Illustrative Math Alignment: Grade 8 Unit 3

Topic

Geometric Models

Topic

Geometric Models

This slide show provides 8 examples of quadratic functions in factored form and analyzes their graphs.

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This slide show provides 18 examples of quadratic functions in standard form and analyzes their graphs.

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This slide show provides 8 examples of quadratic functions in vertex form and analyzes their graphs.

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This set of tutorials provides an overview of the 24 worked-out examples that show how to calculate the surface area of different three-dimensional figures.

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This set of tutorials provides 40 examples of absolute value functions in tabular and graph form. NOTE: The download is a PPT file.

In this drag-and-drop game, match f(x) with f(a) for a given value of a. This game generates thousands of different equation combinations, offering an ideal opportunity for skill review in a game format.

Related Resources

In this drag-and-drop game, match the linear function in standard form with its slope-intercept counterpart. This game generates thousands of different equation combinations, offering an ideal opportunity for skill review in a game format.

Related Resources

Topic

Polynomials

Description

An example showing how to find the side lengths of a square given its area, A = x2+ 2x + 1.

Example 1: Given the area A = x2+ 2x + 1, find the side lengths. Solution: Express the area as a perfect square, (x + 1)2, so the side lengths are x + 1.

Polynomials involve expressions with variables raised to powers, and these examples specifically address perfect squares and cubes. Each example in this collection explores how to derive side lengths or volumes using factorization, demonstrating the practical applications of polynomial expressions.

Topic

Polynomials

Description

Another example of finding the side lengths of a square with area A = x2+ 4x + 4.

Example 2: Given A = x2+ 4x + 4, find the side lengths. Solution: Factor as (x + 2)2, so the side lengths are x + 2.

Polynomials involve expressions with variables raised to powers, and these examples specifically address perfect squares and cubes. Each example in this collection explores how to derive side lengths or volumes using factorization, demonstrating the practical applications of polynomial expressions.

Topic

Polynomials

Description

Shows how to determine the side lengths of a square with area A = x2 - 2x + 1.

Example 3: For A = x2 - 2x + 1, the solution expresses it as (x - 1)2, making the side lengths x - 1.

Polynomials involve expressions with variables raised to powers, and these examples specifically address perfect squares and cubes. Each example in this collection explores how to derive side lengths or volumes using factorization, demonstrating the practical applications of polynomial expressions.

Topic

Polynomials

Description

Example solving for side lengths of a square with area A = x2 - 4x + 4.

Example 4: Given A = x2 - 4x + 4, factor as (x - 2)2 to find side lengths x - 2.

Polynomials involve expressions with variables raised to powers, and these examples specifically address perfect squares and cubes. Each example in this collection explores how to derive side lengths or volumes using factorization, demonstrating the practical applications of polynomial expressions.

Topic

Polynomials

Description

Solves for side lengths of a square with area A = x2 + 6x + 9.

Example 5: With A = x2 + 6x + 9, factor as (x + 3)2, giving side lengths x + 3.

Polynomials involve expressions with variables raised to powers, and these examples specifically address perfect squares and cubes. Each example in this collection explores how to derive side lengths or volumes using factorization, demonstrating the practical applications of polynomial expressions.

Topic

Polynomials

Description

A cubic example where the volume of a cube, A = x3 + 3x2 + 3x + 1, is used to find the side length.

Example 6: Given the volume A = x3 + 3x2 + 3x + 1, factor as (x + 1)3, so the side lengths are x + 1.

Polynomials involve expressions with variables raised to powers, and these examples specifically address perfect squares and cubes. Each example in this collection explores how to derive side lengths or volumes using factorization, demonstrating the practical applications of polynomial expressions.

Topic

Polynomials

Description

Another cubic example solving for side length with volume A = x3 - 3x2 + 3x - 1.

Example 7: For the volume A = x3 - 3x2 + 3x - 1, express as (x - 1)3 to determine side lengths of x - 1.

Polynomials involve expressions with variables raised to powers, and these examples specifically address perfect squares and cubes. Each example in this collection explores how to derive side lengths or volumes using factorization, demonstrating the practical applications of polynomial expressions.

This is part of a collection of math examples that focus on volume.

This is part of a collection of math examples that focus on volume.

This is part of a collection of math examples that focus on volume.

This is part of a collection of math examples that focus on volume.

This is part of a collection of math examples that focus on volume.

This is part of a collection of math examples that focus on volume.

This is part of a collection of math examples that focus on volume.

This is part of a collection of math examples that focus on volume.

This is part of a collection of math examples that focus on volume.

This is part of a collection of math examples that focus on volume.

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.