Illustrative Math Alignment: Grade 8 Unit 3

Topic

Solving Equations

Description

This math example demonstrates solving percent equations by asking "650 is 130% of what number?" The solution involves setting up the equation 650 = 1.3 * x, then solving for x to get x = 650 / 1.3, which equals 500. This example introduces a scenario where we need to find the original value when given a percentage greater than 100%, resulting in a number that is smaller than the given value.

Topic

Solving Equations

Description

This math example focuses on solving percent equations, specifically asking "What is 30% of 9?" The solution involves converting 30% to its decimal form, 0.3, and then multiplying it by 9 to get the result of 2.7. This example introduces a larger percentage, demonstrating how the method applies consistently across various percentage values.

Topic

Solving Equations

Description

This math example demonstrates solving percent equations by asking "What is 28% of 7.2?" The solution involves converting 28% to its decimal equivalent, 0.28, and then multiplying it by 7.2 to obtain the result of 2.016. This example combines a whole number percentage with a decimal base number, further illustrating the versatility of the percent-to-decimal conversion method.

Topic

Solving Equations

Description

This math example focuses on solving percent equations, specifically asking "What is 45% of 68?" The solution involves converting 45% to its decimal form, 0.45, and then multiplying it by 68 to arrive at the answer of 30.6. This example introduces a larger percentage and a larger whole number as the base value, demonstrating the scalability of the percent-to-decimal conversion method.

Topic

Solving Equations

Description

This math example demonstrates solving percent equations by asking "What is 52.3% of 36.9?" The solution involves converting 52.3% to its decimal equivalent, 0.523, and then multiplying it by 36.9 to obtain the result of 19.2987. This example introduces both a decimal percentage and a decimal base number, adding complexity to the calculation and showcasing the versatility of the percent-to-decimal conversion method.

Topic

Solving Equations

Description

This math example focuses on solving percent equations, specifically asking "What is 150% of 8?" The solution involves converting 150% to its decimal form, 1.5, and then multiplying it by 8 to get the result of 12. This example introduces a percentage greater than 100%, demonstrating how the method applies consistently even when dealing with percentages that represent values larger than the whole.

Topic

Volume

Description

A rectangular prism with dimensions labeled: length = 30, width = 10, and height = 8. The image shows how to find the volume of the prism using the formula for volume of a rectangular prism. This image illustrates Example 1: The caption explains how to calculate the volume of the rectangular prism using the formula V = l * w * h. The given dimensions are substituted into the formula: V = 30 * 10 * 8 = 2400..

Topic

Volume

Description

A green cylinder with a general radius y and height x. The radius is marked on the top surface, and the height is marked on the side. This image illustrates Example 10: The task is to find the volume of this cylinder. The volume formula V = πr2h is used, and substituting r = y and h = x, the volume is calculated as V = xy2π.

Topic

Volume

Description

A hollow green cylinder with an outer radius of 10 units, an inner radius of 9 units, and a height of 15 units. The radii are marked on the top surface, and the height is marked on the side. This image illustrates Example 11: The task is to find the volume of this hollow cylinder. The volume formula for a hollow cylinder V = πr12h1 - πr22h2 is used. Substituting values, the result is V = 285π.

Topic

Volume

Description

A hollow green cylinder with an outer radius y, an inner radius y - 1, and a height x. The radii are marked on the top surface, and the height is marked on the side. This image illustrates Example 12: The task is to find the volume of this hollow cylinder. Using V = π(r12h1 - r22h2), substituting values gives: V = πx(y2 - (y - 1)2= πx(2y - 1).

Topic

Volume

Description

A rectangular-based pyramid is shown with dimensions: base length 10, base width 8, and height 30. The image demonstrates how to calculate the volume of this pyramid. This image illustrates Example 13: The caption provides a step-by-step solution for calculating the volume of a pyramid with a rectangular base using the formula V = (1/3) * Area of Base * h. Substituting values: V = (1/3) * 8 * 10 * 30 = 800.

Topic

Volume

Description

A general rectangular-based pyramid is shown with variables x, y, and z representing the base dimensions and height. This example shows how to calculate the volume of a pyramid using variables instead of specific numbers. This image illustrates Example 14: The caption explains how to calculate the volume of a pyramid with a rectangular base using the formula V = (1/3) * Area of Base * h, which simplifies to V = (1/3) * x * y * z.

Topic

Volume

Topic

Volume

Description

A truncated rectangular-based pyramid is shown with variables x, y, and z representing dimensions. The smaller virtual pyramid has reduced dimensions by 3 units for both width and length and reduced height by z - 20. The image demonstrates how to calculate the volume in terms of variables. This image illustrates Example 16: The caption explains how to find the volume of a truncated pyramid using variables for both pyramids' dimensions. Formula: V = (1/3) * xy(z + 20) - (1/3) * (y - 3)(x - 3)(z), which simplifies to V = (1/3) * (xyz + 60x + 60y - 180).

Topic

Volume

Description

A green sphere with a radius labeled as 3. The image is part of a math example showing how to calculate the volume of a sphere. This image illustrates Example 17: The text describes finding the volume of a sphere. The formula used is V = (4/3) * π * r3, where r = 3. After substituting, the result is V = 36π.

Volume is a fundamental concept in geometry that helps students understand the space occupied by three-dimensional objects. In this collection, each example uses various geometric shapes to calculate volume, showcasing real-life applications of volume in different shapes.

Topic

Volume

Description

A green sphere with a radius labeled as x. This image is part of a math example showing how to calculate the volume of a sphere using an unknown radius. This image illustrates Example 18: The text explains how to find the volume of a sphere with an unknown radius x. The formula used is V = (4/3) * π * r3, and substituting r = x gives V = (4/3) * x3 * π.

Topic

Volume

Description

A green cube with side length labeled as 7. The image illustrates how to calculate the volume of a cube with known side length. This image illustrates Example 19: The text describes finding the volume of a cube. The formula used is V = s3, where s = 7. After substituting, the result is V = 343.

Volume is a fundamental concept in geometry that helps students understand the space occupied by three-dimensional objects. In this collection, each example uses various geometric shapes to calculate volume, showcasing real-life applications of volume in different shapes.

Topic

Volume

Description

A rectangular prism with dimensions labeled as x, y, and z. The image shows a general example of calculating the volume of a rectangular prism using variables instead of specific numbers. This image illustrates Example 2: The caption describes how to find the volume of a rectangular prism using variables for length (x), width (y), and height (z). The formula is given as V = x * y * z, but no specific values are provided.

Topic

Volume

Description

A green cube with side length labeled as x. This image is part of a math example showing how to calculate the volume of a cube using an unknown side length. This image illustrates Example 20: The text explains how to find the volume of a cube with an unknown side length x. The formula used is V = s3, and substituting s = x gives V = x3.

Topic

Volume

Description

A hollow cube with an outer edge of 9 and an inner hollow region with an edge of 7. The image shows how to calculate the volume by subtracting the volume of the inner cube from the outer cube. This image illustrates Example 21: Find the volume of a hollow cube. The formula used is V = s13 - s23, where s1 is the outer edge (9) and s2 is the inner edge (7). The solution calculates 9^3 - 7^3 = 386..

Topic

Volume

Description

A hollow cube with an outer edge of x and an inner hollow region with an edge of x - 2. The image shows how to calculate the volume by subtracting the volume of the inner cube from the outer cube. This image illustrates Example 22: Find the volume of a hollow cube. The formula used is V = s13 - s23, where s1 = x and s2 = x - 2. Expanding and simplifying gives V = 6x2 - 12x + 8.

Topic

Volume

Description

A cone with a height of 12 and a radius of 4. The image shows how to calculate its volume using the cone volume formula (V = 1/3 * π * r2 * h). This image illustrates Example 23: Find the volume of a cone. The formula used is V = (1/3) * π * r2 * h, where r = 4 and h = 12. Substituting these values gives V = (1/3) * π * (42) * 12 = 64π.

Topic

Volume

Description

A cone with a height labeled as y and a radius labeled as x. The image shows how to calculate its volume using the cone volume formula (V = 1/3 * π * r2 * h). This image illustrates Example 24: Find the volume of a cone. The formula used is V = (1/3) * π * r2 * h, where r = x and h = y. Substituting these variables gives V = (x^2 * y)/3 * π.

Topic

Volume

Description

A hollow rectangular prism with outer dimensions: length = 60, width = 20, and height = 16. The inner hollow part has dimensions: length = 60, width = 18, and height = 14. The image shows how to subtract volumes to find the hollow volume. This image illustrates Example 3: The caption explains how to calculate the volume of a hollow rectangular prism by subtracting the inner volume from the outer volume. V = (60 * 20 * 16) - (60 * 18 * 14) = 4080.

Topic

Volume

Description

A hollow rectangular prism with outer dimensions labeled as x, y, and z, and inner hollow dimensions labeled as x - 2 and y - 2. The image shows a symbolic calculation for finding the hollow volume using variables. This image illustrates Example 4: The caption describes how to calculate the volume of a hollow rectangular prism by subtracting the inner volume from the outer volume using variables: V = xyz - z(y - 2)(x - 2) = 2z(y + x - 2).

Topic

Volume

Description

The image shows a triangular prism with dimensions labeled as base (7), height (10), and length (25). It is part of an example on how to calculate the volume of a solid triangular prism. This image illustrates Example 5: "Find the volume of this triangular prism." The solution involves substituting the given measurements into the volume formula for a triangular prism: V = 1/2 * b * h * l = 1/2 * 7 * 10 * 25 = 875.

Topic

Volume

Description

The image depicts a triangular prism with dimensions labeled as x, y, and z. The example demonstrates how to calculate the volume using a general formula for a triangular prism. This image illustrates Example 6: "Find the volume of this triangular prism." The solution uses the formula V = 1/2 * b * h * l, which is simplified to V = 1/2 * x * y * z..

Topic

Volume

Description

The image shows a hollow triangular prism with outer dimensions labeled as base (10), height (7), and length (35), and inner dimensions labeled as base (8) and height (5). The example calculates the volume by subtracting the hollow region from the full prism. This image illustrates Example 7: "Find the volume of this hollow triangular prism." The solution calculates the full volume using V = 1/2 * b1 * h1 * l1 - 1/2 * b2 * h2 * l2, which simplifies to V = 1/2 * 10 * 7 * 35 - 1/2 * 8 * 5 * 35 = 525..

Topic

Volume

Description

This image shows a hollow triangular prism with outer dimensions labeled as x, y, and z, and inner dimensions reduced by 2 units each. It demonstrates how to calculate the volume by subtracting the hollow region from the full prism. This image illustrates Example 8: "Find the volume of this hollow triangular prism." The solution uses V = 1/2 * b1 * h1 * l1 - 1/2 * b2 * h2 * l2, which simplifies to V = z(xy - (x - 2)(y - 2)) = z(x + y - 2)..

Topic

Volume

Description

A green cylinder with a radius of 10 units and a height of 8 units. The radius is marked on the top surface, and the height is marked on the side. This image illustrates Example 9: The task is to find the volume of the cylinder. The volume formula V = πr2h is used. Substituting the values r = 10 and h = 8, the volume is calculated as V= 800π.

Topic

Numerical and Algebraic Expressions

Description

This example demonstrates the addition of 7 and 12 using algebra tiles. Two groups of yellow tiles, one with 7 tiles and another with 12 tiles, are combined to show a total of 19 tiles. This visual representation helps students understand the concept of addition with positive integers.

Topic

Numerical and Algebraic Expressions

Description

This example demonstrates the subtraction of 5 from -3 using algebra tiles. The image shows 3 red tiles representing -3, and 5 yellow tiles are added to create zero pairs. After removing the zero pairs, 2 additional red tiles remain, illustrating that -3 - 5 = -8.

Topic

Numerical and Algebraic Expressions

Description

This example illustrates the subtraction of -3 from -8 using algebra tiles. The image shows 8 red tiles representing -8, and 3 red tiles are removed to demonstrate the subtraction of -3. The result is 5 red tiles, visually showing that -8 - (-3) = -5.

Topic

Numerical and Algebraic Expressions

Description

This example demonstrates the subtraction of -6 from -4 using algebra tiles. The image shows 4 red tiles representing -4, and 6 yellow tiles are added to form zero pairs. After removing these pairs, 2 yellow tiles remain, illustrating that -4 - (-6) = 2.

Topic

Numerical and Algebraic Expressions

Description

This example illustrates the subtraction of -7 from -7 using algebra tiles. The image shows 7 red tiles representing -7, and 7 red tiles are removed to demonstrate the subtraction of -7. The result is zero tiles remaining, visually showing that -7 - (-7) = 0.

Topic

Numerical and Algebraic Expressions

Description

This example demonstrates solving the equation x + 3 = 5 using algebra tiles. The image shows green tiles representing the variable x, yellow tiles representing positive integers, and red tiles representing negative integers. By creating zero pairs and isolating the variable, the solution x = 2 is visually represented.

Topic

Numerical and Algebraic Expressions

Description

This example illustrates solving the equation x + 4 = 2 using algebra tiles. The image shows green tiles representing the variable x, yellow tiles representing positive integers, and red tiles representing negative integers. By manipulating the tiles and creating zero pairs, the solution x = -2 is visually demonstrated.

Topic

Numerical and Algebraic Expressions

Description

This example demonstrates solving the equation x + 5 = 5 using algebra tiles. The image shows green tiles representing the variable x, and yellow tiles representing positive integers. By manipulating the tiles, students can visually see that x = 0.

Topic

Numerical and Algebraic Expressions

Description

This example illustrates solving the equation x + (-2) = 5 using algebra tiles. The image shows green tiles representing the variable x, yellow tiles representing positive integers, and red tiles representing negative integers. By strategically moving pieces and creating zero pairs, students can visually determine that x = 7.

Topic

Numerical and Algebraic Expressions

Description

This example demonstrates solving the equation x + 3 = -2 using algebra tiles. The image shows green tiles representing the variable x, yellow tiles representing positive integers, and red tiles representing negative integers. By manipulating the tiles and creating zero pairs, students can visually see that x = -5.

Topic

Numerical and Algebraic Expressions

Description

This example illustrates solving the equation x + (-4) = -1 using algebra tiles. The image shows green tiles representing the variable x, yellow tiles representing positive integers, and red tiles representing negative integers. By manipulating the tiles and creating zero pairs, students can visually determine that x = 3.

Topic

Numerical and Algebraic Expressions

Description

This example illustrates the addition of 6 and -3 using algebra tiles. It shows 6 yellow tiles representing positive integers and 3 red tiles representing negative integers. After removing zero pairs (one yellow tile cancels out one red tile), the result is 3 yellow tiles, demonstrating that 6 + (-3) = 3.

Topic

Numerical and Algebraic Expressions

Description

This example demonstrates solving the equation x + (-2) = -6 using algebra tiles. The image shows green tiles representing the variable x, yellow tiles representing positive integers, and red tiles representing negative integers. By manipulating the tiles and creating zero pairs, students can visually see that x = -4.

Topic

Numerical and Algebraic Expressions

Description

This example demonstrates solving the equation x+(-5) = -5 using algebra tiles. The image depicts green tiles representing the variable x, red tiles representing negative integers, and yellow tiles representing positive integers. By manipulating these tiles, students can visually see that x = 0.

This is part of a collection of math examples that use algebra tiles. This includes modeling integers, sums, differences, and equations.

Topic

Numerical and Algebraic Expressions

Description

This example demonstrates solving the equation 3x + 3 = x + 1 using algebra tiles. The image shows green bars representing the variable x, yellow squares representing positive constants, and red squares representing negative constants. After removing common terms and adding red squares to balance the equation, students can visually see that x = -1.

Topic

Numerical and Algebraic Expressions

Description

This example illustrates solving the equation 3x + 2 = x + 2 using algebra tiles. The image shows green bars representing the variable x and yellow squares representing constants. After removing common terms from both sides, students can visually see that 2x = 0, leading to the solution x = 0.

Topic

Numerical and Algebraic Expressions

Description

This example demonstrates solving the equation 4x + (-2) = x + 3 using algebra tiles. The image shows green bars representing the variable x, yellow squares representing positive constants, and red squares representing negative constants. By eliminating common terms and zero pairs, students can visually determine the solution.

Topic

Numerical and Algebraic Expressions

Description

This example illustrates solving the equation 3x + 4 = x + (-2) using algebra tiles. The image shows green bars representing the variable x, yellow squares representing positive constants, and red squares representing negative constants. By eliminating common terms and zero pairs, students can visually determine the solution.

Topic

Numerical and Algebraic Expressions

Description

This example demonstrates solving the equation 4x + (-3) = x + (-2) using algebra tiles. The image shows green bars representing the variable x, yellow squares representing positive constants, and red squares representing negative constants. By eliminating common terms and zero pairs, students can visually determine the solution.