**What is a Palindrome?**

A palindrome is a word or number that reads backwards the same as it does forwards. Examples of palindromes include "Anna" and "radar," as well as numbers like 1001 and 69896.

**Counting the Palindromes**

All single-digit numbers (1, 2, 3, ..., 9) are palindromes. There are 9 two-digit palindromes (11, 22, 33, ..., 99). For every two-digit number, there is a corresponding three-digit and four-digit palindrome. For example, the two-digit number 34 corresponds to the three-digit palindrome 343 and the four-digit palindrome 3443. This pattern results in 90 three-digit palindromes and 90 four-digit palindromes.

Similarly, for every three-digit number, there is a corresponding five-digit and six-digit palindrome. For instance, the three-digit number 562 corresponds to the five-digit palindrome 56265 and the six-digit palindrome 562265. This results in 900 five-digit palindromes and 900 six-digit palindromes. In total, there are 9 + 9 + 90 + 90 + 900 + 900 = 1998 palindromes up to one million, which is approximately 0.1998% of all numbers up to one million. On average, about every 500th number is a palindrome.

**Position of the Palindromes**

The distribution of palindromes is not uniform across all numbers.

**Products with the Digit 1**

Certain products involving the digit 1 result in palindromes. For example:
- 11 × 11 = 121
- 111 × 111 = 12321
- 1111 × 1111 = 1234321
- ...
- 111 111 111 × 111 111 111 = 12345678987654321

Additionally, products like 11 × 111 = 1221 and 111 × 1111 = 123321 also yield palindromes. It appears that all products with the digit 1, where one factor has at most 9 digits, result in palindromes with the shape 123...321.

**Prime Numbers among the Palindromes**

Among three-digit palindromes, the prime numbers are:
101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929.

There are no four-digit or six-digit palindromic primes, as they all have 11 as a factor. For example, 4554 = 4004 + 550 = 4 × 1001 + 550 = 4 × 91 × 11 + 11 × 50 = 11 × (4 × 91 + 50). There are 93 five-digit palindromic primes and 668 seven-digit palindromic primes.

**196-Problem**

Given a number, add its mirror (the number read from right to left) to the original number. If the sum is not a palindrome, repeat the process with the sum. Nearly all numbers will eventually yield a palindrome. For example, starting with 49:
- 49 + 94 = 143
- 143 + 341 = 484, which is a palindrome.

However, some numbers do not yield a palindrome through this process. The smallest such number is 196, although a formal proof of this is still pending.


Comprehending numbers and early arithmetics can be made easy with tactile and multisensory practices. As students progress, word problems are introduced to provide pragmatic instances. However, some students may encounter difficulties in comprehending and solving these problems on paper and may require alternative approaches.

Word problems can be learned and worked out with engaging pedagogies like games and activities. To assist with practicing word problems, ten interesting and engaging online games have been identified.

**Solving Word Problems with Ease**

Solving word problems involves multiple stages. According to Polya, there are four steps to facilitate solving word problems:

1. **Understand the Problem**: Read the problem carefully to discern what is being asked.
2. **Make a Plan**: Check which plan of action can lead to the answer and devise an equation to proceed.
3. **Carry out the Plan**: Place the given values in the equation and carry out the calculations.
4. **Look Back**: Put the answer back in the equation to check if it satisfies and ensure all steps are properly depicted.

**Online Games for Practicing Word-Based Math Problems**

1. **Thinking Blocks Multiplication**: A game for multiplication and related word problems, where players choose a problem type and enter the game to solve step-by-step.
2. **Jumpy: Multi-Step Word Problems**: A game where players help Jumpy, the cat, jump over obstacles by solving multi-step word problems.
3. **Word Problems Related to Division**: A game that asks players to estimate and choose the right answer among four options, with a voice assistant and no time limit.
4. **2-Steps Money Problems**: A game that deals with money-related word problems, where players reckon the final value based on a price chart.
5. **Subtraction Word Problems**: A game that rewards players with a dice roll for every correct answer, where players solve subtraction-based word problems.
6. **Multiply Fractions**: A game that provides auxiliary support for practicing multiplication of fractions, with a voice assistant and scratchpad area.
7. **Computation Castle**: A visually enriched game where players answer word problems to help mix up a potion and free five innocent people from a curse.
8. **Math Pirate Game**: A game that stipulates players to understand simple sums to solve word problems, with a pirate theme and voice assistant.
9. **George’s Expedition**: A game where a teacher creates a story with word problems, and students solve them to proceed with the story.
10. **Tippy Tippy Tap, Pick a Math Operation to Grab**: An offline activity where players create a paper toy and choose a math operation to solve word problems, which can be made into a time-based challenge.

**Concluding Thoughts**

Word problems may not be easy for all students, and games can be an effective way to assist with learning. The games listed above cover basic operations, and offline ideas can be implemented to any notion by a mentor. By picking a suitable game, students can enjoy study time and improve their math skills.


### Properties of Exponents to Simplify Polynomials

The properties of exponents are essential to simplify polynomials. This concept helps students understand how to add, subtract, multiply, and divide powers used as exponents in algebraic expressions.

### Understanding the Properties of Exponents

The "Algebra 2 properties of exponents worksheet" is designed to teach students the properties of exponents. The worksheet covers the application of exponent properties to simplify complex equations and expressions. By following the conceptual overviews and practicing with math drills, students will gain a deeper understanding of exponent properties.

### Benefits of Learning Exponent Properties

Learning the properties of exponents will enable students to write complex equations or expressions in simplified form. This is achieved by understanding the order of operation of the exponents. With practice and reflection, students will become proficient in solving equations more quickly.

### Applying Exponent Properties

To reinforce learning, students will practice applying exponent properties to various math problems. A reflective section allows students to recreate their thoughts and solidify their understanding. Additionally, students are encouraged to create their own math problems or methods relating to exponent properties, promoting a deeper understanding of the concept.

### Conclusion

By mastering the properties of exponents, students will be able to solve equations more efficiently. This concept is crucial in simplifying polynomials and will benefit students in their algebraic studies.


Swapping variables is a fundamental operation in programming, and JavaScript provides several methods to achieve this. Here are five ways to swap variables in JavaScript:

**Method 1: Temporary Variable**
This approach involves creating a temporary variable to store the value of one variable, then assigning the value of the second variable to the first variable, and finally assigning the value of the temporary variable to the second variable.

**Method 2: Destructuring Assignment**
This method uses array destructuring to swap variable values. An array is created with the variable values in reverse order, and then destructuring assignment is used to assign the array values to the variables.

**Method 3: Addition/Subtraction**
This method uses arithmetic operators to swap variable values. It adds the values, stores the result in the first variable, subtracts the second variable's value from the first, and stores the result in the second variable. Finally, it subtracts the second variable's value from the first variable again and stores the result in the first variable. This method only works with numeric variables.

**Method 4: Multiplication/Division**
This method adapts the addition/subtraction approach using multiplication and division operators. It assigns the product of the variables to the first variable, then assigns the quotient to the second variable, and finally assigns the quotient to the first variable.

**Method 5: Bitwise XOR**
This method uses the bitwise XOR operator to swap variable values. It performs the XOR operation between the variables, stores the result in the first variable, and then performs the XOR operation between the first variable and the second variable, storing the result in the second variable. Finally, it performs the XOR operation between the first variable and the second variable again, storing the result in the first variable.

These five methods demonstrate different approaches to swapping variables in JavaScript. The choice of method depends on personal preference and the specific use case.


To convert fractions to percentages and vice versa, we can follow a few simple steps.

First, let's convert 5/24 to a percentage. To do this, we convert the fraction to a decimal first by dividing the numerator by the denominator: 5 ÷ 24 = 0.2083 repeating.

To express this as a percentage, we can multiply by 100: 0.2083 × 100 = 20.83%. Alternatively, we can think of the decimal as "per 100," so 0.2083 is equivalent to 20.83/100 or 20.83%.

Conversely, to convert a percentage to a fraction, we can divide by 100. For example, 16% = 16/100, which simplifies to 4/25.

Now, let's apply these concepts to some word problems.

Problem 1: A TV is advertised on sale at 35% off, with a new price of $195. What was the pre-sale price?

Let x be the pre-sale price. Then, x - 35% of x = $195. We can rewrite this as x - 0.35x = $195, or 0.65x = $195. To solve for x, we can divide both sides by 0.65: x = $195 / 0.65 = $300.

Problem 2: An employee is paid $9.50 per hour and receives a 12% pay raise. What will be her new hourly rate?

We can calculate the raise as 12% of $9.50, which is 0.12 × $9.50 = $1.14. Adding this to her current hourly rate, we get $9.50 + $1.14 = $10.64. Alternatively, we can think of the raise as multiplying her current rate by 1.12 (1 + 0.12): $9.50 × 1.12 = $10.64.

Problem 3: Two stores, A and B, sell bikes. Store A has a 40% markup, while store B has a 250% markup but sells at 60% off. Which store has a better deal?

Let x be the price at which both stores buy bikes from the supplier. Store A sells bikes for x + 40% of x, which is 1.4x. Store B sells bikes for x + 250% of x, which is 3.5x, but then offers a 60% discount. The effective price at store B is therefore 3.5x - 60% of 3.5x, which is 3.5x - 0.6(3.5x) = 0.4(3.5x) = 1.4x. Thus, both stores sell bikes for the same price, 1.4x. If the supplier price is $100, both stores will sell the bike for $140, but store B will advertise it as a 60% discount.


# Parallelogram

To calculate the area of parallelogram ABCD, we need to know its base and height. Given |AB| = 19 cm, |BC| = 18 cm, and angle BAD = 90°, we can use the formula for the area of a parallelogram: Area = base * height. Since angle BAD is 90°, |AB| can be considered the height and |BC| the base.

## Key Concepts and Formulas

- The area of a parallelogram is given by the formula: Area = base * height.
- When the angle between the base and the side is 90°, the side can be considered the height.

## Similar Examples and Problems

1. **Parallelogram ABCD with Known Area**: The area of parallelogram ABCD is 440 cm². Points M and N are the midpoints of the sides AB and BC. To find the area of quadrilateral MBND, we need to understand that since M and N are midpoints, MBND is also a parallelogram and its area is half the area of ABCD.

2. **Parallelogram with Given Sides and Angle**: Given |AB| = 76 cm, |BC| = 44 cm, and angle BAD = 30°, to find the area, we first need to find the height of the parallelogram using trigonometric functions, specifically the sine function since we have an angle and the side opposite to it.

3. **Rectangle with Changing Dimensions**: If a rectangle's side 'a' is enlarged by 5 cm and side 'b' is decreased by 2 cm, resulting in a reduction of the area by 5 cm², we can express this as an equation: (a+5)*(b-2) = ab - 5. This equation can help us find a relationship between 'a' and 'b'.

4. **Maximum Area of Rhombus**: The area of a rhombus can be maximized when it is formed into a square since all sides of a rhombus are equal. The maximum area occurs when the interior angles are 90°, making it a square.

5. **Acreage of a Diamond-Shaped Plot**: Given the side length of the diamond (rhombus) is 25.6 m and the distance between opposite sides (height) is 22.2 m, we can calculate its area using the formula for the area of a rhombus: Area = side * height.

6. **Rhombus with Given Circumference and Height**: With a circumference of 44 cm and a height of 89 mm (or 8.9 cm), first, find the side length from the circumference (since all sides are equal), then use the height to find the area.

7. **Area of a Rhombus**: Given PQ = 3 cm and the height of the rhombus is 2 cm, the area can be directly calculated using the formula: Area = base * height.

8. **Percentage of Light Entering a Window**: If a window frame is 147×51 cm and the glass is 13.3 cm from the hole in each direction, calculate the area of both the hole and the glass, then find the percentage difference in area to determine the percentage of light entering.

9. **Area of a Triangle within a Square**: Given a square DBLK with side |BL| = 13 cm, to find the area of triangle DKU where vertex U lies on line LB, we need more specific information about the position of U.

10. **Area of a Rectangular Field with Deviation**: If a rectangular field has dimensions 18 x 15 m with a possible deviation of 25 cm at each corner, calculate the maximum and minimum possible area considering the deviations.

11. **Percentage Change in Area of a Hop Garden**: If the length of a rectangular hop garden is increased by 25% and its width by 30%, calculate the new area and then find the percentage change from the original area.

12. **Dimensions of a Rectangle**: Given one side of a rectangle is 10 cm longer than the other, and shortening the longer side by 6 cm while extending the shorter by 14 cm increases the area by 130 cm², set up equations to solve for the original dimensions.

13. **Area of a Square Playground**: If fencing a square playground costs €464 and 1 meter costs €19, first find the perimeter of the playground, then calculate its side length and area.

14. **Number of Tiles Needed**: To pave a square room with a side length of 2.75 meters using tiles with an area of 121 cm², calculate the area of the room in square centimeters, then divide by the area of one tile.

15. **Actual Acreage of a Square Garden**: Given a square garden with an area of 81 cm² on a plan with a scale of 1:1500, first convert the area to square meters using the scale, then find the actual side length and area of the garden.

16. **Percentage Decrease in Cube Surface Area**: If the surface area of a cube decreases by 25%, calculate the new surface area as a percentage of the original.

17. **Cost of Lawn Maintenance**: For two rectangular lawns, each 40 m by 12 m, with circular flowerbeds of diameter 8 meters, calculate the area of the lawns, subtract the area of the flowerbeds, and then calculate the cost of maintenance based on the area and the cost per square meter.


The reliance on pseudorandom numbers poses a potential problem for IT security. If a machine uses a known algorithm to generate a number, an attacker can calculate the same number if they know the algorithm. In strong computer security, it is assumed that an attacker knows the algorithms used. Therefore, understanding how pseudorandom numbers are generated is crucial to avoid inappropriate reliance on them.

The measure of randomness in a number is known as entropy, which refers to the uncertainty about a number. Entropy is not related to the number of bits in a number, but rather to the number of possible values it could have. The number of bits that remain after discarding the certain bits is the number of bits of entropy. For example, a 100-bit number with two possible values would have 1 bit of entropy.

Mathematically, the entropy of a number X is defined as:

H(X) ≔ − ∑P(X=x) log₂P(X=x)

where the summation is from 0 to a number x, and P(X=x) is the probability of X taking the value x.

Generating a number with the greatest possible entropy is a challenging task. Research has focused on using activities like mouse movements or keystrokes to generate random numbers, but these methods have proven to be not entirely random. Another approach is to incorporate physical devices that generate truly random numbers based on chaotic behavior. However, this method has drawbacks, such as taking a long time to generate large numbers and the risk of device failure.

Pseudorandom number generation is currently the most accepted method, as it provides sufficient entropy while being efficient and accessible. The Fortuna approach, an improvement on the Yarrow method, uses a seed, a small but truly random number, and modern encryption techniques like AES to generate pseudorandom numbers. This approach recognizes potential attacks and minimizes threats by incorporating elements like AES and SHA, keeping some secret information local to the machine.

In summary, the best way to use random numbers is to generate pseudorandom numbers while replacing unreliable sources of randomness with a secret store of entropy, keeping the means of generation secret through a secret key. For a comprehensive understanding of random number generation, Luc Devroye's book is a recommended resource.


Introduction to Function
A function is a term used to define a relation between variables. A variable y is called a function of a variable x if for every value of x, there is a definite value of y. Symbolically, y = f(x). We can assign values of x arbitrarily, so x is called the independent variable, whereas y is called the dependent variable, as its values depend upon the value of x.

Definition of Function
A function defines a relation between variables, where for every value of x, there is a definite value of y. This relationship is represented as y = f(x), where x is the independent variable and y is the dependent variable.

Types of Functions
1. Even Function: A function f(x) is said to be an even function if f(x) = f(-x). For example, f(x) = 2x^2 + 4x^4 is an even function because f(-x) = 2(-x)^2 + 4(-x)^4 = 2x^2 + 4x^4 = f(x).
2. Odd Function: A function is said to be an odd function if f(x) = -f(-x). For example, f(x) = 3x + 2x^5 is an odd function because f(-x) = 3(-x) + 2(-x)^5 = -3x - 2x^5 = -f(x).


To assess students' understanding of geometry, a math foldable is used as an alternative assessment method. The lesson begins with a 10-minute introduction, where the teacher explains that students will be assessed on three geometry topics: congruent figures, line of symmetry, and rotational symmetry.

Each student is given a piece of construction paper and instructed to fold it uniformly along the vertical axis. The paper is then divided into three sections, labeled as congruent figures, line of symmetry, and rotational symmetry. The top piece of the paper is cut along the lines to create three equal pieces, forming a foldable.

The assessment is completed independently by students over a period of 30 minutes. A guideline is provided, outlining the expectations and rubric for scoring. Students are allowed to use math resources, such as textbooks or the internet, to aid them in completing the foldable. The teacher monitors progress, making note of students who require additional support, without interrupting their work.

After the assessment, students are given 10 minutes to share their findings on each topic. The teacher listens for key concepts, such as congruent figures being the same size and shape, line of symmetry dividing a figure into mirror images, and rotational symmetry occurring when a figure appears the same after less than a 360-degree turn.

The foldables are collected and assessed based on the provided rubric. Struggling students will receive additional support in small groups to ensure they master the skills. This approach to assessment aims to reduce student stress and promote the use of available resources to achieve success.


**1. Find the area of each circle.**
1. Radius 17 m, ANSWER: 907.92 m²
2. Diameter 7.8 in., ANSWER: 47.78 in²

**2. Find the volume of each solid.**
3. Cylinder, radius 6 ft, height 10 ft, ANSWER: 1130.97 ft³
4. Cone, radius 3.1 cm and height 11.8 cm, ANSWER: 359.29 cm³

**3. Find the circumference of a circle with radius 48 m.**
ANSWER: 301.59 m

**4. Find the surface area of a sphere**
EXAMPLE 1: Find the surface area of a sphere with radius 8, ANSWER: 201.06 square inches

**Standardized Test Practice**
EXAMPLE 2: Find the radius of a sphere with surface area 20.25π, ANSWER: 2.25, diameter: 4.5 cm

**GUIDED PRACTICE**
1. The diameter of a sphere is 40 feet, find the surface area, ANSWER: 5026.55 ft²
2. The surface area of a sphere is 30π square meters, find the radius, ANSWER: 2.71
3. The circumference of the inner ball is 6π feet, find the surface area, ANSWER: 36π ft² or 113.10 ft²

**Find the volume of a sphere**
EXAMPLE 4: The soccer ball has a diameter of 9 inches, find its volume, ANSWER: 381.69 cubic inches

**Find the volume of a composite solid**
EXAMPLE 5: Find the volume of the composite solid, ANSWER: 8.38 cubic inches

**GUIDED PRACTICE**
4. The radius of a sphere is 5 yards, find the volume, ANSWER: 523.6 yd³
5. A solid consists of a hemisphere of radius 1 meter on top of a cone with the same radius and height 5 meters, find the volume, ANSWER: 21.21 m³

**Daily Homework Quiz**
1. r = 6 ft, ANSWER: 452.39 ft²; 904.78 ft³
2. d = 12.4 m, ANSWER: 481.36 m²; 1023.18 m³
3. A great circle of sphere has circumference 16π in., find the volume, ANSWER: 2144.66 in³
4. Find the radius of a sphere with volume 1000 ft³, ANSWER: 6.82 ft
5. Find the surface area and volume of the solid, ANSWER: Not provided


# Sizing Up the Area of a Polygon

Polygons can be classified by the number of sides and angles, as well as by their qualities, including equilateral, equiangular, and regular. An equilateral polygon has equal sides, while an equiangular polygon has equal angles. A regular polygon is both equilateral and equiangular, with total symmetry.

## Key Lines in Polygons

Regular polygons have lines with special meaning, including the radius and apothem. The radius is a line from the center of the polygon to a vertex, splitting the angle evenly. The apothem is a line from the center of the polygon to a side, acting as a perpendicular bisector.

## Theorems for Regular Polygons

Several theorems exist for radii, central angles, and apothems of regular polygons:
* Radii bisect interior angles.
* Central angles are congruent.
* The measure of a central angle is 360° divided by the number of sides.
* An apothem bisects the central angle and is a perpendicular bisector to the side.

## Calculating the Area of a Regular Polygon

The area of a regular polygon can be calculated using the length of its apothem and perimeter. The perimeter is found by multiplying the number of sides by the length of one side. The area formula is A = 1/2 * ap, where a is the apothem and p is the perimeter.

## Example and Postulate

For example, a pentagon with a side length of 5 and an apothem of 6 has a perimeter of 25. Plugging this into the area formula gives A = 1/2 * (6)(25) = 75 square units.

According to Postulate 5-1, the area of a polygon is equal to the sum of the areas of its enclosed regions. This means that the area of a concave polygon can be found by breaking it into non-overlapping regions, finding the area of each, and adding them together.


The recent curriculum renewal has placed a heightened emphasis on counting, set theory, and probability, with almost half of the Grade 12 "Foundations of Mathematics" course covering these topics. This allows teachers to set a tone of curiosity, particularly in probability.

The concept of Grime Dice, pioneered by Dr. James Grime, was introduced as a way to get students engaged with probability before delving into theoretical aspects. Grime Dice are altered to compete at different strengths against varying dice, demonstrating a fascinating phenomenon of non-transitivity. A detailed description of Grime Dice can be found on the PlusMath website, and they are available for purchase from MathGear.co.uk.

To incorporate Grime Dice into a lesson, five novelty dice were purchased and re-worked to match Grime Dice A, C, and E. The task began with an introduction to the altered dice, followed by a graphic and a sheet with three dice charts. Students were asked to determine which dice was the strongest and why.

The initial question was:
* "Which dice is the strongest? Why?"

As students completed the task, they began to notice the circular nature of the dice and how it could be exploited by choosing last in a game. The class discussed probabilities, including the concept of a complement, and students demonstrated a natural understanding of odds and probability.

The lesson then entered Phase Two, which focused on building student authorship. Students were given a fresh set of three blank dice charts and posed the following problems:
* "Can you design three unique dice that all have an equal chance of beating one another, but none are numbered the standard one through six?"
* "Can you design the dice only using the numbers one through five?"

This phase aimed to encourage students to create mathematics, rather than just solving problems, and to think critically about probability and non-transitivity.


Multiplication Property of Exponents
Exponents have several properties that simplify expressions. In these properties, each variable represents a real number.

Multiplication Property
The Multiplication Property of Exponents states that to multiply two exponential expressions with the same base, add their exponents. The base stays the same. Algebraically, this is represented as:
x^{m} Â· x^{n} = x^{m + n}
where m and n are positive integers.

Example: 5^{4} Â· 5^{2} = 5^{4 + 2} = 5^{6}

Example 1:
a. Use the Multiplication Property of Exponents to simplify 2^{3} Â· 2^{4}.
b. Use the definition of exponential notation to justify your answer.
Solution:
a. Since the operation is multiplication and the bases are the same, add the exponents and use 2 as the base: 2^{3} Â· 2^{4} = 2^{3 + 4} = 2^{7}
b. Rewrite the product to show the factors, then simplify: 2^{3} Â· 2^{4} = 2^{3 + 4} = 2^{7} = 128

Note that adding exponents is different from adding the expressions themselves: 2^{3} Â· 2^{4} = 2^{3 + 4} = 2^{7} = 128, whereas 2^{3} + 2^{4} = 8 + 16 = 24.

When dealing with negative bases, remember that a negative sign is part of the base only when it is inside the parentheses that enclose the base. For example:
- In (3)^{2}, the base is 3, so (3)^{2} = (3) Â· (3) = +9
- In 3^{2}, the base is 3, and 3^{2} = (3 Â· 3) = 9

Example 2:
If possible, use the Multiplication Property of Exponents to simplify each expression:
a. (2)^{2} Â· (2)^{4}
b. 2^{2} Â· 2^{4}
c. 2^{2} Â· (2)^{4}
Solutions:
a. Since the base is 2, (2)^{2} Â· (2)^{4} = (2)^{2 + 4} = (2)^{6} = 64
b. Since the base is 2, 2^{2} Â· 2^{4} = (2^{2}) Â· (2^{4}) = (2^{2 + 4}) = (2^{6}) = 64
c. Although the bases of 2^{2} and (2)^{4} are both 2, they are not the same due to the parentheses. Therefore, the Multiplication Property of Exponents cannot be applied. However, the expression can still be evaluated: 2^{2} Â· (2)^{4} = 4 Â· 16 = 64


### Holiday Math Pennants

The Holiday Math section of the store offers a variety of math pennants for different holidays and math concepts. These pennants are designed to be engaging and fun for students to learn and practice various math skills.

#### Back to School and Fall Themes

* A Golden Spiral themed pennant for back to school activities
* Pumpkin rounding decimals pennants, where students round numbers to the nearest ten, whole, and tenth (30 pennants with 2 problems each)
* Rounding numbers pennant for younger kids, where students round whole numbers to the nearest ten and hundred
* Multiplication ghost pennants, where students multiply 2-digit numbers by 2-digit numbers with varying difficulty levels
* Domain and range cornucopia pennants, where students find the domain and range of graphs and identify them as functions or not
* Solving equations ghost pennants, with 40 pennants (20 for 1-step equations and 20 for 2-step equations)
* Proportion math pennants with a fall theme
* Plotting points snowflake pennants, introducing students to plotting functions from tables
* Graphing proportional relationship snowflake pennant

#### Winter and Holiday Themes

* Solving equations ornaments activity, where students solve 2-step and multi-step equations (great for decorating an Algebra class)
* Graphing linear equations ornaments, where students graph from slope-intercept and point-slope forms
* Christmas-themed fraction pennants, where students convert between mixed numbers and improper fractions
* Fraction operations holiday ornaments (mini-bundle)
* Area and perimeter ornaments for 3rd grade students
* Decimals ornaments for adding, subtracting, and multiplying decimals (120 ornaments, differentiated with 4 blank pennants for customization)
* Holiday multiplication ornaments

#### Valentine's Day and Spring Themes

* Inequalities hearts pennants, where students practice graphing linear inequalities
* Linear equations hearts pennants, where students graph slope-intercept form equations and find equations of given graphs
* Add and subtract mixed numbers Valentine's Day heart pennants
* Multiplying and dividing mixed numbers hearts pennants
* Graphing exponential functions hearts pennants

#### Pi Day and Saint Patrick's Day Themes

* Pi Day pennants, where students work with circle formulas to find area, circumference, radius, and diameter
* Pi Day pennants for younger kids, where students find areas and circumferences of circles with a coloring activity
* Dividing fractions and whole numbers pennant for Saint Patrick's Day
* Shamrocks for graphing exponential functions

#### Easter and Earth Day Themes

* Easter egg functions pennants, a fun way to celebrate Spring or Easter
* Earth Day math pennant, combining science and math with word problems related to Earth facts

These holiday math pennants offer a fun and engaging way for students to practice and learn various math concepts throughout the year.


**SSC BOARD EXAM ICT Practical P8 - Using Geogebra to Draw a Regular Polygon with Six Sides**

To construct a regular polygon with six sides using the Polygon tool in Geogebra, follow these steps:

**Tools/Software:** Computer with Geogebra Software installed.

**Procedure:**

### Method 1

1. Start the Geogebra Software and click on the down arrow of Tool No. 5 to select “Circle with Centre through point.” Draw a circle {c} with centre at point A and passing through point B.
2. Draw a circle {d} with centre as point B and passing through point A.
3. Use the Intersect two objects tool to intersect circles {c} and {d}, obtaining vertices C and D.
4. Draw circles {e} and {f} with centre as point C and centre as point D, both passing through point A.
5. Use the New point tool to get vertices E and F at the intersections of circles {c} and {e}, and {c} and {f}, respectively.
6. Draw a circle {g} with centre as point F and passing through A, and use the New point tool to get vertex G at the intersection of circles {c} and {g}.
7. In the Algebra area, right-click the conic group, select Object properties, and change the Line Style to dotted.
8. Select the Polygon tool and click on vertices from E in anticlockwise direction to form a closed polygon.
9. Hide point A by right-clicking it and turning off the show object option.
10. Use the Angle tool to verify the interior angles of the hexagon, which should all be 120°.
11. Save the construction with a suitable file name.

### Method 2

1. Start the Geogebra Software and create three points A, O, and D in a straight line with a distance difference of 5 cm, using the input bar to set their coordinates (e.g., A=(1,6), O=(6,6), D=(11,6)).
2. Select the Circle with Centre through point tool and draw a circle with centre as O and passing through A.
3. Draw a circle with centre as point A and point D passing through O.
4. Intersect the new circles with the middle circle using the Intersect two objects tool to get new vertices C, B, E, and F.
5. In the Algebra area, right-click the conic group, select Object properties, and change the Line Style to dotted.
6. Select the Polygon tool and click on vertices from E in anticlockwise direction to form a closed polygon.
7. Hide point O by right-clicking it and turning off the show object option.
8. Use the Angle tool to verify the interior angles of the hexagon, which should all be 120°.
9. Save the construction with a suitable file name.

**Result:** A regular hexagon with a side length of 5 cm is drawn using both methods.


### Shell Sort

Introduction
Shell Sort is an optimization of the insertion sort algorithm, a comparison-based sorting algorithm that operates in-place. It improves upon insertion sort, which compares each element with the previous one and inserts it at the appropriate position, but is slow for larger lists.


# Running Time of Binary Search

Binary search makes fewer guesses than linear search, and the difference becomes more striking as the array length increases. To analyze the maximum number of guesses that a binary search makes, we consider that when binary search makes an incorrect guess, the portion of the array that contains reasonable guesses is reduced by at least half.

Starting with an array of length n, if we make an incorrect guess, we can eliminate at least half of the array. For example, with an array of 16 elements, the first guess would eliminate at least 8 elements, leaving at most 8 elements to search. This means we need at most five guesses for an array of 16 elements.

As we double the size of the array, we need at most one more guess. This pattern suggests a relationship between the size of the array and the number of guesses needed. In general, for an array of length n, the maximum number of guesses is related to the base-2 logarithm of n.

The base-2 logarithm of n represents the power to which 2 must be raised to produce n. For example, the base-2 logarithm of 16 is 4, because 2^4 = 16. This means that for an array of 16 elements, we need at most 4 guesses, plus one more for the final comparison, making a total of 5 guesses.

The following table illustrates the relationship between the size of the array and the base-2 logarithm:

| Array Length | Base-2 Logarithm |
| --- | --- |
| 16 | 4 |
| 32 | 5 |
| 64 | 6 |
| 128 | 7 |
| 256 | 8 |

This relationship shows that the maximum number of guesses needed for binary search grows much more slowly than the size of the array, making binary search a more efficient algorithm than linear search for large arrays.


A truck of mass $M$ is at rest on a frictionless road when a monkey of mass $m$ starts moving forward on the truck. The truck recoils with a speed $v$ backward. To find the velocity of the monkey with respect to the truck, we consider the conservation of momentum.

The initial momentum of the system is zero since both the truck and the monkey are initially at rest. When the monkey moves forward and the truck recoils backward, the total momentum of the system remains zero. Thus, the momentum gained by the monkey equals the momentum lost by the truck.

Let $v_m$ be the velocity of the monkey relative to the ground and $v$ be the velocity of the truck relative to the ground. The velocity of the monkey with respect to the truck is $v_m - (-v)$, because the truck is moving in the opposite direction.

From the conservation of momentum: $0 = mv_m - Mv$. Solving for $v_m$, we get $v_m = \frac{Mv}{m}$. The velocity of the monkey with respect to the truck is $v_m + v = \frac{Mv}{m} + v = v\left(1 + \frac{M}{m}\right)$.

Therefore, the velocity of the monkey with respect to the truck is $\left(1 + \frac{M}{m}\right)v$.

A man of mass $60\ kg$ is standing on a platform of mass $40\ kg$. To accelerate upward with the platform at $2\ m/s^2$, the total force required is given by $F = (m + M)a$, where $m$ is the mass of the man, $M$ is the mass of the platform, and $a$ is the acceleration.

Thus, $F = (60 + 40) \times 2 = 100 \times 2 = 200\ N$. This is the force the man should apply to the rope.

Three blocks $A, B$, and $C$ are suspended, with $A$ and $C$ each having mass $m$ and $B$ having mass $M$. When $B$ is pulled with velocity $V$, and considering the system's equilibrium, the velocity of $A$ can be determined by the relationship of the pulleys and the conservation of momentum.

Since $A$ and $C$ are connected through a pulley system with $B$, the velocity of $A$ will be related to the velocity of $B$. For every unit of distance $B$ moves, $A$ and $C$ move half the distance but in opposite directions due to the pulley system. Thus, if $B$ moves with velocity $V$, $A$ moves with velocity $\frac{V}{2}$ in the opposite direction of $B$'s motion relative to the pulley system connecting them.

Given $A$ has a velocity of $0.6\ m/s$ to the right, and considering the pulley system, $B$'s velocity can be determined by the inverse relationship of their masses and the mechanical advantage of the pulley system. However, without explicit details on the pulley system's mechanical advantage, we assume a direct relationship for velocity based on the question's implication.

If block $B$ moves down with a velocity $10\ m/s$, the velocity of $A$ can be determined by the pulley system's configuration. Assuming a 1:1 correspondence for the sake of this explanation (which might not fully align with the specifics of pulley systems without further details), $A$ would move upward with half the velocity of $B$ due to the typical mechanical advantage in such setups, resulting in $5\ m/s$ for $A$. However, the exact relationship depends on the specifics of the pulley system and how the blocks are connected.

For multiple-choice questions provided:
- The correct answer for the velocity of the monkey with respect to the truck is $\left(1 + \frac{M}{m}\right)v$, which corresponds to option B.
- The force the man should apply to the rope to accelerate upward with the platform is $200\ N$.
- The velocity of block $A$ when $B$ is pulled with velocity $V$ would be $\frac{V}{2}$, considering a basic pulley system.
- The velocity of block $B$ when $A$ has a velocity of $0.6\ m/s$ would depend on the specifics of the pulley system and the masses involved, which is not directly calculable from the information given.
- The velocity of $A$ when $B$ moves down with $10\ m/s$ would be $5\ m/s$ if assuming a direct and simple pulley relationship, but this requires clarification on the system's specifics.


### True Talent Levels for NHL Team Shooting Percentage

The difference in team shooting percentage (SH%) can be attributed to both luck and talent. To determine the extent of each factor, we can use the Palmer/Tango method.

In 5-on-5 tied situations in the 2013-14 season, the average team took 721 shots with an 8% SH%. The standard deviation (SD) of binomial luck is approximately one percentage point, which can significantly impact a team's standings. For example, a team with an average SH% of 7.50 (16th) could move up to 8.50 (4th) with a one percentage point increase.

In comparison, the SD of binomial luck for Corsi (CF%) is also about one percentage point. However, this would only move a team with a 50.0% CF% to 51.0%, resulting in a change of only three places in the standings. This suggests that SH% is not as reliable an indicator as Corsi, as a run of luck can greatly impact a team's SH%.

The observed SD of actual team SH% is also around one percentage point. Using the formula SD(talent)^2 = SD(observed)^2 - SD(luck)^2, we get SD(talent)^2 = 1.0 - 1.0, which equals zero. This implies that there is no variance in talent, and SH% is completely random.

However, this result may not be entirely accurate. The observed SD is itself random and based on a single season's data. A simulation of 5,000 seasons with 30 teams, each with 700 shots and an 8.00% SH%, showed that the average SD was around 1.00, but with a range of 0.83 to 1.27. This confidence interval suggests that the true SD of talent could be substantial, with an SD of 0.52 percentage points at the top end of the interval.

The wide range of the confidence interval is due to the observed spread being similar to the binomial luck. In other sports, such as MLB team wins, the SD of actual wins is almost double the theoretical SD from luck. In the NHL, it is about one-and-a-half times as big.

To narrow down the estimate, we can use data from multiple seasons. However, out of five seasons, four resulted in impossible situations, with teams being closer together than expected even if they were identical. This could be due to random chance, incorrect assumptions about talent and luck being independent, or other factors.

One possible explanation is that the assumption of all shots being the same is not accurate. A more realistic assumption, such as a combination of high-probability and low-probability shots, can result in a lower SD of luck. For example, if 56 goals come from a combination of shots with different probabilities, the luck SD drops from 1.0% to 0.91%. This would put the SD of team SH% talent at 0.4 points, which is meaningful and equivalent to about five places in the standings.

Other data from the hockey research community, such as year-to-year SH% correlations, can also provide insight. Two studies found correlations in the range of 0.15, which would put the talent SD at around 0.15. This is consistent with the confidence intervals and guesses made earlier.

In comparison, the estimate of SD(talent) for Corsi is around 4 percentage points, which is much higher than for SH%. Converting this to goals, one talent SD in SH% is equivalent to 1 goal, while one talent SD in CF% is equivalent to 10 goals. This suggests that Corsi is 10 times more useful than SH% in predicting future performance, although this may be misleading since CF% is based on both offense and defense.

The analysis suggests that teams with a very high SH% are demonstrating a couple of 5-on-5 tied goals worth of talent. Instead of using Corsi alone, combining CF% and SH% can provide a more accurate estimate of a team's talent. This is particularly useful in 5-on-5 tied situations, where score effects are minimal.

For example, the Toronto Maple Leafs had an absurdly high SH% in 5-on-5 tied situations in the 2012-13 season, but an absurdly low CF%. By combining these metrics, we can project the Leafs' scoring talent more accurately, taking into account both their SH% and CF%. This approach can help to better understand a team's true talent level and make more informed predictions about their future performance.


**Numbers and Place Value**

Expressing numbers from 41 to 49:
41 - forty-one,
42 - forty-two,
43 - forty-three,
44 - forty-four,
45 - forty-five,
46 - forty-six,
47 - forty-seven,
48 - forty-eight,
49 - forty-nine.

When two figures are written side by side, the figure at the right represents ones, and the figure at the left represents tens.

Expressing numbers in words:
24 - twenty-four,
32 - thirty-two,
38 - thirty-eight,
47 - forty-seven,
18 - eighteen,
29 - twenty-nine,
10 - ten,
31 - thirty-one,
49 - forty-nine,
30 - thirty.

**Tens and Ones**

Expressing 5 tens or fifty: 50.
The figure at the right is 0, representing no ones, and the figure at the left is 5, representing 5 tens.
There are 5 tens in fifty.

Expressing 5 tens and 1, or fifty-one: 51.
Numbers from fifty-one to fifty-nine:
51 - fifty-one,
52 - fifty-two,
53 - fifty-three,
54 - fifty-four,
55 - fifty-five,
56 - fifty-six,
57 - fifty-seven,
58 - fifty-eight,
59 - fifty-nine.

Expressing 6 tens or sixty: 60.
Numbers from sixty to sixty-nine:
60 - sixty,
61 - sixty-one,
62 - sixty-two,
63 - sixty-three,
64 - sixty-four,
65 - sixty-five,
66 - sixty-six,
67 - sixty-seven,
68 - sixty-eight,
69 - sixty-nine.

Expressing seventy: 70.
There are 7 tens in seventy, and no ones.

Numbers from seventy to seventy-nine:
70 - seventy,
71 - seventy-one,
72 - seventy-two,
73 - seventy-three,
74 - seventy-four,
75 - seventy-five,
76 - seventy-six,
77 - seventy-seven,
78 - seventy-eight,
79 - seventy-nine.

Numbers from eighty to eighty-nine:
80 - eighty,
81 - eighty-one,
82 - eighty-two,
83 - eighty-three,
84 - eighty-four,
85 - eighty-five,
86 - eighty-six,
87 - eighty-seven,
88 - eighty-eight,
89 - eighty-nine.

**Hundreds**

Expressing 9 tens or ninety: 90.
Numbers from ninety to ninety-nine:
90 - ninety,
91 - ninety-one,
92 - ninety-two,
93 - ninety-three,
94 - ninety-four,
95 - ninety-five,
96 - ninety-six,
97 - ninety-seven,
98 - ninety-eight,
99 - ninety-nine.

Ninety-nine and one more make one hundred: 100.
The figure 0 at the right shows no ones, the next 0 shows no tens, and the figure 1 shows 1 hundred.

**Multiplication**

If you pay 1 cent for a pear, you pay 2 cents for 2 pears.
1 cent + 1 cent = 2 cents.
If you pay 1 cent for a slate-pencil, you pay 3 cents for 3 pencils.
The price of 3 pencils is 3 times the price of 1 pencil.

If a fig costs 1 cent, 4 figs cost 4 times 1 cent.
5 times 1 chair is 5 chairs.
The sign X represents the word "times".
2 times 1 is written as 2x1=2.
If 1 marble costs 1 cent, 6 marbles cost 6 cents: 6x1=6.

Seven boys each own a sled, so they own 7 sleds: 7x1=7.
Each of 8 boys has a pair of skates, so they have 8 pairs: 8x1=8.
Nine men each own an acre of land, so they own 9 acres: 9x1=9.
10 times 1 is 10: 10x1=10.
3 times 1 is 3: 3x1=3.

If you pay 1 cent for an apple, you pay 7 cents for 7 apples.
If each pupil in a class of 9 pupils answers 1 question, they answer 9 questions.

**Multiplication Tables**

1x1=1, 2x1=2, 3x1=3, 4x1=4, 5x1=5, 6x1=6, 7x1=7, 8x1=8, 9x1=9, 10x1=10.
1x2=2, 2x2=4, 3x2=6, 4x2=8, 5x2=10, 6x2=12, 7x2=14, 8x2=16, 9x2=18, 10x2=20.

**Real-World Applications**

If a boy walks 3 miles in 1 hour, he walks 6 miles in 2 hours: 2x3=6.
If 3 engravings cost 3 dollars each, they cost 9 dollars: 3x3=9.
Martha put 3 tulip bulbs in each of 4 pots, so she planted 12 bulbs: 4x3=12.
Each of 5 boys threw 3 stones into the water, so they threw 15 stones: 5x3=15.
If pupils learn 3 songs in a month, they learn 18 songs in 6 months: 6x3=18.
If there are 3 cherries in 1 bunch, there are 21 cherries in 7 bunches: 7x3=21.
If there are 3 globes on 1 chandelier, there are 27 globes on 9 chandeliers: 9x3=27.


## Measures of Network Centralisation

The following functions calculate centralisation measures for a network:

* `network_degree(object, normalized = TRUE, direction = c("all", "out", "in"))`: calculates degree centralisation
* `network_closeness(object, normalized = TRUE, direction = c("all", "out", "in"))`: calculates closeness centralisation
* `network_betweenness(object, normalized = TRUE, direction = c("all", "out", "in"))`: calculates betweenness centralisation
* `network_eigenvector(object, normalized = TRUE)`: calculates eigenvector centralisation

## Arguments

* `object`: an object of a compatible class, including:
	+ matrix (adjacency or incidence) from `{base}`
	+ edgelist, a data frame from `{base}` or tibble from `{tibble}`
	+ igraph, from the `{igraph}` package
	+ network, from the `{network}` package
	+ tbl_graph, from the `{tidygraph}` package
* `normalized`: a logical scalar indicating whether centrality scores should be normalised
* `direction`: a character string specifying the direction of ties, with options "out", "in", or "all"

## Functions

* `network_degree()`: calculates degree centralisation for a graph
* `network_closeness()`: calculates closeness centralisation for a graph
* `network_betweenness()`: calculates betweenness centralisation for a graph
* `network_eigenvector()`: calculates eigenvector centralisation for a graph

## Examples

* `network_degree(ison_southern_women, direction = "in")`: calculates degree centralisation for the "ison_southern_women" network with incoming ties
* `network_closeness(ison_southern_women, direction = "in")`: calculates closeness centralisation for the "ison_southern_women" network with incoming ties
* `network_betweenness(ison_southern_women, direction = "in")`: calculates betweenness centralisation for the "ison_southern_women" network with incoming ties
* `network_eigenvector(mpn_elite_mex)`: calculates eigenvector centralisation for the "mpn_elite_mex" network
* `network_eigenvector(ison_southern_women)`: calculates eigenvector centralisation for the "ison_southern_women" network


A test taker reported that 50% of the test items were duplicates when they took the test for the second time. The test selects 10-20 items randomly from a total of 80 items. To calculate the likelihood of this occurrence, we can start with some back-of-the-envelope calculations.

Assuming an average of 15 test items, the average quantity of test items taken is 18.75% of the total. If the same person takes the test again, we would expect around 2.8 items to be duplicates. However, we want to know the probability of 50% of the items being duplicates in the second test.

To calculate this, we would need to consider all possible combinations of overlapping items. For example, if we have a 10-item exam, the probability of the first 5 items overlapping and the last 5 not overlapping is calculated as the product of the individual probabilities: 1/80 * 1/79 * 1/78 * 1/77 * 1/76 * 74/75 * 73/74 * 72/73 * 71/72 * 70/71. This process becomes tedious, so we can use a simulation instead.

We simulated the process 100,000 times, with the following parameters:
- Nsim = 10^5 (number of simulations)
- testMin = 10 (minimum number of test items)
- testAdd = 10 (range of possible test items)

The simulation results showed a mean overlap of 0.1877303, which is close to the expected average. We can use the empirical cumulative distribution function (ecdf) to calculate the likelihood of a certain level of overlap. For example, the likelihood of 25% or less overlap is 0.76224, and the likelihood of 18.75% or less overlap is 0.53936.

The likelihood of 50% or more overlap between exams is 0.00212, which seems small. However, considering that around 1,250 people have taken the exam, with 625 attempts (assuming each person takes the test twice), the probability that someone received 50% or more of the same items is 0.7482862. If we assume 100 people took the test twice, the probability is 0.1912173, indicating a 20% chance that one person would see 50% of the same items on the second test.

Conversely, the likelihood that someone would not see any of the same items is around 5%. These numbers suggest that it is not unlikely for someone to see 50% of the same items on the second test, especially if multiple people take the test multiple times.

Multiple choice questions and answers can be refined as follows:
What is the average quantity of test items taken?
A) 10%
B) 18.75%
C) 25%
D) 50%

Answer: B) 18.75%

What is the likelihood of 50% or more overlap between exams?
A) 0.01
B) 0.1
C) 0.00212
D) 0.5

Answer: C) 0.00212

What is the probability that someone would not see any of the same items?
A) 1%
B) 5%
C) 10%
D) 20%

Answer: B) 5%


A horizontal angle is an angle between lines on a horizontal plane. The angle formed by a horizontal line is 180°. Horizontal lines are lines that may be at an angle of 0°, 180°, or 360°. If vertical lines stand on horizontal lines, it is at an angle of 90°.

A horizontal surface is level and parallel to the ground, while an angled surface slopes or tilts in relation to a horizontal plane. Horizontal surfaces are typically used for stability and flatness, while angled surfaces may be inclined for specific purposes such as drainage or aesthetics.

The angle of depression is the angle between a horizontal line and the line joining the observer's eye to some object beneath the horizontal line. The launch angle is the angle that the velocity makes with the horizontal and can be calculated using trigonometry.

To draw an angle, start by drawing a horizontal line, then place a dot near the middle of it where you want the vertex. Draw an angle upward and to the right using your dot as the vertex, and the horizontal line as one leg of your angle.

A vertical angle runs up and down or from top to bottom or vice versa. A horizontal angle runs from left to right or vice versa. Key concepts include horizontal lines, vertical lines, angles of depression, and launch angles.

Related concepts include the angle of inclination, which is the angle between a horizontal line and an inclined line. Horizontal lines and vertical lines are perpendicular, forming a 90° angle.


A quart is a unit of measurement for liquid volume equal to nearly 946 milliliters, or just slightly more than 1/4 of a gallon. There are 4 quarts in 1 gallon. A cup, on the other hand, is a unit of measurement for either liquid or dry volume equal to 8 fluid ounces or 236.59 milliliters.

4 cups are equal to 32 fluid ounces or a little over 946 milliliters, which is slightly more than 1 quart. Therefore, it would take slightly more than 4 cups to equal 1 quart.

## Does 4 gallons equal 1 quart?
No, 4 gallons does not equal 1 quart. There are 4 quarts in a gallon, so 4 gallons are equal to 16 quarts.

## How many quarts does 4 cups have?
4 cups is equal to 2 pints. 2 pints is equal to 1 quart. Therefore, 4 cups has 1 quart.

## Is 4 cups of milk a quart?
No, 4 cups of milk is not a quart. A quart is a unit of measurement equal to 32 ounces or 2 pints. That is the equivalent of 4 cups of milk plus 4 additional tablespoons.

## What size of milk is 4 cups?
4 cups of milk is equal to 32 fluid ounces, or 946 milliliters. That’s roughly 1 US quart, or just shy of 1 liter.

## Does 4 cups make a gallon?
No, 4 cups do not make a gallon. A gallon is a unit of volume measurement found primarily in the United States, but also in the United Kingdom, Canada, and other countries. One U.S. gallon is equal to 16 U.S. cups. Therefore, 4 cups would be equal to 1/4 of a gallon.

## What is 4 cups the same as?
4 cups is equal to 32 fluid ounces, which is the same as 2 pints, or 1 quart. In terms of dry measurements, 4 cups is equal to 16 ounces or 1 pound.

## What measurement is 4 cups?
4 cups is equal to 32 fluid ounces, or 2 pints. In terms of metric measurements, 4 cups is equal to 950 milliliters.

## How many cups are in a quarter?
There are 2 cups in one quarter when measuring dry ingredients, and 1 cup is equal to 8 fluid ounces when measuring liquid ingredients.

## What are the 4 sizes of measuring cups from smallest to largest?
The four sizes of measuring cups, from smallest to largest, are:
1. Tablespoon: equal to 1/16 of a cup,
2. Fluid Ounce: equal to 1/8 of a cup,
3. Cup: equal to 1 cup,
4. Pint: equal to 2 cups.

## What does 1 quart equal to in cups?
One quart is equal to 4 cups. A quart is a unit of measurement for volume, and is traditionally used for measuring liquids.

## What makes up 1 quart?
One quart is equivalent to approximately 0.946352946 liters, or 32 fluid ounces. It is also equal to 4 cups or 2 pints.

## Is 1 quart more than 8 cups?
No, 1 quart is not more than 8 cups. To be exact, 1 quart is equal to 4 cups, which is half the amount of 8 cups.

## Is 32 oz the same as 1 quart?
Yes, 32 oz is the same as 1 quart in the United States. However, in the United Kingdom, one quart is equal to 40 fluid ounces.

## Is 8 oz always 1 cup?
No, 8 oz is not always equivalent to 1 cup. This measurement varies depending on what you are measuring. For example, when measuring a liquid, 1 cup is equal to 8 fluid ounces. But when measuring a dry ingredient such as flour or sugar, 8 ounces normally equate to 1.5 cups.


Chapter 7: Infinite Sets
Section 7.2: Countable and Uncountable Sets

Summary:
- If A and B are countable sets, then $A \times B$ and $A \cup B$ are countable.
- The union of countably many countable sets is countable.
- A function $f : I_n \to A$ is called a finite sequence of elements of A, where n is the length of the sequence.
- If A is a countable set, the set of all finite sequences of elements of A is also countable.
- Cantor's theorem states that $\mathcal P(Z^+)$ is uncountable.
- $\mathbb R$ is uncountable.

Solution 1:
(a) We prove that $\mathbb R \setminus \mathbb Q$ is uncountable by contradiction. If it were countable, then $\mathbb R = (\mathbb R \setminus \mathbb Q) \cup \mathbb Q$ would be countable, which is not the case.
(b) To prove $\mathbb R \setminus \mathbb Q \sim \mathbb R$, we define a set A and show that $A \sim \mathbb Z^+$ and $A \cup \mathbb Q \sim \mathbb Z^+$. Then, we use these equivalences to show that $\mathbb R \setminus \mathbb Q \sim \mathbb R$.

Solution 2:
We prove that the function F is one-to-one and onto. F is one-to-one because if $F(f_1,a) = F(f_2,b)$, then $f_1 \cup \{(n+1,a)\} = f_2 \cup \{(n+1,b)\}$, which implies $f_1 = f_2$ and $a = b$. F is onto because for any $f \in S_{n+1}$, we can find $g \in S_n$ and $a \in A$ such that $F(g,a) = f$.

Solution 3:
We define $P = \{ X \in P(Z^+) \vert X \text{ is finite } \}$ and $P_n = \{ X \subseteq Z^+ \vert \vert X \vert \le n \}$. Then, $P = \cup_{i \in \mathbb N} P_n$ is a set of finite sequences of the countable set $Z^+$. By theorem 7.2.4, P is countable. We also show that P is not finite, so it must be denumerable.

Solution 4:
We prove that there is no onto function $f: A \to P(A)$. Suppose there were such a function. Then, for any $a \in A$, we have two cases: $a \in f(a)$ or $a \notin f(a)$. In both cases, we can show that $f(a) \ne D$, where $D = \{a \vert a \notin f(a)\}$. This contradicts the assumption that f is onto.

Solution 5:
(a) We define $\mathcal H: {^A B} \times {^A C} \to {^{A} {(B \times C)} }$ and show that it is one-to-one and onto.
(b) We define $\mathcal H: ^A {(^B C)} \to ^{A \times B} C$ and show that it is one-to-one and onto.
(c) We define $\mathcal H : \mathcal P(A) \to ^A { \{ yes, no\} }$ and show that it is one-to-one and onto.
(d) We need to prove $^{Z^+} {\mathcal P(Z^+)} \sim \mathcal P(Z^+)$.

Solution 6:
We prove that if A is denumerable, then there exists a denumerable family of sets $\mathcal F$ that partitions A. We define $\mathcal F = \{ X_p \vert p \text{ is prime } \}$, where $X_p = \{ a_i \in A \vert \text{ the smallest prime divisor of } i+1 \text{ is } p \}$. We show that $\mathcal F$ is a partition of A and that each $X_p$ is denumerable.

Solution 7:
We define $f: (\mathcal P(A) \times \mathcal P(B)) \to \mathcal P(A \cup B)$ and show that it is one-to-one and onto.

Solution 8:
TODO

Solution 9:
We prove that if $\mathcal F \subseteq \{ f \vert f: \mathbb Z^+ \to \mathbb R \}$ and $\mathcal F$ is countable, then there exists a function g such that $\mathcal F \subseteq \mathbb O(g)$.

Solution 10:
We prove that the set of all grammatical sentences in English is denumerable. We consider the set A of all words in English, which is finite and therefore countable. We then consider the set S of all finite sequences of A, which is countable by theorem 7.2.4. Since the set G of all grammatical sentences is a subset of S, it is also countable.

Solution 11:
(a) We prove that the set S of mathematical symbols defined by English sentences is denumerable. We define a function $f: S \to G$, where G is the set of English sentences, and show that it is one-to-one.
(b) We prove that there exist real numbers that cannot be defined by English sentences. We assume that all real numbers can be defined by English sentences and show that this leads to a contradiction.


### Video Transcript

Harry conducted a survey of 100 people to determine their favorite colors among blue, green, and red. The results showed that 38 people liked all three colors, 49 people liked green and blue, 9 people liked blue and red but not green, and 41 people liked green and red. Additionally, 14 people liked only blue, 67 people liked red, and 2 people did not like any of these colors.

To solve this problem, a Venn diagram is used to visualize the overlapping events. The diagram consists of three circles representing the three colors, enclosed in a rectangle called the universal set, which represents all 100 participants.

Starting with the intersection of all three circles, 38 people like all three colors. For the intersection of green and blue, 49 people like both, but this includes those who like all three colors, so 49 - 38 = 11 people like green and blue but not red. The intersection of blue and red has 9 people who like both but not green.

For the intersection of green and red, 41 people like both, but this includes those who like all three colors, so 41 - 38 = 3 people like green and red but not blue. The 14 people who like only blue are placed outside the intersections. Since 67 people like red, and this includes those who like red and green, red and blue, and all three colors, subtracting 38, 9, and 3 from 67 gives 17 people who like red but not blue or green.

The 2 people who do not like any of these colors are placed outside the circles. To find the number of people who like only green, subtract all values in the Venn diagram from 100: 100 - 14 - 2 - 11 - 9 - 38 - 3 - 17 = 6. So, 6 people like green but not blue or red.

The probability that a person chosen at random likes the color green is the total number of people who like green, which is 58 (11 + 3 + 38 + 6), out of 100. This probability is 58/100 = 0.58.

Given that the person Harry chooses likes the color green, the probability that they also like either blue or red, but not both, is calculated by narrowing down to the 58 people who like green. The number of people who like blue or red but not both is 11 + 3 = 14. The probability is 14/58, which simplifies to 7/29.


**Math Tricks**

### Simple Multiplication Tricks

To multiply numbers easily, try these tricks:

*   To multiply by 10, add a 0 behind the number.
*   To multiply by 11, use the 11 rule: separate the two digits, add them together, and put the result in the middle.

Example: 54 × 11 = 5 _ 4, 5 + 4 = 9, so 54 × 11 = 594.

### The 11 Rule Expanded

To multiply any number by 11, write the number with a 0 in front of it, draw a line under it, and add the neighboring numbers.

Example: 51236 × 11 = ?

Write 051236 with a line under it, then add the neighboring numbers:

*   6 (units) + 0 = 6
*   3 (tens) + 6 = 9
*   2 (hundreds) + 3 = 5
*   1 (thousands) + 2 = 3
*   5 (ten thousands) + 1 = 6
*   0 (hundred thousands) + 5 = 5

So, 51236 × 11 = 563596.

### Multiply Up to 20×20 in Your Head

To multiply numbers up to 20×20 in your head, use this trick:

*   Take 15 × 13 as an example.
*   Place the larger number on top (15) and the smaller number on the bottom (13).
*   Cover the 15 and the 3 with a mental image of Africa.
*   Add 15 + 3 = 18, then add a 0 to get 180.
*   Multiply the covered numbers: 3 × 5 = 15.
*   Add 180 + 15 = 195.

### Finger Math: 9X Rule

To multiply by 9, use your fingers:

1.  Spread your hands out and place them on a desk.
2.  Fold down the finger corresponding to the number you want to multiply by 9 (e.g., fold down the 3rd finger for 3 × 9).
3.  Count the fingers on either side of the folded finger to get the answer.

Example: 3 × 9 = ? Fold down the 3rd finger, count 2 fingers on the left and 7 fingers on the right, so 3 × 9 = 27.

### Square a 2-Digit Number Ending in 5

To square a 2-digit number ending in 5, use this trick:

*   Take 25 as an example.
*   Add 1 to the tens digit (2 + 1 = 3).
*   Multiply the original tens digit by the new tens digit (2 × 3 = 6).
*   Put 25 behind the result to get 625.

### Square 2-Digit Number: UPDOWN Method

To square a 2-digit number, use the UPDOWN method:

*   Take 37 as an example.
*   Find the nearest 10 boundary (40).
*   Go up 3 from 37 to 40 and down 3 from 37 to 34.
*   Multiply 34 × 40 = 1360.
*   Square the distance from the 10 boundary (3 × 3 = 9).
*   Add the two results to get 1360 + 9 = 1369.

### Multiply By 4

To multiply by 4, double the number and then double it again.

### Multiply By 5

To multiply by 5, divide the number in half and multiply by 10.

### Magic Addition Trick #1

Ask a friend to pick three different numbers between 1 and 9, write them down in order, and form a new number by reversing the digits. Then, subtract the smaller number from the larger number and ask your friend to add up the digits of the result. The sum will always be 18.

### Magic Square #15

Every row and column in this magic square sums to 15:

8  3  4
1  5  9
6  7  2

### Magic Square #34

Every row and column in this magic square sums to 34:

1  15  14  4
12  6  7  9
8  10  11  5
13  3  2  16

### A Recipe for Your Own 3 X 3 Magic Square

To create a 3x3 magic square, follow these rules:

*   Let a, b, and c be integers, where a > b + c.
*   Do not let 2b = c.
*   Use the formulas in the table to create the magic square.

### Upside-Down Magic Square

This magic square adds up to 264 in all directions, even when turned upside down:

96  11  89  68
88  69  91  16
61  86  18  99
19  98  66  81

### Antimagic Square

This magic square has as many different totals as possible:

5  1  3
4  2  6
8  7  9

### Lightning Calculator

Have someone write down their Social Security number, scramble it, and copy it next to the original. Change the second digit to 7 and the eleventh digit to 7. The result will always be divisible by 7 with no remainder.

### Fun Number Tables

*   3 × 37 = 111, 1 + 1 + 1 = 3
*   6 × 37 = 222, 2 + 2 + 2 = 6
*   9 × 37 = 333, 3 + 3 + 3 = 9
*   1 × 1 = 1
*   11 × 11 = 121
*   111 × 111 = 12321
*   1 × 9 + 2 = 11
*   12 × 9 + 3 = 111
*   123 × 9 + 4 = 1111
*   9 × 9 + 7 = 88
*   98 × 9 + 6 = 888
*   1 × 8 + 1 = 9
*   12 × 8 + 2 = 98
*   7 × 7 = 49
*   67 × 67 = 4489
*   4 × 4 = 16
*   34 × 34 = 1156

### Did You Know…

*   Each 2-digit number ending in 9 is the sum of the product of its digits plus their sum.
*   40 is the only number whose letters are in alphabetical order when spelled out.
*   A prime number is an integer greater than 1 that cannot be divided evenly by any other integer but itself and 1.
*   139 and 149 are the first consecutive primes differing by 10.
*   69 is the only number whose square and cube use all the digits 0-9 once each.
*   One pound of iron contains an estimated 4,891,500,000,000,000,000,000,000 atoms.
*   There are 318,979,564,000 possible ways to play the first four moves in a game of chess.
*   The Earth travels over 1.5 million miles every day.
*   The Eiffel Tower has 2,500,000 rivets.
*   If all the blood vessels in the human body were laid end to end, they would stretch for 100,000 miles.

### A Math Trick for This Year

This trick works for any year:

1.  Pick a number of days to go out per week (e.g., 17).
2.  Multiply by 2.
3.  Add 5.
4.  Multiply by 50.
5.  Add 1748 (or 1747 if you haven't had your birthday yet).
6.  Subtract your birth year (e.g., 19XX).

The result will be a 3-digit number where the first digit is the number of days to go out per week, and the last two digits are your age.

### The Mystery of 6174

6174 is a special number known as the Kaprekar constant. When you arrange its digits to form the largest and smallest possible numbers and subtract the smaller from the larger, you get 6174. This process will always converge to 6174 for any 4-digit number that is not composed of the same digit.

### Mental Math Tricks to Impress Your Friends

*   To multiply any 2-digit number by 11, write the first digit, add the first and second digits, and write the second digit.
*   To square a 2-digit number ending in 5, add 1 to the tens digit, multiply by the new tens digit, and put 25 behind the result.
*   To multiply two 2-digit numbers with the same first digit and second digits that sum to 10, multiply the first digit by the next higher number and append the product of the second digits.
*   To multiply by 9, multiply by 10 and subtract the original number.
*   To quickly find percentages, use the left-to-right approach for addition and subtraction.

### Addition

To add numbers mentally, use the left-to-right approach:

*   Start with the largest place value (hundreds, thousands, etc.).
*   Add the numbers from left to right.

Example: 45 + 30 + 4 = ?

Add 30 to 45 to get 75, then add 4 to get 79.

### Subtraction

To subtract numbers mentally, use the left-to-right approach:

*   Start with the largest place value (hundreds, thousands, etc.).
*   Subtract the numbers from left to right.

Example: 64 - 17 = ?

Subtract 10 from 64 to get 54, then subtract 7 to get 47.

### Using Complements to Simplify Subtractions

To simplify subtractions, use complements:

*   Find the difference between the two numbers.
*   Add the complement to the result.

Example: 674 - 58 = ?

Subtract 400 from 674 to get 274, then add 42 (the complement of 58) to get 316.

### Multiplication

To multiply numbers mentally, use the left-to-right approach:

*   Start with the largest place value (hundreds, thousands, etc.).
*   Multiply the numbers from left to right.

Example: 37 × 27 = ?

Multiply 30 × 20 to get 600, then multiply 30 × 7 to get 210, and add 7 × 20 to get 140. Add the results to get 950.

### Multiply by 5

To multiply by 5, cut the number in half and multiply by 10.

Example: 17 × 5 = ?

Cut 17 in half to get 8.5, then multiply by 10 to get 85.

### Multiply Numbers with Multiple Digits

To multiply numbers with multiple digits, use the distribution property:

*   Multiply the numbers digit by digit.
*   Add the results.

Example: 23 × 14 = ?

Multiply 2 × 1 to get 2, then multiply 2 × 4 to get 8, and add 3 × 1 to get 3, and 3 × 4 to get 12. Add the results to get 32.

### Square a Number Close to 10^2

To square a number close to 10^2, use Vedic mathematics:

*   Find the difference between the number and 10^2.
*   Square the difference.
*   Add the square to the number.

Example: 87^2 = ?

Find the difference between 87 and 100, which is 13. Square 13 to get 169. Add 169 to 87 to get 7569, but since we are working with a 2-digit number, we only need the last two digits, which are 69. Add 1 to the tens digit of 87 (8 + 1 = 9), then multiply by the new tens digit (9 × 8 = 72), and put 25 behind the result to get 7625, but we are working with 87, so we put 69 behind 72 to get 7269.

### Square Two Digits Ending with 5

To square a 2-digit number ending with 5, use this trick:

*   Take the first digit and multiply it by the next higher number.
*   Put 25 behind the result.

Example: 75^2 = ?

Multiply 7 × 8 to get 56, then put 25 behind the result to get 5625.

### Squaring Any Number

To square any number, use this trick:

*   Find the nearest 10 boundary.
*   Go up and down from the boundary to find the difference.
*   Square the difference.
*   Add the square to the number.

Example: 13^2 = ?

Find the nearest 10 boundary, which is 10. Go up 3 from 10 to 13, and down 3 from 13 to 7. Multiply 10 × 13 to get 130, then square 3 to get 9. Add 130 + 9 to get 139.

### Squaring Two-Digit Numbers

To square a 2-digit number, use this trick:

*   Take the first digit and multiply it by itself plus 2 times the second digit.
*   Take the second digit and multiply it by itself.
*   Add the results.

Example: 35^2 = ?

Multiply 3 × 3 to get 9, then multiply 3 × 5 to get 15, and multiply 5 × 5 to get 25. Add the results to get 1225.

### Mental Math Tricks

*   To multiply any two digits by 11, write the first digit, add the first and second digits, and write the second digit.
*   To square a two-digit number ending in 5, add 1 to the tens digit, multiply by the new tens digit, and put 25 behind the result.
*   To multiply two two-digit numbers with the same first digit and second digits that sum to 10, multiply the first digit by the next higher number and append the product of the second digits.
*   To multiply by 9, multiply by 10 and subtract the original number.
*   To quickly find percentages, use the left-to-right approach for addition and subtraction.


**Using Math4Me: A Step-by-Step Guide**

1. **Select a Topic**: Choose a topic you need to practice.
2. **Watch Video Lessons**: Watch and listen to the video lessons for the selected topic.
3. **Complete Assignments**: Do the IXL assignments and printable assignments that accompany the video lesson. To access IXL lessons, write down the lesson code (e.g., F.1, G.5, B.2), go to IXL.com, log in, and enter the code in the search box. Work on the IXL lesson until you achieve a score of 100.
4. **Take the GAP Test**: Take the GAP test in class, which is scheduled periodically throughout each marking period. The test will cover topics you have been practicing.

**Level 1 Topics**

* **Basic Multiplication**:
    + Basic multiplication facts
    + Exit test: basic_multiplication_facts_exit_test.pdf (24 kb)
* **Basic Division**:
    + Objective: Solve 30 basic division problems in less than 2 minutes
    + Lessons: H.1-H.12 (divide by 1-12)
    + Exit test: basic_division_facts_exit_test.pdf (33 kb)
* **Multiplying Larger Numbers**:
    + Objective: Multiply 1-digit numbers with larger numbers
    + Examples: 47 x 843
    + Lessons: D.5-D.7 (multiply 1-digit numbers by 2-digit, 3-digit, or 4-digit numbers), F.14-F.16 (multiply 2-digit numbers by 3-digit numbers or larger)
    + Exit tests: multiplying_by_1_digit_numbers_test.pdf (90 kb), multiplying_by_2_digit_numbers_test.pdf (91 kb)
* **Multiplying Numbers Ending with Zeroes**:
    + Objective: Quickly multiply numbers ending with zeroes
    + Lessons: E.8, D.8, D.24, D.25
    + Exit test: multiplying_numbers_ending_in_zeros_test.pdf (92 kb)
* **Dividing: 1-Digit Divisors**:
    + Objective: Divide large numbers with 1-digit divisors
    + Example: 498 divided by 6
* **Place Value and Number Sense**:
    + Objective: Read and write large numbers in standard form, word form, and expanded form; identify the value of digits in large numbers
    + Lessons: B.1-B.4
    + Exit test: place_value_test.pdf (95 kb)
* **Rounding Numbers**:
    + Objective: Round numbers to a given place value
    + Exit test: rounding_test.pdf (92 kb)
* **Factors, Primes, and Composites**:
    + Objective: Find the factors of a number; explain the difference between prime and composite numbers
    + Lessons: A.5, D.3, D.4, N.8
    + Exit test: factors_and_multiples_test.pdf (92 kb)
* **Basic Fractions: Parts and Wholes**:
    + Objective: Name fractions using visual models
    + Lessons: S.1-S.4
* **Comparing Fractions**:
    + Objective: Compare fractions with the same numerator or denominator
    + Lessons: U.3-U.11
    + Video lessons: Learnzillion.com (LZ1739, LZ1740, LZ1268)
* **Fractions that Equal 1 Whole**:
    + Objective: Write fractions that are equal to 1 whole and more than 1 whole
    + Lessons: T.4, T.8
    + Video lessons: Learnzillion.com (LZ1736, LZ1737, LZ1738)
* **Equivalent Fractions**:
    + Objective: Make equivalent fractions using models and/or multiplication
    + Lessons: T.1-T.5
    + Video lessons: Learnzillion.com (LZ1521, LZ1522, LZ531, LZ532, LZ533, LZ1731-LZ1735)
* **Decimals**:
    + Objective: Learn about decimals through IXL tasks (video lessons not yet available)
    + Lessons: W.1-W.9

**Note:** The section below is under construction.

**Level 2-3 Topics**

* **Equivalent Fractions with Denominators of 10 and 100**
* **Adding and Subtracting Fractions with Unlike Denominators**
* **Adding and Subtracting Fractions with Like Denominators**
* **Comparing Fractions using Benchmarks**
* **Adding and Subtracting Mixed Numbers with Different Denominators**


A set of vectors is linearly independent if the only way to express the zero vector as a linear combination of the vectors is with all coefficients equal to zero.

Key facts about linear independence include:
- If 31,23,. iTn are linearly independent vectors in X, then there are TA -İ, in X" such that A(x)=6'.
- If X is infinite dimensional, then so is X.
- Every finite dimensional vector subspace of X has a complement.

For a set S of m vectors in Rn with m>n, the set S is linearly dependent. This is because in a vector space of dimension n, there can be at most n linearly independent vectors.

To determine whether a set of vectors is linearly independent, one can use the following method:
- Write the vectors as columns of a matrix.
- Transform the matrix to its row echelon form.
- Count the number of non-zero rows. This is the number of linearly independent vectors.

For example, to determine whether the vectors (1 2), (1 3) are linearly independent in R^2, one can write them as columns of a matrix, transform the matrix to its row echelon form, and count the number of non-zero rows.

In general, a set of vectors is linearly independent if and only if the determinant of the matrix formed by the vectors is non-zero. If the determinant is zero, then the vectors are linearly dependent.

To determine whether a vector ~uis in the span of another given set of vectors, one can use the following method:
- Write the vectors as columns of a matrix.
- Transform the matrix to its row echelon form.
- Determine whether the vector ~u can be expressed as a linear combination of the columns of the matrix.

Theorem 4 (page 37) relates to solving linear systems and can be used to determine whether a set of vectors is linearly independent.

Some key cases to consider when determining linear independence include:
- If there are more vectors than the dimension of the vector space, then the vectors are linearly dependent.
- If the vectors are scalar multiples of each other, then they are linearly dependent.
- If the vectors are orthogonal to each other, then they are linearly independent.

In summary, linear independence is an important concept in linear algebra that can be determined using various methods, including row echelon form and determinants. It is essential to understand the key facts and theorems related to linear independence to solve problems in linear algebra.


Now that you've been introduced to averages and Doppler equations, let's combine these two ideas and explore Einstein's theory of relativity. We'll begin with a few key questions and answers:

* Do you need relativity theory to find the average of two numbers?
Answer: No, you do not need relativity theory to find an average.

* What is the meaning of vx' / (c^2 - v^2) in Section 3 of Einstein's 1905 derivation?
Answer: The expression vx' / (c^2 - v^2) is the half-difference, not a non-specific "adjustment to time" as it is commonly explained.

To begin demystifying relativity theory, let's find the average Doppler wavelength equation. The Doppler wavelength equations for an approach light source is x' / (1 + v/c) and a receding light source is x' / (1 - v/c), where x' is the original wavelength, v is the velocity of the light source, and c is the propagation speed of the light wave. The average Doppler equation, ξ, is easily found using the addition mean equation ξ = (U1 + U2) / 2.

We found this equation without using relativity theory. What's important to notice is that this exact equation appears in Einstein's paper. However, Einstein does not realize that he arrives at an average, probably because he does not use the addition mean equation.

Einstein uses the receding and approaching Doppler equations, x' / (c - v) and x' / (c + v), and finds their half-difference using the subtraction mean equation τ = U1 - (U1 - U2) / 2. The half-difference of these expressions is vx' / (c^2 - v^2). When subtracted from the larger operand, x' / (c - v), it produces the average of the two expressions.

Einstein performs this subtraction in Section 3 of his paper and immediately multiplies by c to arrive at the average ξ. Although his approach is complicated, Einstein finds the average ξ using the non-obvious subtraction mean equation rather than with the familiar addition mean equation. Failing to recognize ξ as an average, Einstein interprets it to represent the x-axis coordinate of a space-time point.

Summarizing, we've shown that:
- You don't need relativity to find an average.
- Einstein's ξ equation (prior to his final adjustments) is identical to the average Doppler equation.
- The expression vx' / (c^2 - v^2) is the half-difference, not a non-specific time adjustment.

This is only the beginning of our look at Einstein's theory, which we'll come back to in later tutorials.


# ISTE Student Standard: 5d - Automation & Algorithmic Thinking

The "Students Compute like Computers to Solve Real Problems" lesson series, designed for K-2 students, focuses on developing computational thinking skills to analyze and interpret numerical data. The series consists of 7 lessons, each building on the previous one to help students derive insights and offer solutions to a real-world problem of supply and demand for marbles in a classroom.

### Lesson 1: Organize Categorical Data
Students collect and organize data into numerical values using free educational technology, preparing to visualize numerical data.

### Lesson 2: Visualize Numerical Data
Students visualize numerical data into bar graphs and learn a computational procedure to analyze and interpret the data.

### Lesson 3: Analyze Data in Bar Charts
Students interpret visualized numerical data, deriving insights about the supply and demand problem.

### Lesson 4: (Not Provided)
No information is available for this lesson.

### Lesson 5: Brainstorm Solutions to a Problem
Using a digital collaboration board, students brainstorm solutions to the identified problem, applying computational procedures to analyze and interpret numerical data.

### Lesson 6: Test Solutions
Students test and evaluate their brainstormed solutions, using data to inform their decisions.

### Lesson 7: Recite Steps for Algorithmic Thinking
Students review and codify the steps taken to analyze and interpret numerical data, solidifying their understanding of computational procedures.


To find the inscribed square $ABCD$, follow these steps:

1. Choose a point $H$ on side $PQ$, near $P$, and draw a line $HG$ perpendicular to $PR$.
2. Take a segment $HE$ of the same length as $HO$ on a line through $H$ parallel to $PR$.
3. Draw the square $HEFO$ by drawing $EF$ perpendicular to $PR$.
4. The line $PE$ intersects $QR$ at $A$.
5. From $A$, draw a parallel to $PR$ (intersecting $PQ$ at $D$) and a perpendicular to $PR$ (intersecting $PR$ at $B$).
6. Find the vertex $C$ on $PR$ such that $CD$ is perpendicular to $PR$, completing the rectangle $ABCD$.

The rectangle $ABCD$ is a square because triangles $PEF$ and $PAB$ are similar, with a common angle, and triangles $PEH$ and $PAD$ are similar, with equal angles $PEH$ and $PAD$ and a common angle $HPE$. The similarity ratios are:

$$\frac{PE}{PA} = \frac{EF}{AB}$$
$$\frac{PE}{PA} = \frac{HE}{AD}$$

Combining these ratios yields:

$$\frac{AB}{EF} = \frac{AD}{HE}$$

Since $HE = EF$ (sides of a square), $AB = AD$, making $ABCD$ a rectangle with two adjacent sides equal, and thus a square.

The construction of the inscribed square can be done in the following steps:

1. Choose a point $H$ on side $PQ$, near $P$.
2. Draw line $HG$, perpendicular to $PR$.
3. Take the distance $HG$ as the compass distance and draw a circle arc with center $G$, intersecting $PR$ at $F$.
4. Construct two circle arcs with centers $F$ and $H$ (with the same radius as before), intersecting at point $E$, forming a square $EFGH$.
5. Draw line $PE$, intersecting side $QR$ at $A$.
6. Draw from $A$ parallels to $EF$ and $HE$, intersecting $PR$ and $QP$ at points $B$ and $D$, respectively.
7. Draw from $D$ a parallel to $AB$, intersecting $PR$ at $C$.
8. $ABCD$ is the inscribed square.

Note that the choice of point $H$ such that $E$ is interior to triangle $PQR$ is not restrictive, as the construction remains the same if $E$ is exterior to triangle $PQR$.


# Selecting 6 People for a Flight to Hawaii

There are 14 standbys hoping to get on a flight to Hawaii, but only 6 seats are available. The task is to determine the number of different ways to select 6 people from the 14 standbys.

## Solution

The answer is given by the combination formula, denoted as "14 choose 6". This can be calculated using the formula:

nCk = n! / (k!(n-k)!)

where n is the total number of items, k is the number of items to select, and "!" denotes factorial, the product of all natural numbers from 1 to the given number.

## Calculation

For this problem, n = 14 and k = 6. Applying the formula:

14C6 = 14! / (6!(14-6)!)
= 14! / (6!8!)
= 3003

Therefore, there are 3003 different ways to select 6 people from the 14 standbys.


## Gujarat Board Textbook Solutions Class 12 Maths Chapter 8 Application of Integrals Ex 8.1

### Question 1
Find the area of the region bounded by the curve y^2 = x and the lines x = 1, x = 4, and the x-axis.

The curve y^2 = x is a parabola with its vertex at the origin. The area of the region bounded by the curve, x = 1, x = 4, and the x-axis can be calculated by finding the area under the curve between x = 1 and x = 4.

### Question 2
Find the area of the region bounded by y^2 = 9x, x = 2, x = 4, and the x-axis in the first quadrant.

The given curve is y^2 = 9x, which is a parabola with its vertex at (0, 0) and axis along the x-axis. The area of the region bounded by the curve, x = 2, x = 4, and the x-axis can be calculated by finding the area under the curve between x = 2 and x = 4.

### Question 3
Find the area of the region bounded by x^2 = 4y, y = 2, y = 4, and the y-axis in the first quadrant.

The given curve x^2 = 4y is a parabola with its vertex at (0, 0). The area of the region bounded by the curve, y = 2, y = 4, and the y-axis can be calculated by finding the area under the curve between y = 2 and y = 4.

### Question 4
Find the area of the region bounded by the ellipse (x^2)/16 + (y^2)/9 = 1.

The equation of the ellipse is (x^2)/16 + (y^2)/9 = 1. The area bounded by the ellipse can be calculated by finding the area of the ellipse in the first quadrant and multiplying it by 4.

### Question 5
Find the area of the region bounded by the ellipse (x^2)/4 + (y^2)/9 = 1.

The given ellipse has its center at (0, 0), with a semi-major axis of length 3 and a semi-minor axis of length 2. The area bounded by the ellipse can be calculated using the formula for the area of an ellipse.

### Question 6
Find the area of the region in the first quadrant enclosed by the x-axis, the line x = √3y, and the circle x^2 + y^2 = 4.

The circle x^2 + y^2 = 4 has its center at (0, 0) and a radius of 2. The line x = √3y intersects the circle at the point (√3, 1). The required area is the shaded region, which can be calculated by finding the area of the sector of the circle and subtracting the area of the triangle.

### Question 7
Find the area of the smaller part of the circle x^2 + y^2 = a^2 cut off by the line x = a/√2.

The equation of the given circle is x^2 + y^2 = a^2. The line x = a/√2 is a straight line parallel to the y-axis at a distance of a/√2 units to the right of the y-axis. The point of intersection of the circle and the line in the first quadrant is (a/√2, a/√2).

### Question 8
The area between x = y^2 and x = 4 is divided into two equal parts by the line x = a. Find the value of a.

The graph of the curve x = y^2 is a parabola with its vertex at (0, 0) and axis along the x-axis. The area of the region bounded by the curve and the ordinate x = a can be calculated by finding the area under the curve between x = 0 and x = a.

### Question 9
Find the area of the region bounded by the parabola y = x^2 and the lines y = |x|.

The curve y = x^2 represents a parabola with its vertex at (0, 0) and axis along the y-axis. The lines y = |x|, i.e., y = x and y = -x, are two lines passing through the origin and making an angle of 45° and 135° with the positive direction of the x-axis.

### Question 10
Find the area bounded by the curve x^2 = 4y and the straight line x = 4y - 2.

The given curve x^2 = 4y is a parabola with its vertex at (0, 0) and axis along the y-axis. The equation of the line is x = 4y - 2. The points of intersection of the curve and the line are A(-1, 1/4) and B(2, 1).

### Question 11
Find the area of the region bounded by the curve y^2 = 4x and the line x = 3.

The curve y^2 = 4x is a parabola with its vertex at (0, 0) and axis along the x-axis. The area of the region bounded by the curve and the line x = 3 can be calculated by finding the area under the curve between x = 0 and x = 3.

### Question 12
Area lying in the first quadrant and bounded by the circle x^2 + y^2 = 4 and the lines x = 0 and x = 2 is

A) π
B) π/2
C) π/3
D) π/4

The area of the required region can be calculated by finding the area of the sector of the circle and subtracting the area of the triangle.

### Question 13
Area of the region bounded by the curve y^2 = 4x, y-axis, and the line y = 3 is

A) π
B) 9/4
C) 9/3
D) 9/2

The curve y^2 = 4x is a parabola with its vertex at (0, 0) and axis along the x-axis. The line y = 3 intersects the parabola at the point (9/4, 3). The area of the region bounded by the curve, y-axis, and the line y = 3 can be calculated by finding the area under the curve between y = 0 and y = 3.


### All SSAT Upper Level Math Resources

## Example Questions

### Example Question #1 : Arithmetic Sequences

The angle measures of a pentagon form an arithmetic sequence. The smallest angle measures 108 degrees. What does the largest angle measure?

The measures of the five angles of a pentagon add up to 540 degrees. If the measures of the five angles form an arithmetic sequence, the angles have measures increasing by a common difference. Given this pattern and the total sum, we can solve for the common difference. The greatest of the angle measures is given by 108 + 4d, where d is the common difference.

### Example Question #2 : Arithmetic Sequences

Hunter makes $15 for the first hour of work, $18 for his second hour of work, $21 for his third hour of work, and so on. How much money will he make if he works for 11 hours?

To find the sum of the first terms of the arithmetic sequence, use the formula: S = n/2 * (a + l), where S is the sum, n is the number of terms, a is the first term, and l is the last term. To find the sum, we need to first find the 11th term of the sequence using the formula: an = a + (n-1)d, where an is the nth term, a is the first term, n is the term number, and d is the common difference.

### Example Question #3 : Arithmetic Sequences

Find the 8th term of the arithmetic sequence: 2, 5, 8, 11, 14, ...

To find any term in an arithmetic sequence, use the formula: an = a + (n-1)d, where an is the nth term, a is the first term, n is the term number, and d is the common difference. For this sequence, a = 2 and d = 3. Now, plug in the information to find the value of the 8th term.

### Example Question #4 : Arithmetic Sequences

Find the 16th term of the following arithmetic sequence: 3, 7, 11, 15, ...

To find any term in an arithmetic sequence, use the formula: an = a + (n-1)d, where an is the nth term, a is the first term, n is the term number, and d is the common difference. For this sequence, a = 3 and d = 4. Now, plug in the information to find the value of the 16th term.

### Example Question #5 : Arithmetic Sequences

Find the 9th term of the following arithmetic sequence: 2, 5, 8, 11, 14, ...

To find any term in an arithmetic sequence, use the formula: an = a + (n-1)d, where an is the nth term, a is the first term, n is the term number, and d is the common difference. For this sequence, a = 2 and d = 3. Now, plug in the information to find the value of the 9th term.

### Example Question #6 : Arithmetic Sequences

Find the 12th term of the following arithmetic sequence: 1, 4, 7, 10, ...

To find any term in an arithmetic sequence, use the formula: an = a + (n-1)d, where an is the nth term, a is the first term, n is the term number, and d is the common difference. For this sequence, a = 1 and d = 3. Now, plug in the information to find the value of the 12th term.

### Example Question #7 : Arithmetic Sequences

Find the 7th term for the following arithmetic sequence: 3, 6, 9, 12, ...

To find any term in an arithmetic sequence, use the formula: an = a + (n-1)d, where an is the nth term, a is the first term, n is the term number, and d is the common difference. For this sequence, a = 3 and d = 3. Now, plug in the information to find the value of the 7th term.

### Example Question #8 : Arithmetic Sequences

Find the 10th term of the following arithmetic sequence: 2, 5, 8, 11, 14, ...

To find any term in an arithmetic sequence, use the formula: an = a + (n-1)d, where an is the nth term, a is the first term, n is the term number, and d is the common difference. For this sequence, a = 2 and d = 3. Now, plug in the information to find the value of the 10th term.

### Example Question #9 : Arithmetic Sequences

Brandon is improving very quickly at math. He improves at a rate of 2 points per test. If he scored a 70 on his first test, 72 on his second test, and 74 on his third test, what score will he get on his 8th test?

You should recognize this as an arithmetic sequence. The question is asking you to find the 8th term in this particular sequence. To find any term in an arithmetic sequence, use the formula: an = a + (n-1)d, where an is the nth term, a is the first term, n is the term number, and d is the common difference. Using the information given from the question, a = 70 and d = 2. Now, plug in the information to find the value of the 8th term.

### Example Question #10 : Arithmetic Sequences

Julia gets better every time she plays basketball. In her first game, she scored 10 points. In her second game, she scored 12 points, and in her third game, she scored 14 points. If she continues to improve her basketball skills at this same pace, how many points should she be scoring by her 12th game?

You should recognize this as an arithmetic sequence. The question is asking you to find the 12th term in this particular sequence. To find any term in an arithmetic sequence, use the formula: an = a + (n-1)d, where an is the nth term, a is the first term, n is the term number, and d is the common difference. Using the information given from the question, a = 10 and d = 2. Now, plug in the information to find the value of the 12th term.


I've been working with a time-domain function to calculate the Ricker (Mexican Hat) wavelet for $-0.2 \leq t \leq 0.2$. The function is given as $s(t) = \left( 1 - \frac{1}{2} \omega_0^2 t^2 \right) \mbox{exp} \left ( -\frac{1}{4} \omega_0^2 t^2 \right )$, where $t$ is time (s) and $\omega_0$ is the angular frequency (1/s).

The goal is to find an analytical expression to determine the width of this wavelet, which is the time between the two side-lobes. The wavelet has a maximum value at $t = 0$, with side-lobes on either side.

A paper states that the width $w$ of the Ricker wavelet is $w = \sqrt{6} / \pi / f_0$, where $f_0$ is the peak frequency (Hz) of the wavelet. However, the derivation of this expression is unclear.

Given values are:
- 76
- 0
- #1

Note: The provided values (#1, 76, 0) do not seem to be directly related to the problem description. If they are meant to be part of a multiple-choice question or have a specific relevance, please clarify for proper refinement.


KEMS: GRADE 4 Homework
Problem Solving
1. SOLVE - S and O: Homework for students to develop problem-solving skills.
2. SOLVE - L: Homework for students to apply problem-solving strategies.
3. SOLVE - V and E: Homework for students to enhance problem-solving abilities.

Number & Operations in Base Ten
4. Multi-Digit Place Value and Rounding: Homework to understand place value and rounding concepts.
5. Multi-Digit Expanded Form: Homework to apply expanded form in multi-digit numbers.
6. Add and Subtract Multi-Digit Whole Numbers: Homework to practice addition and subtraction of multi-digit numbers.
7. Multiply Multi-Digit Whole Numbers with Property Application: Homework to apply multiplication properties to multi-digit numbers.
8. Divide Multi-Digit Whole Numbers with Property Application: Homework to apply division properties to multi-digit numbers.

Operations & Algebraic Thinking
9. Multiplication and Division Equations: Homework to solve equations involving multiplication and division.
10. Multiplication and Division Situations: Homework to apply multiplication and division in real-world situations.
11. Multi-Step Word Problems with Remainders: Homework to solve multi-step word problems with remainders.
12. Least Common Multiple (LCM): Homework to understand and apply LCM concepts.
13. Greatest Common Factor (GCF), Prime and Composite Numbers: Homework to understand GCF, prime, and composite numbers.
14. Numeric Patterns: Homework to identify and extend numeric patterns.

Numbers & Operations – Fractions
15. Concept of Fractions: Homework to introduce fraction concepts.
16. Add Fractions - Like Denominators: Homework to practice adding fractions with like denominators.
17. Subtract Fractions - Like Denominators: Homework to practice subtracting fractions with like denominators.
18. Add Mixed Fractions - Like Denominators: Homework to practice adding mixed fractions with like denominators.
19. Subtract Mixed Fractions - Like Denominators: Homework to practice subtracting mixed fractions with like denominators.
20. Multiply Fractions: Homework to practice multiplying fractions.
21. Fraction and Decimal Equivalence: Homework to understand fraction and decimal equivalence.
22. Compare Decimals: Homework to compare and order decimals.

Measurement & Data
23. Measurement Equivalence: Homework to understand measurement equivalence.
24. Measurement with Word Problems: Homework to apply measurement concepts to word problems.
25. Area and Perimeter of Rectangles with Word Problems: Homework to find area and perimeter of rectangles in word problems.
26. Line Plots: Homework to create and interpret line plots.
27. Angle Discovery: Homework to introduce angle concepts.
28. Additive Angle Measures: Homework to apply additive angle measures.

Geometry
29. Classify Two Dimensional Figures: Homework to classify two-dimensional figures.
30. Line Symmetry: Homework to understand line symmetry concepts.

Additional
Appendix A - Fact Masters - Multiplication: Homework to master multiplication facts.
Appendix B - Fact Masters - Division: Homework to master division facts.