitable for comparison. We will then plot histograms of the original and log-transformed data to visualize the effect of the transformation.

### Simple Linear Regression

To compare the means of the two groups, we will use simple linear regression. We will create a Pandas dataframe with the log-transformed data and use the `statsmodels` library to perform the regression. The regression output will provide us with the estimate of the between-group difference and its precision, which is more informative than a p-value.

### Interpreting Regression Output

The key pieces of information from the regression output are the intercept and slope. The intercept represents the mean of the control group, while the slope represents the effect of treatment. We can use these values to calculate the mean of the treated group and the between-group difference.

### Back-Transforming Log-Transformed Data

When we back-transform the log-transformed data, we need to be careful in interpreting the results. The back-transformed between-group difference and 95% CI represent the ratio of the anti-logged means of the groups, not the difference in days.

### Pros and Cons of Log-Transforming

The advantage of comparing two means by log-transforming non-normal data is that it helps to normalize the data. However, the disadvantage is that the between-group difference and 95% CI may be difficult to interpret, either in logged units or as a ratio of anti-logged means.

### Example Code

The following Python code demonstrates the process of comparing two means using log-transformed data:
```python
import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import lognorm
import pandas as pd
import statsmodels.formula.api as smf

# Generate data
np.random.seed(15)
days_con = lognorm.rvs(s=0.8, scale=4, loc=0, size=100)
days_trt = lognorm.rvs(s=0.4, scale=4, loc=0, size=100)

# Log-transform data
lndays_con = np.log(days_con)
lndays_trt = np.log(days_trt)

# Create dataframe
treated = np.concatenate([np.zeros(len(days_con), dtype=int), np.ones(len(days_trt), dtype=int)])
days = np.concatenate([days_con, days_trt])
lndays = np.concatenate([lndays_con, lndays_trt])
df = pd.DataFrame({'treated': treated, 'days': days, 'lndays': lndays})

# Perform simple linear regression
md = smf.ols('lndays ~ treated', data=df)
print(md.fit().summary())

# Back-transform results
effect = md.fit().params.treated
ci_ll = md.fit().conf_int()[0].treated
ci_ul = md.fit().conf_int()[1].treated
print(f'\nExponentiated values: '
      f'\nEffect = {np.exp(effect):.2f} '
      f'\n95% CI = {np.exp(ci_ll):.2f} to {np.exp(ci_ul):.2f}')
```
This code generates data, log-transforms it, performs simple linear regression, and back-transforms the results. The output provides the estimate of the between-group difference and its precision, as well as the back-transformed results.


**Demonstration of Math Applets**

The demonstration module illustrates the use of two applets: the Java Math Pad Applet and the PlotSolve Applet. Each applet comes with instructions and a detailed help menu.

**Java Math Pad Applet**

This applet allows users to evaluate mathematical expressions, which are made up of numbers, variables, built-in and user-defined functions, and operations.

* **Example 1: Entering and Evaluating Mathematical Expressions**
Evaluate the expression 3ab + 4c^3 when a = 2, b = 3, and c = 5.
	1. Assign values to variables: type a = 2, press enter; type b = 3, press enter; type c = 5, press enter.
	2. Type the expression using * to indicate multiplication and ^ for exponents, and press enter.
* **Example 2: Defining and Evaluating Functions**
Evaluate the function f(x) = 3x^2 + 17x - 33 when x = 2.4.
	1. Click the Clear button to clear the screen.
	2. Type f(x) = 3*x^2 + 17*x - 33 and press enter to define the function.
	3. Type f(2.4) and press enter to evaluate the function.
* **Example 3: Evaluating Trigonometric Functions**
Find the value of each trigonometric expression:
	(a) sin(2.14)
	(b) cos(π/4)
	(c) arcsin(0.5)
	1. Use built-in trigonometric functions in Java Math Pad.
	2. Ensure arguments are enclosed in parentheses and expressed in radians.
* **Example 4: Combining Functions to Form a New Function**
Given functions f(x) = 2x + 5, g(x) = 3x^2 + 2, and h(x) = √(x-1), evaluate:
	(a) f(2) + g(2)
	(b) f(1)/g(1)
	(c) h(f(6))
	1. Define each function using the syntax similar to the Java Math Engine.
	2. Use the defined functions to evaluate the given expressions.

**PlotSolve Applet**

This applet plots functions and solves equations numerically. It can plot up to 10 different functions at the same time.

* **Viewing Window Parameters**
Specify the desired values for XMin, XMax, YMin, and YMax. The applet performs basic error checking.
* **Function Information**
Enter functions to be plotted using the New Function button. The syntax is similar to the Java Math Engine.
* **Zoom and Trace**
Zoom in by left-clicking and dragging the mouse to form a rectangular region. Trace by single left-clicking in the graph area.
* **Control Buttons**
Clear All: erases all function definitions. Plot: updates the plot area after changes.
* **Messages**
The applet communicates with the user through error messages and information displayed in the Messages area.

**Example 1: Graphing a Function**
Graph the function f(x) = 3x^3 - 2x^2 + x + 5.
	1. Click the New Function button and type the expression.
	2. Set the viewing window: xmin = -5, xmax = 5, ymin = -5, ymax = 12.
	3. Click the Plot button.
* **Example 2: Solving an Equation Graphically**
Solve the equation x = x^3 + 1 graphically.
	1. Graph both sides of the equation on the same coordinate axes.
	2. Click the New Function button and enter x, then x^3 + 1.
	3. Click the Plot button.
	4. Estimate the solution by zooming in on the point of intersection and using the Point Motion buttons.


# Be thorough with ratio and proportion problems

This article provides practice questions and solutions to help prepare for State government recruitment tests.

**1. A person spends an amount equal to his 3 months income in 4 months. After one year, he distributes his savings among his three sons in a specific ratio. Find the share of his youngest son?**

a) Rs 198
b) Rs 196
c) Rs 195
d) Rs 192

Solution: Income of 1 month = Rs 363. Income of 3 months = Rs 1089. Expense of 4 months = Rs 1089. Savings of 4 months = Rs 363. Savings of 12 months = Rs 1089. Ratio of share of sons = 6 : 3 : 2. Share of youngest son = Rs 198.

**2. Meena, Chaitanya, and Priyanka rented a car for Rs 2,100, using it for 24, 36, and 66 hours respectively. How much rent did Chaitanya pay?**

a) Rs 650
b) Rs 625
c) Rs 600
d) Rs 615

Solution: Ratio of rent = 4 : 6 : 11. Chaitanya's rent = Rs 600.

**3. The ratio of three numbers is 3 : 5 : 4, and the sum of their squares is 200. Find the difference between the largest and smallest numbers.**

a) 1
b) 2
c) 3
d) 4

Solution: (3x)² + (5x)² + (4x)² = 200. x² = 4. x = 2. Difference = 2x = 4.

**4. The difference between two numbers is 30% of their sum. Find the ratio of the larger to the smaller number.**

a) 7 : 5
b) 13 : 7
c) 11 : 7
d) 9 : 5

Solution: (x - y) = 0.3(x + y). x : y = 13 : 7.

**5. If a³b³ : a³ - b³ = 185 : 158, then find a : b.**

a) 7 : 3
b) 2 : 3
c) 10 : 2
d) 5 : 4

Solution: a³/b³ = 343/27. a/b = 7/3.

**6. Two numbers are 40% and 60% more than a third number. Find the ratio of the first to the second number.**

a) 5 : 6
b) 6 : 7
c) 7 : 8
d) 8 : 9

Solution: First number = 140, second number = 160. Ratio = 7 : 8.

**7. Income and expenditure are in the ratio 9 : 5. If income is Rs 27,000, find the savings.**

a) Rs 10,000
b) Rs 12,000
c) Rs 11,500
d) Rs 12,700

Solution: Savings = Rs 12,000.

**8. If a : b = 3 : 4 and d : b = 4 : 3, find the ratio of a to d.**

a) 9 : 16
b) 16 : 9
c) 4 : 3
d) 3 : 4

Solution: a/b = 3/4, b/d = 3/4. a/d = 9/16.

**9. If 1/3 of A = 3/4 of B = 1/6 of C, find the ratio of A : B : C.**

a) 18 : 4 : 9
b) 9 : 18 : 4
c) 9 : 4 : 18
d) 18 : 9 : 4

Solution: A : B : C = 9 : 4 : 18.


## Evaluating Graphs and Equations Using Function Notation

The objective of this lesson is for students to evaluate f(x) and x using given information and the equation or graph of the function.

### Warm Up (10 minutes)
The Warm Up is designed to reinforce students' fluency with the vocabulary of independent variable, dependent variable, and function. It introduces how these concepts relate to function notation f(x).

### Guided Practice (15 minutes)
The Guided Practice consists of two parts. The first part provides a quadratic function and a t-table for students to complete by evaluating the function for a set of input values. The primary objective is for students to understand when to find x and when to find f(x) when evaluating functions.

The second part gives students practice evaluating the graph of a function using function notation. Students have 5 minutes to complete this section, after which the teacher will randomly call on students to provide answers to different problems.

### Independent Practice (15 minutes)
The Independent Practice provides more opportunities for students to evaluate functions, focusing on the following skills:
- Evaluating a function for x and f(x) using an equation
- Graphing a function using a t-table
- Evaluating a function for x and f(x) using a graph

One of the problems involves a multi-step scenario about a bathtub draining, where students must write an equation for the situation and use it or the graph to answer questions about G(s) and s.

### Exit Slip (5 minutes)
The Exit Slip is a quick formative assessment to check students' understanding of evaluating a function from a graph using function notation. Students complete it on their own without assistance, and it is collected before the end of class.

### Resources and Similar Lessons
Additional resources and similar lessons are available, including Graphing Radical Functions Day 2, The Function Game, and Graphing Linear Functions Using Given Information. These lessons are part of a larger unit on functions, which includes topics such as expressions, equations, inequalities, linear functions, systems of equations, radical expressions, exponential functions, polynomial operations, and quadratic functions.


This tutorial covers the steps to add polynomials vertically. To begin, it's essential to understand the definition of a polynomial and common names for certain polynomials, which are fundamental concepts in algebra.

When adding polynomials, the key is to combine like terms. This involves grouping terms with the same variable and exponent, then adding or subtracting their coefficients. To add polynomials of any size, simply group like terms and combine them.

For example, when combining integers, a number line can be used to help find the answer, especially when adding integers with the same sign or opposite signs. The same principle applies to polynomials, where like terms are combined.

To subtract polynomials, distribute the negative sign to the second polynomial, then combine like terms. This process involves changing the sign of each term in the second polynomial, then adding the resulting terms to the first polynomial.

By following these steps and understanding how to combine like terms, you can easily add and subtract polynomials. This tutorial provides a step-by-step guide on how to do it.


#### Answer

Option ($D$).
Simplified form: $\dfrac{\sqrt3}{2}$

#### Work Step by Step

A radical expression is not yet simplified when:
(1) there is a radical in the denominator; or
(2) the radicand still has a factor that is a perfect root of the index.
The radical expression in Option (D) has a denominator inside the radical sign, so it is not yet in simplified form.
To simplify the expression in $D$, factor each radicand to obtain:
$=\sqrt{\dfrac{3}{2^2}}$
Using the rule $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$ ($a\gt0, b\gt0$), the expression simplifies to:
$=\dfrac{\sqrt{3}}{\sqrt{2^2}}$
$=\dfrac{\sqrt{3}}{2}$


**OP Malhotra Class 9 ICSE Maths Solutions Chapter 18 Surface Area and Volume of 3D Solids**

This chapter provides solutions to exercises 18a and 18b, covering surface area and volume of 3D solids. The solutions are based on the ICSE latest syllabus.

**Surface Area and Volume of 3D Solids Exercise 18(a) Solution**

**Question 1**
Find the surface area of a cuboid with dimensions:
(a) 3m, 4m, 5m
Surface area = 2(3 × 4 + 4 × 5 + 5 × 3) m² = 94 m²
(b) 4cm, 1.7cm, 2.3cm
Surface area = 2(4 × 1.7 + 1.7 × 2.3 + 2.3 × 4) cm² = 39.82 cm²

**Question 2**
Find the surface area of a cube with edge:
(a) 7cm
Surface area = 6 × (7)² cm² = 294 cm²
(b) 10m
Surface area = 6 × (10)² m² = 600 m²
(c) Find the length of the edge of a cube with surface area 96 cm²
Edge = √(96/6) = 4 cm
(d) Find the length of the edge of a cube with surface area 726 cm²
Edge = √(726/6) = 11 cm

**Question 3**
Find the volume of a cuboid with dimensions:
(a) 4cm, 3cm, 5cm
Volume = 4 × 3 × 5 cm³ = 60 cm³
(b) 2cm, 6cm, 8cm
Volume = 2 × 6 × 8 cm³ = 96 cm³
(c) 7cm, 3cm, 4cm
Volume = 7 × 3 × 4 cm³ = 84 cm³
(d) 12cm, 9cm, 12cm
Volume = 12 × 9 × 12 cm³ = 1296 cm³
(e) 16cm, 14cm, 18cm
Volume = 16 × 14 × 18 cm³ = 4032 cm³
(f) Find the length of a cuboid with volume 25984 cm³ and dimensions x, 28cm, 24cm
x = 25984 / (28 × 24) = 7 cm
(g) Find the width of a cuboid with volume 2400 cm³ and dimensions 40cm, 24cm, x
x = 2400 / (40 × 24) = 2.5 cm

**Question 4**
Find the length of the diagonal of a cuboid with dimensions:
(a) 2cm, 3cm, 4cm
Diagonal = √(2² + 3² + 4²) = √29 = 5.38 cm
(b) 3cm, 4cm, 5cm
Diagonal = √(3² + 4² + 5²) = √50 = 7.07 cm

**Question 5**
Find the length of the diagonal and volume of a cube with edge:
(a) 2m
Diagonal = √3 × 2 = 3.46 m, Volume = (2)³ = 8 m³
(b) 5m
Diagonal = √3 × 5 = 8.66 m, Volume = (5)³ = 125 m³
(c) 8cm
Diagonal = √3 × 8 = 13.86 cm, Volume = (8)³ = 512 cm³
(d) Find the volume and surface area of a cube with edge 12cm
Volume = (12)³ = 1728 cm³, Surface area = 6 × (12)² = 864 cm²

**Question 6**
Find the length of the edge of a cube with volume:
(a) 216 m³
Edge = ∛216 = 6 m
(b) 2197 m³
Edge = ∛2197 = 13 m
(c) Verify if the volume of a cuboid with dimensions 100cm, 50cm, 0.5m is 250000 cm³
Volume = 100 × 50 × 50 = 250000 cm³, so the statement is true
(d) Find the height of a cuboid with diagonal 5√2 and dimensions 5cm, 4cm, h
h = √(50 - 41) = 3 dm

**Question 7**
Find the breadth of a rectangular solid with volume 3600 cm³, length 20cm, and height 9cm
Breadth = 3600 / (20 × 9) = 20 cm

**Question 8**
Find the height of a rectangular solid with volume 7000 cm³, length 25cm, and breadth 20cm
Height = 7000 / (25 × 20) = 14 cm

**Question 9**
Find the total surface area of a cube with edge 5cm
Surface area = 6 × (5)² = 150 cm²

**Question 10**
Find the cost of covering a playground with gravel, given the area 4800 m² and depth 1 cm
Volume = 4800 × 1/100 = 48 m³, Cost = 48 × 4.80 = 230.40 Rs

**Question 11**
Find the volume of water in a tank with base area 42 m² and height 5m
Volume = 42 × 5 = 210 m³

**Question 12**
Find the number of 4cm cubes that can fit in a box with volume 7680 cm³
Number of cubes = 7680 / (4)³ = 120

**Question 13**
Find the dimensions of a cuboid with surface area 1216 cm² and ratio of length, breadth, and height 5:4:2
Length = 5x, Breadth = 4x, Height = 2x, Surface area = 2(5x × 4x + 4x × 2x + 2x × 5x) = 76x²
76x² = 1216, x² = 16, x = 4
Length = 5x = 20 cm, Breadth = 4x = 16 cm, Height = 2x = 8 cm

**Question 14**
Find the volume of water that runs out of a tank with dimensions 40m, 7m, and 1.20m
Volume = 40 × 7 × 1.20 = 336 m³

**Question 15**
Find the number of rectangular blocks that can fit in a box with volume 90 × 78 × 42 cm³ and dimensions of each block 5/2 cm, 2cm, 3/2 cm
Number of blocks = (90 × 78 × 42) / (15/2) = 39312

**Question 16**
Find the rise in water level when a volume of 25920 cm³ is added to a tank with base area 72 × 60 cm²
Rise in water level = 25920 / (72 × 60) = 6 cm

**Question 17**
Find the volume of wood used to make a cuboid with outer dimensions 22.5cm, 15cm, 12cm and inner dimensions 20cm, 12.5cm, 9.5cm
Volume of wood = (22.5 × 15 × 12) - (20 × 12.5 × 9.5) = 4050 - 2375 = 1675 cm³

**Question 19**
Find the rise in level of a field when a pit of dimensions 6m, 4m, and 3m is dug
Rise in level = (6 × 4 × 3) / (540 - 24) × 100 = 13.9 cm

**Question 20**
Find the rise in level of a field when a pit of dimensions 6m, 3m, and 2.5m is dug
Rise in level = (6 × 3 × 2.5) / (280 - 18) = 17.18 cm

**Question 22**
Find the volume of a cube that can fit in a container with water level 6 × 6 × 1 cm³ and 2 cm³ of water
Volume of cube = 238 cm³

**Question 23**
Find the surface area of a cuboid formed by joining two cubes with edge 8cm
Surface area = 2(16 × 8 + 8 × 8 + 8 × 16) = 640 cm²

**Question 24**
Find the relation between the volume and surface area of a cuboid
xy/z = (lbh)² = v²

**Question 26**
Find the breadth and height of a room with volume 5400 m³, length 25m, and floor area 450 m²
Breadth = 450 / 25 = 18 m, Height = 5400 / (25 × 18) = 12 m

**Question 27**
Find the volume of a box formed by cutting a rectangular sheet with dimensions 42cm, 36cm, and 6cm
Volume = (30 × 24 × 6) = 4320 cm³

**Question 28**
Find the surface area of a cuboid formed by joining two cubes with edge 7cm
Surface area = 2(14 × 7 + 7 × 7 + 7 × 14) = 490 cm²


This lesson unit is intended to help teachers assess how well students working with square numbers are able to choose an appropriate, systematic way to collect and organize data, examining the data for patterns, describe and explain findings clearly and effectively, generalize using numerical, geometrical, graphical and/or algebraic structure, and explain why certain results are possible/impossible, moving towards a proof.

In a 15-day module, students learn the concept of a function and why functions are necessary for describing geometric concepts and occurrences in everyday life. Once a formal definition of a function is provided, students consider functions of discrete and continuous rates and understand the difference between the two. They apply their knowledge of linear equations and their graphs to graphs of linear functions, inspect the rate of change of linear functions, and conclude that the rate of change is the slope of the graph of a line. Students learn to interpret the equation y=mx+b as defining a linear function whose graph is a line and compare linear functions and their graphs, gaining experience with non-linear functions as well.

In the second topic of this module, students extend what they learned in Grade 7 about solving real-world and mathematical problems related to volume from simple solids to include problems that require the formulas for cones, cylinders, and spheres. The intent of clarifying statements is to provide additional guidance for educators to communicate the intent of the standard to support the future development of curricular resources and assessments aligned to the 2021 math standards.

Students will explore phase change at a molecular level, tracing the path of an atom to view intermolecular interactions and investigating how temperature relates to phase change. Upon activity completion, students will be able to give examples of phase change, explain how the input of energy into a system affects the state of matter, and describe how both latent heat and evaporative cooling play a role in changes of phase.

This course will place an emphasis on the continued study of integers, order of operations, variables, expressions, equations, and polynomials. Students will solve equations, write and solve proportions, explore polynomials, and build an understanding of important mathematical properties. They will measure relative humidity in the air using a simple device and electronically plot the data to analyze and learn from it.

Students will study the Treaty of 1868 between the Eastern Shoshone Tribe and the United States Government, learn about the Eastern Shoshone people, and understand the history of the Shoshone people in the Wind River Mountains. They will understand the indigenous perspective of interconnectedness, how bison populations were devastated by western expansion, and how to construct, read, compare, and analyze different population graphs. Students will acquire knowledge of the Wind River Reservation communities and be able to identify these locations on a map, further describing how their culture has shaped them and defining the concept of culture.

EngageNY Mathematics resources are available at https://archive.org/details/engageny-mathematics. The resource is available under Fair Use and the TEACH Act for US-based educators or other custom arrangements.


**Measuring and Apportioning Dilution in ESOPs**

The dilution to the seller is calculated as the pre-transaction fair market value (FMV) of shares sold minus the price paid. This can be expressed as:

pDE (13-3o)

1 (1 t)pDE

The post-transaction value of the firm can be calculated using the formula:

(1 0.04)

$1 million

[1 (0.6) (0.3) (0.98)]

$1 million 0.816049

The value of the ESOP is: 0.3 0.98 0.816049 $1,000,000

$239,918

This amount is also the payment made to the owner in cash. The after-tax cost of the payment to the owner is $143,951. The final value of the firm is $816,049, which is the same result obtained using the direct calculation by formula.

**Sharing the Dilution**

The dilution can be shared between the ESOP and the owner. The amount of dilution is calculated by subtracting the post-transaction value of the ESOP from the cash paid to the owner. The formula to calculate the payment to the owner when dilution is shared is:

t)p 2D 2

x [pDE(1 e) (1 t)pDEx] k[(1 pDE e] (13-4)

E

Collecting terms, we get:

t)p 2D 2

x[1 (1 t)pDE] pDE(1 e) k[(1 pDE e]

E

Dividing both sides by [1 (1 t)pDE], we solve to:

t)p 2D E

2

pDE(1 e) k[(1 pDE e]

x (13-4a)

1 (1 t)pDE

**Type 2 Dilution**

Type 2 dilution is equal to pDE, the pre-transaction selling price adjusted for control and marketability, minus the engineered selling price, x. Substituting equation (13-4a) for x, we get:

t)p 2D 2

pDE(1 e) k[(1 pDE e]

E

D2 pDE (13-4b)

1 (1 t)pDE

**Adjusting Dilution to Desired Levels**

Table 13-3 is a numerical example using equation (13-4a). We let p = 30%, DE = 98%, k = 2/3, t = 40%, and e = 4%. The calculation of x, the payment to the seller, is 27.6%. The value of the ESOP post-transaction is 23.36%. Subtracting the post-transaction value of the ESOP from the payment to the owner gives us the amount of type 1 dilution, which is 4.24%.

The default type 1 dilution, where the ESOP bears all of the dilution, is 6.36%. The actual dilution divided by the default dilution is 66.67%, which is the same as k. This proves the accuracy of equation (13-4a).

**Summary of Dilution Tradeoffs**

Table 13-3B summarizes the dilution options. In Table 13-2, the ESOP bears all dilution, which is 6.36%. In Table 13-3, the seller reduces type 1 dilution by 2.12% and undertakes type 2 dilution of 1.80%. In Table 13-3A, the seller bears all dilution, eliminating the 6.36% type 1 dilution and accepting 5.41% type 2 dilution.

The results suggest that when the seller takes on a specific level of type 2 dilution, the decrease in type 1 dilution is greater than the corresponding increase in type 2 dilution. This is proven in the Mathematical Appendix.

**The Iterative Approach**

The iterative approach involves calculating the payment to the owner, which becomes the ESOP loan, and the post-transaction fair market values of the firm and the ESOP. The first iteration calculates the payment to the owner without regard for the ESOP loan. The second iteration calculates an engineered payment to the owner that will attempt to equal the post-transaction value at the end of the first iteration.


# Statistical Lies I: Coronavirus

This post analyzes the misuse of statistical models and data analysis in the context of the COVID-19 pandemic.

First, we examine forecasting attempts, such as the cubic fit model, which lacks theoretical motivation. In contrast, the Institute for Health Metrics and Evaluation (IHME) presented a statistical model for cumulative COVID-19 deaths in March. However, their model was based on a Gaussian distribution, which is not suitable for this type of data.

Gaussian distributions appear in many statistical distributions due to the central limit theorem, but they require an averaging process to be applicable. For instance, the number of photons received from a star follows a Gaussian distribution when collected over a large time scale, but not when collected over a short time scale. Similarly, the income distribution is not Gaussian, and there is no clear reason why COVID-19 death dates should follow a Gaussian distribution.

A review of the IHME model posted on arxiv in April found that it failed to correctly predict the number of deaths for the next day, even within the reported confidence interval. Since then, IHME has changed its methods. In contrast, the COVID-19 Forecast Hub provides a more thorough analysis, comparing different forecasts to build an honest picture of uncertainty.

Another example of flawed analysis is a study that claims to have found a correlation between vegetable consumption and COVID-19 mortality rates. The study used quasi-Poisson regression models to analyze mortality counts and found a statistically significant correlation between head cabbage and cucumber consumption and lower COVID-19 death rates. However, the plot showing this correlation is heavily dominated by two or three data points and should be viewed with skepticism.

Finally, we examine the use of data to evaluate the effectiveness of Sweden's COVID-19 policies. Different plots can be used to support opposing views, depending on the chosen comparison cohort. A more honest plot reveals that Sweden's performance is actually relatively average on a global scale.

In conclusion, statistical models and data analysis can be misused to support flawed conclusions. It is essential to carefully evaluate the underlying assumptions and methods used in these analyses to ensure that they are valid and reliable.


# Adding and Subtracting Fractions
Addition and subtraction of algebraic fractions involve three key steps:
(i) rewriting fractions with a common denominator, ideally the simplest one;
(ii) combining numerators while retaining the common denominator;
(iii) simplifying the resulting fraction as much as possible.

To determine the simplest common denominator, factor the denominators into numerical and symbolic factors. The simplest common denominator is the product of the highest powers of all factors involved.

**Example 1:**
Combine and simplify:
The denominators are 15x = (3)(5)(x) and 12x^2 = (2^2)(3)(x^2). The simplest common denominator is (2^2)(3)(5)(x^2) = 60x^2.
Simplifying the numerator yields:
The final answer is .

**Example 2:**
Simplify:
The denominators are 6xy^2 = (2)(3)(x)(y^2) and 15x^2y = (3)(5)(x^2)(y). The simplest common denominator is (2)(3)(5)(x^2)(y^2) = 30x^2y^2.
The final answer is .

**Example 3:**
Simplify:
To add these fractions, convert the first term into a fraction with denominator 1:
The final answer is .

**Example 4:**
Combine and simplify:
Convert the problem into a two-fraction problem:
The final answer is .

When simplifying, always try to factor the numerator and denominator to achieve the simplest form possible. Be cautious of cancelling terms that are not part of the same fraction. By following these steps and being mindful of potential pitfalls, you can successfully add and subtract algebraic fractions.


The standard error of the mean, also known as the standard deviation of the mean, determines the differences between multiple samples of information, accounting for variations in the data. For example, measuring the weight of multiple samples of men can result in substantial variations, but the mean of these samples will vary by only a few pounds. The standard error of the mean illustrates how much the different weights vary from the mean.

To calculate the standard error of the mean, it is essential to keep the sets of numbers clearly labeled, as confusing them can lead to errors. The formula to determine the standard error of the mean is σM = σ/√N, where σM is the standard error of the mean, σ is the standard deviation of the original distribution, and √N is the square root of the sample size.

To apply the formula, first determine the standard deviation of the original distribution, which indicates how far apart the numbers are on the number line. If the standard deviation is provided, replace σ in the formula with this value. If not, calculate it by finding the mean of the set of numbers, subtracting the mean from each number, squaring the results, and then finding the average of these squared results, known as the variance. The standard deviation is the square root of the variance.

Next, determine the sample size, which is the number of items or observations being worked with, and replace N in the formula with this value. Find the square root of the sample size using a calculator. Finally, divide the standard deviation by the square root of the sample size to find the standard error of the mean.

Key steps to calculate the standard error of the mean include:
1. Determine the standard deviation of the original distribution.
2. Calculate the sample size.
3. Find the square root of the sample size.
4. Divide the standard deviation by the square root of the sample size.

By following these steps and using the formula σM = σ/√N, the standard error of the mean can be accurately calculated, providing insight into the variability of the sample means.


# Compact Spaces as BW Spaces

A topological space $X$ is a Bolzano Weierstrass space (BW space) if every infinite subset of $X$ has an accumulation point. Previously, it was shown that a Hausdorff space $X$ is a BW space if and only if it is countably compact. However, not all topological spaces are Hausdorff, so we will explore a new theorem that does not require this condition.

Theorem 1 states that every compact space is a BW space.

**Proof:** Let $X$ be a compact space. Every open cover of $X$ has a finite subcover. Assume $X$ is not a BW space; then there exists an infinite subset $A$ of $X$ without an accumulation point, meaning $A' = \emptyset$. The closure of $A$ is $\bar{A} = A \cup A' = A \cup \emptyset = A$, so $\bar{A} = A$, implying $A$ is closed.

Since $X$ is compact and $A$ is a closed subset of $X$, $A$ is compact in $X$. With no accumulation points, for each $a \in A$, there exists an open neighborhood $U_a$ of $a$ such that $A \cap U_a = \{a\}$. The set $\{U_a : a \in A\}$ is an open cover of $A$. As $A$ is compact, there exists a finite subcover $\{U_{a_1}, U_{a_2}, ..., U_{a_n}\} \subset \{U_a : a \in A\}$ with $A \subseteq \bigcup_{i=1}^{n} U_{a_i}$.

However, $\bigcup_{i=1}^{n} U_{a_i}$ contains only $n$ points from $A$, implying $A$ is finite, which is a contradiction. Therefore, the assumption that $X$ was not a BW space is false. If $X$ is compact, then $X$ is also a BW space.


A codimension one foliation of a 3-manifold is considered taut if every leaf in its leaf space has a circle transverse to it, intersecting the leaf. Taut foliations significantly contribute to topology, along with incompressible surfaces, in revealing topological and geometric information about 3-manifolds, as noted by Gabai and Oertel in 1989. Consequently, substantial research has focused on studying taut foliations on 3-manifolds.

Key findings include that every taut foliation is Reebless, and in non-taut Reebless foliations, leaves without a closed transversal are tori. Additionally, the closed leaves of a taut foliation are homologically nontrivial. Classification results for taut foliations have also been established. A notable result, attributed to Novikov and Sullivan by Eliashberg and Thurston, states that a foliation on a closed 3-manifold is taut if it differs from the foliation on  and meets one of the following conditions:

1. Each leaf is intersected by a transversal closed curve.
2. There exists a vector field transversal to the foliation that preserves a volume form on the manifold.
3. The foliation admits a Riemannian metric where all leaves are minimal surfaces.

Furthermore, a foliation is taut if and only if it contains no generalized Reeb components, as established by Goodman in 1975. This condition serves as both necessary and sufficient for determining tautness.


The logic gates described in this section are seldom used alone, but are used in combinations to form gating/logic networks, known as combinational logic circuits. For these logic circuits, the Boolean Algebra expression can be derived by systematically progressing from input to output on the gates.

Combinational logic circuits are formed by interconnecting logic gates. The Boolean expression for the output of a logic circuit can be found by analyzing the gates and their connections.

Example 1:
Find the Boolean expression for the output of the given logic circuit.
The circuit is constructed using the following gates:
- A NOT gate with input A, producing output `A`.
- An OR gate with inputs B and C, producing output B + C.
- An AND gate with inputs `A` and B + C, producing output `A`.(B+C).
Hence, the Boolean expression for the output D is `A`.(B+C).

Example 2:
Find the Boolean expression for the output of the given logic circuit.
The circuit is constructed using the following gates:
- An OR gate with inputs A and B, producing output A + B.
- An AND gate with inputs A and B, producing output A . B.
- A NOT gate with input A . B, producing output `A.B`.
- An AND gate with inputs A + B and `A.B`, producing output (A + B) . (`A.B`).
Hence, the Boolean expression for the output C is (A + B) . (`A.B`).


## Decimal Number

A decimal number is formed by combining a whole number, a decimal point, and a fractional value. The decimal point separates the whole number part from the fractional part. Each digit can be any number from 0 to 9. Values less than 1 are written to the right of the decimal point. Decimal numbers are also known as base-10 numbers or counting numbers. The place value of a decimal number varies as the whole number powers of 10, starting from the left of the decimal point.

## Gray Code

Gray code, also known as reflected binary code, is a code consisting of digits 0 and 1. It does not have a place value for its digits. Any successive codes in the Gray code system have only one bit change.

## Converting Decimal to Gray Code

To convert a decimal number to Gray code, first convert the decimal number to binary, then convert the binary to Gray code.

### Example: Convert 25_{10} to Gray Code

#### Step 1: Convert 25_{10} to Binary

25_{10} = 11001_{2}

#### Step 2: Convert 11001_{2} to Gray Code

The Most Significant Bit (MSB) of the Gray code and binary code will be the same. The next digit of the Gray code is the EXOR of the MSB and the digit to its right in the binary code. Repeat this process for each digit.

Alternatively, a simpler method can be used:

- int(25/02+1/2) = 13 (odd) -> 1
- int(25/04+1/2) = 6 (even) -> 0
- int(25/08+1/2) = 3 (odd) -> 1
- int(25/16+1/2) = 2 (even) -> 0
- int(25/32+1/2) = 1 (odd) -> 1

The Gray code for 25 is 10101.


I have a CS background with minimal statistics knowledge and a question about internet gaming statistics. I want to determine the average win rate for a single player in a 15 vs 15 game over thousands of games without playing that many.

Assuming randomized teams with players of different skill levels, and the game outcome is only affected by individual player skills, how can I calculate a large enough sample size? For example, if a player wins 55% of 100 games, what is the margin of error compared to playing 10,000 games?

This translates to a statistics problem where I need to find the sample size required to estimate the average win rate. I am looking for a way to represent this problem in notation and solve it. If this is a basic problem, I would appreciate a resource to help me figure it out.

The key question is: what sample size is sufficient to determine the average win rate, and is 100 games enough?


# Nonrecursive Movement Formulas

Arithmetico-geometric sequences enable the creation of non-recursive functions to calculate key results, such as a player's height on a given tick or jump distance based on initial speed and duration.

**Definitions:**

- Initial speed (speed before jumping)
- Number of ticks considered (e.g., jump duration on flat ground is t=12)
- "Jump bonus" (e.g., 0.3274 for sprintjump, 0.291924 for strafed sprintjump, 0.1 for 45° no-sprint jump)
- Movement multiplier after jumping (e.g., 1.3 for 45° sprint, 1.274 for normal sprint, 1.0 for no-sprint 45°)

## Vertical Movement (jump) [1.8]

Vertical speed after jumping:
Relative height after jumping:

## Vertical Movement (jump) [1.9+]

Vertical speed after jumping:
Relative height after jumping:

## Horizontal Movement (instant jump)

Assuming the player was airborne before jumping:
- Horizontal speed after sprintjumping:
- Sprintjump distance:

## Horizontal Movement (delayed jump)

Assuming the player is on the ground before jumping (at least 1 tick since landing):
- Horizontal speed after sprintjumping:
- Sprintjump distance:

## Advanced Formulas

Horizontal speed after consecutive sprintjumps on a momentum of period (, ):
If the first sprintjump is delayed, multiply by 0.6

**Note:** These formulas are accurate for most values of , but negative values can activate the speed threshold, resetting the player's speed and rendering the formulas inaccurate.


# How to Calculate the Coefficient of Determination (R²)

## Introduction to R²

The coefficient of determination, denoted as R², is a statistical measure used in regression models to determine the proportion of variance in the dependent variable that can be explained by the independent variable. In the case of linear correlation, R² is equal to the square of the correlation coefficient (r²), where r is the linear correlation coefficient.

## Calculating R²

To calculate R², you first need to determine the correlation coefficient (r) and then square the result. The formula for R² is r². The correlation coefficient (r) can be calculated using the formula:

r = [formula not fully provided in the original text, but it involves the number of data points (n) and the variables (x)].

## Understanding R²

R² measures the strength of the relationship between the model and the dependent variable, indicating the percentage of variance in the dependent variable that the independent variables explain collectively. It provides a convenient percentage scale to evaluate the model's fit.

## Formula for R²

The formula for calculating R² involves the sum of squares due to regression (SSregression) and the total sum of squares (SStotal). Although the original text does not fully elaborate on these components, understanding them is crucial for calculating R² accurately.

## Interpreting R²

R² (or the coefficient of determination) is crucial in statistics as it indicates how well the independent variables in a model explain the dependent variable. A higher R² value suggests a stronger relationship between the variables, meaning more of the variance is explained by the model.

Note: The sections regarding "How to find the fit?" and "How do you know if you have TSS?" seem unrelated to the calculation and understanding of the coefficient of determination (R²) and have been omitted for clarity and relevance.


The laws of inequality can be outlined as follows:

**I. Inequalities with Addition and Subtraction**
1. If $m > n$, then:
   (i) $m + k > n + k$
   (ii) $m - k > n - k$, where $k$ is any positive or negative number.
Examples:
- If $x + 5 > 25$, then $x + 5 + 3 > 25 + 3$, i.e., $x + 8 > 28$.
- If $x + 3 > 12$, then $x + 3 - 3 > 12 - 3$, i.e., $x > 9$.
- If $7x + 3 > 6x + 5$, then $7x + 3 - 6x - 3 > 6x + 5 - 6x - 3$, i.e., $x > 2$.

2. If $m > n$, then:
   (i) $km > kn$, where $k$ is a positive number.
   (ii) $km < kn$, where $k$ is a negative number.
Examples:
- If $7x > 5$, then $\frac{1}{7} \cdot 7x > \frac{1}{7} \cdot 5$, i.e., $x > \frac{5}{7}$.
Note: If $m > n$, then multiplying both sides by $-1$ gives $-m < -n$, or equivalently, $m < n$ when multiplied by $-1$.

3. If $m > n$, then:
   (i) $\frac{m}{k} > \frac{n}{k}$, where $k$ is a positive number.
   (ii) $\frac{m}{k} < \frac{n}{k}$, where $k$ is a negative number.
Examples:
- If $7x > 35$, then $\frac{7x}{7} > \frac{35}{7}$, i.e., $x > 5$.
- If $5x > 20$, then $\frac{5x}{5} > \frac{20}{5}$ is incorrect because it was given as an example for $k$ being negative, which does not apply here. The correct interpretation should be based on the sign of $k$.

Similar laws hold for the inequality "$\geq$".

**II. Inequalities with Less Than**
1. If $m < n$, then $m + k < n + k$ and $m - k < n - k$, where $k$ is any positive or negative number.
2. If $m < n$, then:
   (i) $km < kn$, where $k$ is a positive number.
   (ii) $km > kn$, where $k$ is a negative number.
3. If $m < n$, then:
   (i) $\frac{m}{k} < \frac{n}{k}$, where $k$ is a positive number.
   (ii) $\frac{m}{k} > \frac{n}{k}$, where $k$ is a negative number.
Similar laws hold for the inequality "$\leq$".

**III. Signs of Variables in Inequalities**
1. If $mn > 0$, then either $m > 0$ and $n > 0$, or $m < 0$ and $n < 0$.
2. If $mn < 0$, then either $m > 0$ and $n < 0$, or $m < 0$ and $n > 0$.
Note: $p < q$ and $q > p$ represent the same inequality. Additionally, $p < q$ and $q < r$ can be combined as $p < q < r$.


## Introduction to Variance in Excel

Variance is a statistical measure of variation, indicating the difference between values. A higher variance means more diverse values.

## Calculating Variance in Excel

There are two primary functions to calculate variance in Excel: **VAR** and **VAR.S**.

*   **VAR**: A common function available in every Excel version, calculating variance. Although still available, it has been improved and largely replaced by **VAR.S** since Excel 2010.
*   **VAR.S**: Calculates the sample variance of a given set of values. Introduced in Excel 2010, it is not compatible with earlier versions.

To calculate variance for an entire population, use the **VAR.P** function.

## Syntax of Variance Functions

The syntax for **VAR** and **VAR.S** is as follows:

*   **=VAR(number1,[number2],...])**
*   **=VAR.S(number1,[number2],...])**

## Examples of VAR Function in Excel

### Example 1: Simplest Syntax

Use numbers as arguments:

**=VAR(2,3,4,5,6,7,8,125)**

### Example 2: Cells as Arguments

Use cells containing numbers:

**=VAR(A2,A3,A4,A5)**

### Example 3: Range of Cells as Arguments

Use a range of cells:

**=VAR(A2:A10,B4:C10)**

### Example 4: Mixed Arguments

Combine different types of arguments:

**=VAR(A2:A10,B4,5,7)**

The syntax for **VAR.S** is identical to **VAR**.


## Module 3: Mathlets in Homework

This module consists of sample homework assignments created around MIT Mathlets, followed by discussion opportunities and remarks on the features and use of the Mathlets. The module includes exercises and resources.

## 1. Outline

The module outline serves as a guide, including:
- Outline
- Learning Objectives
- Mathlets in Homework, Segment 1
- Discussion Questions
- Discussion of Segment 1
- Conclusion
- Mathlets in Homework, Segment 2
- Discussion Questions
- Module Conclusion
- Exercises
- Resources

## 2. Learning Objectives

After completing this module, participants will be able to use Mathlets to:
- Mix experiment with computation in homework
- Improve student understanding of computations in homework
- Increase student enjoyment of homework exercises

## 3. Mathlets in Homework, Segment 1

This segment introduces using Mathlets in homework, focusing on the *Wheel* and *Taylor Polynomials* Mathlets.

### First Assignment: Vector Addition

Using the *Wheel* Mathlet:
- Adjust sliders and describe their representations
- Give parametric formulas for the center of the wheel and the light as functions of time
- Identify settings for zero velocity and verify mathematically
- Experiment with the Mathlet to identify situations where the horizontal component of the velocity vector vanishes and verify mathematically

### Second Assignment: Taylor Polynomials

Using the *Taylor Polynomials* Mathlet:
- Compute the full MacLaurin series and verify using the Mathlet
- Animate the second-order Taylor series and find values of x where the transitions occur
- Animate the third-order Taylor series and observe where the coefficient of the third-order term vanishes
- Predict and prove the entire Taylor polynomial for a given function

## 4. Mathlets in Homework, Segment 2

This segment focuses on the *Secant Approximation* and *Amplitude and Phase: Second Order IV* Mathlets.

### First Assignment: Secant Approximation

Using the *Secant Approximation* Mathlet:
- Explain why the yellow secant line disappears when x = a
- Find the slope of the tangent line at a given point
- Move the slider and observe the relationship between the secant and tangent lines
- Gather empirical data to show how small Δx must be to guarantee a certain accuracy

### Second Assignment: Frequency Response

Using the *Amplitude and Phase: Second Order IV* Mathlet:
- Describe the system response as the angular frequency is swept
- Estimate the near-resonant frequency and the amplitude of the system response
- Compute the near-resonant frequency and amplitude exactly
- Find the angular frequency for which the phase lag is exactly π/2

## Discussion Questions

- Module 3 Discussion Question 1: Wheel Project Ideas
- Module 3 Discussion Question 2: Taylor Polynomial Project Ideas
- Module 3 Discussion Question 3: Secant Approximation Project Ideas
- Module 3 Discussion Question 4: Frequency Response Project Ideas

## Module 3 Resources

- Mathlets-in-Homework-Slides.pdf
- Module-3-Mathlets-in-Homework-Conclusion-Transcript.pdf
- Module-3-Mathlets-in-Homework-Segment-1-Transcript.pdf