Title: Chapter 13
Course: quiz duplicate error

Q: Find the volume of the prism. Round to the nearest tenth if necessary.

[Image]
- 273 in^ [Image]3 (The volume of a rectangular prism is found by the formula [Image]. For example, if length is 10, width is 6, and height is 3, then the volume is [Image].)
- 50 in^ [Image]3 (The volume of a rectangular prism is found by the formula [Image]. For example, if length is 10, width is 6, and height is 3, then the volume is [Image].)
- 260 in^ [Image]3 (The volume of a rectangular prism is found by the formula [Image]. For example, if length is 10, width is 6, and height is 3, then the volume is [Image].)
- 176 in^ [Image]3 (The volume of a rectangular prism is found by the formula [Image]. For example, if length is 10, width is 6, and height is 3, then the volume is [Image].)
A: 260 in^ [Image]3

Q: Find the volume of the prism. Round to the nearest tenth if necessary.

[Image]
- 1120.0 cm^ [Image]3 (The volume of a triangular prism is found by the formula [Image]. To find the area of the base, use the triangle area formula [Image]. In this figure, the height is not given. To find it, use the Pythagorean Theorem. For example, if the base of the triangle is 6 and the slanted side is 4, the height is [Image]. Then if the height is 10, the volume is [Image].)
- 560.0 cm^ [Image]3 (The volume of a triangular prism is found by the formula [Image]. To find the area of the base, use the triangle area formula [Image]. In this figure, the height is not given. To find it, use the Pythagorean Theorem. For example, if the base of the triangle is 6 and the slanted side is 4, the height is [Image]. Then if the height is 10, the volume is [Image].)
- 660.4 cm^ [Image]3 (The volume of a triangular prism is found by the formula [Image]. To find the area of the base, use the triangle area formula [Image]. In this figure, the height is not given. To find it, use the Pythagorean Theorem. For example, if the base of the triangle is 6 and the slanted side is 4, the height is [Image]. Then if the height is 10, the volume is [Image].)
- 437.1 cm^ [Image]3 (The volume of a triangular prism is found by the formula [Image]. To find the area of the base, use the triangle area formula [Image]. In this figure, the height is not given. To find it, use the Pythagorean Theorem. For example, if the base of the triangle is 6 and the slanted side is 4, the height is [Image]. Then if the height is 10, the volume is [Image].)
A: 437.1 cm^ [Image]3

Q: Find the volume of the cylinder. Use 3.14 for p. Round to the nearest tenth.

[Image]
- 7877.6 in^ [Image]3 (The volume of a cylinder is found by the formula [Image]. In this figure, the height and diameter are given. To find the radius, divide the diameter by 2. For example, if the height of the cylinder is 8 and the diameter is 12, the radius is 6, and the volume is [Image].)
- 627.2 in^ [Image]3 (The volume of a cylinder is found by the formula [Image]. In this figure, the height and diameter are given. To find the radius, divide the diameter by 2. For example, if the height of the cylinder is 8 and the diameter is 12, the radius is 6, and the volume is [Image].)
- 281.3 in^ [Image]3 (The volume of a cylinder is found by the formula [Image]. In this figure, the height and diameter are given. To find the radius, divide the diameter by 2. For example, if the height of the cylinder is 8 and the diameter is 12, the radius is 6, and the volume is [Image].)
- 1969.4 in^ [Image]3 (The volume of a cylinder is found by the formula [Image]. In this figure, the height and diameter are given. To find the radius, divide the diameter by 2. For example, if the height of the cylinder is 8 and the diameter is 12, the radius is 6, and the volume is [Image].)
A: 1969.4 in^ [Image]3

Q: Find the volume of the cylinder. Use 3.14 for p. Round to the nearest tenth.

[Image]
- 888.3 in^ [Image]3 (The volume of a cylinder is found by the formula [Image]. In this figure, the height and radius are given. For example, if the height of the cylinder is 8 and the radius is 6, the volume is [Image].)
- 12,258.6 in^ [Image]3 (The volume of a cylinder is found by the formula [Image]. In this figure, the height and radius are given. For example, if the height of the cylinder is 8 and the radius is 6, the volume is [Image].)
- 49,034.5 in^ [Image]3 (The volume of a cylinder is found by the formula [Image]. In this figure, the height and radius are given. For example, if the height of the cylinder is 8 and the radius is 6, the volume is [Image].)
- 3904.0 in^ [Image]3 (The volume of a cylinder is found by the formula [Image]. In this figure, the height and radius are given. For example, if the height of the cylinder is 8 and the radius is 6, the volume is [Image].)
A: 12,258.6 in^ [Image]3

Q: Find the volume of the solid formed by the net. Use 3.14 for p when needed. Round to the nearest tenth.

[Image]
- 43 units^ [Image]3 (This net results in a rectangular solid. The volume of a rectangular solid is computed using the formula [Image]. These three dimensions are shown in the figure by the different rectangles that are there. For example, if the first rectangle at left shows dimensions that are 3 and 4, then those are two of your dimensions. Then look at the rectangle next to the previous one. The length of its base is the third dimension you need.)
- 45 units^ [Image]3 (This net results in a rectangular solid. The volume of a rectangular solid is computed using the formula [Image]. These three dimensions are shown in the figure by the different rectangles that are there. For example, if the first rectangle at left shows dimensions that are 3 and 4, then those are two of your dimensions. Then look at the rectangle next to the previous one. The length of its base is the third dimension you need.)
- 41 units^ [Image]3 (This net results in a rectangular solid. The volume of a rectangular solid is computed using the formula [Image]. These three dimensions are shown in the figure by the different rectangles that are there. For example, if the first rectangle at left shows dimensions that are 3 and 4, then those are two of your dimensions. Then look at the rectangle next to the previous one. The length of its base is the third dimension you need.)
- 22 units^ [Image]3 (This net results in a rectangular solid. The volume of a rectangular solid is computed using the formula [Image]. These three dimensions are shown in the figure by the different rectangles that are there. For example, if the first rectangle at left shows dimensions that are 3 and 4, then those are two of your dimensions. Then look at the rectangle next to the previous one. The length of its base is the third dimension you need.)
A: 45 units^ [Image]3

Q: Find the volume of the solid formed by the net. Use 3.14 for p when needed. Round to the nearest tenth.

[Image]
- 12.9 ft^ [Image]3 (This net represents a cylinder. The volume formula for a cylinder is [Image], where r is the radius of the circular base and h is the height of the cylinder. In the figure, the radius and height are given, so enter those numbers in the formula and compute. For example, if the given radius is 1.3 and the height is 2.5, then the volume would be [Image].)
- 18.5 ft^ [Image]3 (This net represents a cylinder. The volume formula for a cylinder is [Image], where r is the radius of the circular base and h is the height of the cylinder. In the figure, the radius and height are given, so enter those numbers in the formula and compute. For example, if the given radius is 1.3 and the height is 2.5, then the volume would be [Image].)
- 9.2 ft^ [Image]3 (This net represents a cylinder. The volume formula for a cylinder is [Image], where r is the radius of the circular base and h is the height of the cylinder. In the figure, the radius and height are given, so enter those numbers in the formula and compute. For example, if the given radius is 1.3 and the height is 2.5, then the volume would be [Image].)
- 4.1 ft^ [Image]3 (This net represents a cylinder. The volume formula for a cylinder is [Image], where r is the radius of the circular base and h is the height of the cylinder. In the figure, the radius and height are given, so enter those numbers in the formula and compute. For example, if the given radius is 1.3 and the height is 2.5, then the volume would be [Image].)
A: 12.9 ft^ [Image]3

Q: Find the volume of the pyramid. Round to the nearest tenth if necessary.

[Image]
- 11,520 m^ [Image]3 (The volume formula for a pyramid is [Image], where B is the area of the base and h is the height of the pyramid. In the figure, the length and width of the base are given. Since the base is a rectangle, the area is found by multiplying the length and width. To find the height of the pyramid, it is necessary to use the Pythagorean Theorem. As shown in the figure, there is a right triangle formed with legs that are the height and half the width. The hypotenuse is the slant height of the pyramid’s face. So the height is equal to [Image].)
- 14,400 m^ [Image]3 (The volume formula for a pyramid is [Image], where B is the area of the base and h is the height of the pyramid. In the figure, the length and width of the base are given. Since the base is a rectangle, the area is found by multiplying the length and width. To find the height of the pyramid, it is necessary to use the Pythagorean Theorem. As shown in the figure, there is a right triangle formed with legs that are the height and half the width. The hypotenuse is the slant height of the pyramid’s face. So the height is equal to [Image].)
- 4800 m^ [Image]3 (The volume formula for a pyramid is [Image], where B is the area of the base and h is the height of the pyramid. In the figure, the length and width of the base are given. Since the base is a rectangle, the area is found by multiplying the length and width. To find the height of the pyramid, it is necessary to use the Pythagorean Theorem. As shown in the figure, there is a right triangle formed with legs that are the height and half the width. The hypotenuse is the slant height of the pyramid’s face. So the height is equal to [Image].)
- 3840 m^ [Image]3 (The volume formula for a pyramid is [Image], where B is the area of the base and h is the height of the pyramid. In the figure, the length and width of the base are given. Since the base is a rectangle, the area is found by multiplying the length and width. To find the height of the pyramid, it is necessary to use the Pythagorean Theorem. As shown in the figure, there is a right triangle formed with legs that are the height and half the width. The hypotenuse is the slant height of the pyramid’s face. So the height is equal to [Image].)
A: 3840 m^ [Image]3

Q: Find the volume of the pyramid. Round to the nearest tenth if necessary.

[Image]
- 26,520 yd^ [Image]3 (The volume formula for a pyramid is [Image], where B is the area of the base and h is the height of the pyramid. Since the base of the pyramid is an isosceles triangle, the area is found by multiplying one-half times the measure of the base of the triangle times the height of the triangle. In the figure, the base of the triangle is given, but not the height. To find the height, it is necessary to use the Pythagorean Theorem. There is a right triangle formed with legs that are the height and half the base. The hypotenuse is the slant height of the triangle. So the height of the triangle is equal to [Image]. Next, find the height of the pyramid. Again, it is necessary to use the Pythagorean Theorem. There is a right triangle formed with legs that are the height of the pyramid and the slanted side of the triangular base. The hypotenuse is the slant height of the pyramid. So the height of the pyramid is equal to [Image].)
- 1874.6 yd^ [Image]3 (The volume formula for a pyramid is [Image], where B is the area of the base and h is the height of the pyramid. Since the base of the pyramid is an isosceles triangle, the area is found by multiplying one-half times the measure of the base of the triangle times the height of the triangle. In the figure, the base of the triangle is given, but not the height. To find the height, it is necessary to use the Pythagorean Theorem. There is a right triangle formed with legs that are the height and half the base. The hypotenuse is the slant height of the triangle. So the height of the triangle is equal to [Image]. Next, find the height of the pyramid. Again, it is necessary to use the Pythagorean Theorem. There is a right triangle formed with legs that are the height of the pyramid and the slanted side of the triangular base. The hypotenuse is the slant height of the pyramid. So the height of the pyramid is equal to [Image].)
- 5623.7 yd^ [Image]3 (The volume formula for a pyramid is [Image], where B is the area of the base and h is the height of the pyramid. Since the base of the pyramid is an isosceles triangle, the area is found by multiplying one-half times the measure of the base of the triangle times the height of the triangle. In the figure, the base of the triangle is given, but not the height. To find the height, it is necessary to use the Pythagorean Theorem. There is a right triangle formed with legs that are the height and half the base. The hypotenuse is the slant height of the triangle. So the height of the triangle is equal to [Image]. Next, find the height of the pyramid. Again, it is necessary to use the Pythagorean Theorem. There is a right triangle formed with legs that are the height of the pyramid and the slanted side of the triangular base. The hypotenuse is the slant height of the pyramid. So the height of the pyramid is equal to [Image].)
- 3983.5 yd^ [Image]3 (The volume formula for a pyramid is [Image], where B is the area of the base and h is the height of the pyramid. Since the base of the pyramid is an isosceles triangle, the area is found by multiplying one-half times the measure of the base of the triangle times the height of the triangle. In the figure, the base of the triangle is given, but not the height. To find the height, it is necessary to use the Pythagorean Theorem. There is a right triangle formed with legs that are the height and half the base. The hypotenuse is the slant height of the triangle. So the height of the triangle is equal to [Image]. Next, find the height of the pyramid. Again, it is necessary to use the Pythagorean Theorem. There is a right triangle formed with legs that are the height of the pyramid and the slanted side of the triangular base. The hypotenuse is the slant height of the pyramid. So the height of the pyramid is equal to [Image].)
A: 1874.6 yd^ [Image]3

Q: Find the volume of the cone. Use 3.14 for p. Round to the nearest tenth.

[Image]
- 5576.6 m^ [Image]3 (The volume of a cone is found by the formula [Image]. In this figure, the diameter is given but not the radius. To find the radius, divide the diameter by 2. To find the height it is necessary to use the Pythagorean Theorem. A right triangle is formed with the radius and the height as legs. The hypotenuse is the slant height of the cone. So the height is [Image].)
- 5275.2 m^ [Image]3 (The volume of a cone is found by the formula [Image]. In this figure, the diameter is given but not the radius. To find the radius, divide the diameter by 2. To find the height it is necessary to use the Pythagorean Theorem. A right triangle is formed with the radius and the height as legs. The hypotenuse is the slant height of the cone. So the height is [Image].)
- 15,825.6 m^ [Image]3 (The volume of a cone is found by the formula [Image]. In this figure, the diameter is given but not the radius. To find the radius, divide the diameter by 2. To find the height it is necessary to use the Pythagorean Theorem. A right triangle is formed with the radius and the height as legs. The hypotenuse is the slant height of the cone. So the height is [Image].)
- 16,729.9 m^ [Image]3 (The volume of a cone is found by the formula [Image]. In this figure, the diameter is given but not the radius. To find the radius, divide the diameter by 2. To find the height it is necessary to use the Pythagorean Theorem. A right triangle is formed with the radius and the height as legs. The hypotenuse is the slant height of the cone. So the height is [Image].)
A: 5275.2 m^ [Image]3

Q: Find the volume of the cone. Use 3.14 for p. Round to the nearest tenth.

[Image]
- 1884.0 m^ [Image]3 (The volume of a cone is found by the formula [Image]. In this figure, the height is given but not the radius. To find the radius it is necessary to use the Pythagorean Theorem. A right triangle is formed with the diameter and the height as legs. The hypotenuse is the slant height of the cone. So the diameter is [Image]. To find the radius, divide the diameter by 2.)
- 7536.0 m^ [Image]3 (The volume of a cone is found by the formula [Image]. In this figure, the height is given but not the radius. To find the radius it is necessary to use the Pythagorean Theorem. A right triangle is formed with the diameter and the height as legs. The hypotenuse is the slant height of the cone. So the diameter is [Image]. To find the radius, divide the diameter by 2.)
- 628.0 m^ [Image]3 (The volume of a cone is found by the formula [Image]. In this figure, the height is given but not the radius. To find the radius it is necessary to use the Pythagorean Theorem. A right triangle is formed with the diameter and the height as legs. The hypotenuse is the slant height of the cone. So the diameter is [Image]. To find the radius, divide the diameter by 2.)
- 2512.0 m^ [Image]3 (The volume of a cone is found by the formula [Image]. In this figure, the height is given but not the radius. To find the radius it is necessary to use the Pythagorean Theorem. A right triangle is formed with the diameter and the height as legs. The hypotenuse is the slant height of the cone. So the diameter is [Image]. To find the radius, divide the diameter by 2.)
A: 628.0 m^ [Image]3

Q: Find the volume of a sphere that has a radius of 9.5 meters. Use 3.14 for p. Round to the nearest tenth.
- 3589.5 m^ [Image]3 (The volume of a sphere is found by the formula [Image]. In this problem, the radius or diameter is given. To find the radius from the diameter, simply divide by 2. For example, if the diameter is given as 18, the radius is 9 and the volume is [Image]. If the solid is a hemisphere, then it has one-half the volume of a sphere. So divide the volume of the sphere by 2 to get the volume of a hemisphere.)
- 2692.2 m^ [Image]3 (The volume of a sphere is found by the formula [Image]. In this problem, the radius or diameter is given. To find the radius from the diameter, simply divide by 2. For example, if the diameter is given as 18, the radius is 9 and the volume is [Image]. If the solid is a hemisphere, then it has one-half the volume of a sphere. So divide the volume of the sphere by 2 to get the volume of a hemisphere.)
- 377.8 m^ [Image]3 (The volume of a sphere is found by the formula [Image]. In this problem, the radius or diameter is given. To find the radius from the diameter, simply divide by 2. For example, if the diameter is given as 18, the radius is 9 and the volume is [Image]. If the solid is a hemisphere, then it has one-half the volume of a sphere. So divide the volume of the sphere by 2 to get the volume of a hemisphere.)
- 897.4 m^ [Image]3 (The volume of a sphere is found by the formula [Image]. In this problem, the radius or diameter is given. To find the radius from the diameter, simply divide by 2. For example, if the diameter is given as 18, the radius is 9 and the volume is [Image]. If the solid is a hemisphere, then it has one-half the volume of a sphere. So divide the volume of the sphere by 2 to get the volume of a hemisphere.)
A: 3589.5 m^ [Image]3

Q: Find the volume of the sphere. Use 3.14 for p. Round to the nearest tenth.
[Image]
- 302.5 m^3 (The volume of a sphere is found by the formula [Image]. In this problem, the radius or diameter is given. To find the radius from the diameter, simply divide by 2. For example, if the diameter is given as 18, the radius is 9, and the volume is [Image].)
- 642.8 m^3 (The volume of a sphere is found by the formula [Image]. In this problem, the radius or diameter is given. To find the radius from the diameter, simply divide by 2. For example, if the diameter is given as 18, the radius is 9, and the volume is [Image].)
- 1,928.4 m^3 (The volume of a sphere is found by the formula [Image]. In this problem, the radius or diameter is given. To find the radius from the diameter, simply divide by 2. For example, if the diameter is given as 18, the radius is 9, and the volume is [Image].)
- 2,571.1 m^3 (The volume of a sphere is found by the formula [Image]. In this problem, the radius or diameter is given. To find the radius from the diameter, simply divide by 2. For example, if the diameter is given as 18, the radius is 9, and the volume is [Image].)
A: 2,571.1 m^3

Q: Astronomers have used different methods for estimating the size of the planets. One estimate of the diameter of the planet Pluto is 2274 kilometers. Based on that diameter, find the volume of Pluto, assuming that it is a sphere. Use 3.14 for p. Round to the nearest tenth.
- 1,538,472,676.4 km^3 (The volume formula for a sphere is [Image], where r is the radius of the sphere. The problem gives the diameter of Pluto. The radius is half the diameter, so it is 1137 km. The volume of Pluto is [Image] km^3.)
- 49,231,125,636.5 km^3 (The volume formula for a sphere is [Image], where r is the radius of the sphere. The problem gives the diameter of Pluto. The radius is half the diameter, so it is 1137 km. The volume of Pluto is [Image] km^3.)
- 6,153,890,704.6 km^3 (The volume formula for a sphere is [Image], where r is the radius of the sphere. The problem gives the diameter of Pluto. The radius is half the diameter, so it is 1137 km. The volume of Pluto is [Image] km^3.)
- 5,412,392.9 km^3 (The volume formula for a sphere is [Image], where r is the radius of the sphere. The problem gives the diameter of Pluto. The radius is half the diameter, so it is 1137 km. The volume of Pluto is [Image] km^3.)
A: 6,153,890,704.6 km^3

Q: Baseballs and softballs come in different sizes for different types of leagues. If the diameter of a baseball is 3 inches and a softball has a diameter of 3.8 inches, find the difference between the volumes of the two balls. Use 3.14 for p. Round to the nearest tenth.
- 10.9 in^3 (The volume formula for a sphere is [Image], where r is the radius of the sphere. The problem gives the diameter of each ball. The radius is half the diameter, so for the baseball it is 1.5 inches and for the softball it is 1.9 inches. So the volume of the baseball is [Image] in^3. The volume of the softball is [Image] in^3. The difference in their volumes is [Image] in^3.)
- 14.6 in^3 (The volume formula for a sphere is [Image], where r is the radius of the sphere. The problem gives the diameter of each ball. The radius is half the diameter, so for the baseball it is 1.5 inches and for the softball it is 1.9 inches. So the volume of the baseball is [Image] in^3. The volume of the softball is [Image] in^3. The difference in their volumes is [Image] in^3.)
- 4.3 in^3 (The volume formula for a sphere is [Image], where r is the radius of the sphere. The problem gives the diameter of each ball. The radius is half the diameter, so for the baseball it is 1.5 inches and for the softball it is 1.9 inches. So the volume of the baseball is [Image] in^3. The volume of the softball is [Image] in^3. The difference in their volumes is [Image] in^3.)
- 87.5 in^3 (The volume formula for a sphere is [Image], where r is the radius of the sphere. The problem gives the diameter of each ball. The radius is half the diameter, so for the baseball it is 1.5 inches and for the softball it is 1.9 inches. So the volume of the baseball is [Image] in^3. The volume of the softball is [Image] in^3. The difference in their volumes is [Image] in^3.)
A: 14.6 in^3

Q: Suppose a snow cone has a paper cone that is 8 centimeters deep and has a diameter of 5 centimeters. The flavored ice comes in a spherical scoop with a diameter of 5 centimeters and rests on top of the cone. If all the ice melts into the cone, will the cone overflow? Explain.
- No. The volume of the ice is less than the volume of the cone. (The volume of the cone is [Image] cm^3. The volume of the ice (a sphere) is [Image] cm^3. So the volume of the ice is greater than the volume of the cone, causing it to overflow the cone.)
- No. The volume of the ice is exactly the same as the volume of the cone. (The volume of the cone is [Image] cm^3. The volume of the ice (a sphere) is [Image] cm^3. So the volume of the ice is greater than the volume of the cone, causing it to overflow the cone.)
- Yes. The volume of the ice is greater than the volume of the cone. (The volume of the cone is [Image] cm^3. The volume of the ice (a sphere) is [Image] cm^3. So the volume of the ice is greater than the volume of the cone, causing it to overflow the cone.)
- There is not enough information given to solve this problem. (The volume of the cone is [Image] cm^3. The volume of the ice (a sphere) is [Image] cm^3. So the volume of the ice is greater than the volume of the cone, causing it to overflow the cone.)
A: Yes. The volume of the ice is greater than the volume of the cone.

Q: Suppose a spherical snowball with a diameter of 6 centimeters is melted in a large bowl. The resulting water is then poured into a cone-shaped paper cup that is 10 centimeters deep and has a diameter of 6 centimeters. The water overflows the paper cup, as the volume of the snowball turns out to be more than that of the paper cup. How much greater is the snowball’s volume than that of the cone-shaped cup? Use 3.14 for p and round your answer to the nearest tenth.
- 527.5 cm^3 (The volume of the cone-shaped cup is [Image] cm^3. The volume of the snowball (a sphere) is [Image] cm^3. So the volume of the snowball is greater than the volume of the cone by [Image] cm^3.)
- 18.8 cm^3 (The volume of the cone-shaped cup is [Image] cm^3. The volume of the snowball (a sphere) is [Image] cm^3. So the volume of the snowball is greater than the volume of the cone by [Image] cm^3.)
- 29.3 cm^3 (The volume of the cone-shaped cup is [Image] cm^3. The volume of the snowball (a sphere) is [Image] cm^3. So the volume of the snowball is greater than the volume of the cone by [Image] cm^3.)
- 9.2 cm^3 (The volume of the cone-shaped cup is [Image] cm^3. The volume of the snowball (a sphere) is [Image] cm^3. So the volume of the snowball is greater than the volume of the cone by [Image] cm^3.)
A: 18.8 cm^3

Q: Suppose that the pie graph below is a sphere with a radius of 12 centimeters.
[Image]

What is the volume of this sphere, rounded to the nearest tenth?
- 7234.6 cm^3 (The formula for the volume of a sphere is [Image]. So for this sphere, the volume is [Image] cm^3.)
- 1808.6 cm^3 (The formula for the volume of a sphere is [Image]. So for this sphere, the volume is [Image] cm^3.)
- 602.9 cm^3 (The formula for the volume of a sphere is [Image]. So for this sphere, the volume is [Image] cm^3.)
- 150.7 cm^3 (The formula for the volume of a sphere is [Image]. So for this sphere, the volume is [Image] cm^3.)
A: 7234.6 cm^3

Q: Suppose that the pie graph below is a sphere with a radius of 12 centimeters.
[Image]

What is the volume of the portion of the sphere in which the students’ favorite color is red? Round your answer to the nearest tenth.
- 2025.7 cm^3 (The formula for the volume of a sphere is [Image]. Since the percentage of students that like red best is 28%, the volume of that portion of the sphere will be 28% of the entire sphere’s volume. The volume of the sphere is [Image] cm^3. The red part is [Image] cm^3.)
- 506.4 cm^3 (The formula for the volume of a sphere is [Image]. Since the percentage of students that like red best is 28%, the volume of that portion of the sphere will be 28% of the entire sphere’s volume. The volume of the sphere is [Image] cm^3. The red part is [Image] cm^3.)
- 168.8 cm^3 (The formula for the volume of a sphere is [Image]. Since the percentage of students that like red best is 28%, the volume of that portion of the sphere will be 28% of the entire sphere’s volume. The volume of the sphere is [Image] cm^3. The red part is [Image] cm^3.)
- 42.2 cm^3 (The formula for the volume of a sphere is [Image]. Since the percentage of students that like red best is 28%, the volume of that portion of the sphere will be 28% of the entire sphere’s volume. The volume of the sphere is [Image] cm^3. The red part is [Image] cm^3.)
A: 2025.7 cm^3

Q: Suppose that the pie graph below is a sphere with a radius of 12 centimeters.
[Image]

What is the surface area of the portion of the sphere in which the students’ favorite color is green? Round your answer to the nearest tenth.
- 1808.6 cm^2 (The formula for the surface area of a sphere is [Image]. Since the percentage of students that like green best is 34%, the surface area of that portion of the sphere will be 34% of the entire sphere’s surface area. The surface area of the sphere is [Image] cm^2. The green part is [Image] cm^2.)
- 614.9 cm^2 (The formula for the surface area of a sphere is [Image]. Since the percentage of students that like green best is 34%, the surface area of that portion of the sphere will be 34% of the entire sphere’s surface area. The surface area of the sphere is [Image] cm^2. The green part is [Image] cm^2.)
- 687.3 cm^2 (The formula for the surface area of a sphere is [Image]. Since the percentage of students that like green best is 34%, the surface area of that portion of the sphere will be 34% of the entire sphere’s surface area. The surface area of the sphere is [Image] cm^2. The green part is [Image] cm^2.)
- 506.4 cm^2 (The formula for the surface area of a sphere is [Image]. Since the percentage of students that like green best is 34%, the surface area of that portion of the sphere will be 34% of the entire sphere’s surface area. The surface area of the sphere is [Image] cm^2. The green part is [Image] cm^2.)
A: 614.9 cm^2

Q: Suppose that the pie graph below is a sphere with a radius of 12 centimeters.
[Image]

What is the area of the two-dimensional sector of the circle in which the students’ favorite color is blue? Round your answer to the nearest tenth.
- 687.3 cm^2 (The formula for the area of a circle is [Image]. Since the percentage of students that like blue best is 38%, the area of that portion of the circle will be 38% of the entire circle’s area. The area of the circle is [Image] cm^2. The blue part is [Image] cm^2.)
- 452.2 cm^2 (The formula for the area of a circle is [Image]. Since the percentage of students that like blue best is 38%, the area of that portion of the circle will be 38% of the entire circle’s area. The area of the circle is [Image] cm^2. The blue part is [Image] cm^2.)
- 2,749.1 cm^2 (The formula for the area of a circle is [Image]. Since the percentage of students that like blue best is 38%, the area of that portion of the circle will be 38% of the entire circle’s area. The area of the circle is [Image] cm^2. The blue part is [Image] cm^2.)
- 171.8 cm^2 (The formula for the area of a circle is [Image]. Since the percentage of students that like blue best is 38%, the area of that portion of the circle will be 38% of the entire circle’s area. The area of the circle is [Image] cm^2. The blue part is [Image] cm^2.)
A: 171.8 cm^2

Q: A sphere that has a radius of 8 centimeters is inscribed in a cylinder. Find the probability of choosing a point at random inside the cylinder that is also inside the sphere. Use 3.14 for p. Round to the nearest hundredth.
- 0.72 (The volume of a cylinder is [Image]. Since the sphere has a radius of 8, so does the cylinder. Also, the height of the cylinder will be equal to the diameter of the sphere which is 16 (double the radius). The volume of the cylinder is [Image] cm^3. The formula for the volume of a sphere is [Image]. The volume of this sphere is [Image] cm^3. The percent of the cylinder’s volume is occupied by the inscribed sphere is [Image] = 67%. That is the probability of choosing a point at random inside the cylinder that is also inside the sphere.)
- 0.67 (The volume of a cylinder is [Image]. Since the sphere has a radius of 8, so does the cylinder. Also, the height of the cylinder will be equal to the diameter of the sphere which is 16 (double the radius). The volume of the cylinder is [Image] cm^3. The formula for the volume of a sphere is [Image]. The volume of this sphere is [Image] cm^3. The percent of the cylinder’s volume is occupied by the inscribed sphere is [Image] = 67%. That is the probability of choosing a point at random inside the cylinder that is also inside the sphere.)
- 0.82 (The volume of a cylinder is [Image]. Since the sphere has a radius of 8, so does the cylinder. Also, the height of the cylinder will be equal to the diameter of the sphere which is 16 (double the radius). The volume of the cylinder is [Image] cm^3. The formula for the volume of a sphere is [Image]. The volume of this sphere is [Image] cm^3. The percent of the cylinder’s volume is occupied by the inscribed sphere is [Image] = 67%. That is the probability of choosing a point at random inside the cylinder that is also inside the sphere.)
- 0.76 (The volume of a cylinder is [Image]. Since the sphere has a radius of 8, so does the cylinder. Also, the height of the cylinder will be equal to the diameter of the sphere which is 16 (double the radius). The volume of the cylinder is [Image] cm^3. The formula for the volume of a sphere is [Image]. The volume of this sphere is [Image] cm^3. The percent of the cylinder’s volume is occupied by the inscribed sphere is [Image] = 67%. That is the probability of choosing a point at random inside the cylinder that is also inside the sphere.)
A: 0.67

Q: Some racquet balls are sold in cylindrical cans of three balls. Each ball has a diameter of 2.25 inches. The can has a diameter of 2.25 inches and is 6.75 inches tall. Find the volume of the empty space in the can. Use 3.14 for p. Round to the nearest hundredth.
- 8.94 in^3 (To answer this question, subtract the volume of the three balls (spheres) from the can’s volume. The volume of a can is [Image]. The volume of this can is [Image] in^3. The formula for the volume of a sphere is [Image]. The volume of each ball is [Image] in^3. Since there are 3 balls, the total volume occupied by the balls is [Image] in^3. The volume of the empty space in the can is [Image] in^3.)
- 26.82 in^3 (To answer this question, subtract the volume of the three balls (spheres) from the can’s volume. The volume of a can is [Image]. The volume of this can is [Image] in^3. The formula for the volume of a sphere is [Image]. The volume of each ball is [Image] in^3. Since there are 3 balls, the total volume occupied by the balls is [Image] in^3. The volume of the empty space in the can is [Image] in^3.)
- 22.35 in^3 (To answer this question, subtract the volume of the three balls (spheres) from the can’s volume. The volume of a can is [Image]. The volume of this can is [Image] in^3. The formula for the volume of a sphere is [Image]. The volume of each ball is [Image] in^3. Since there are 3 balls, the total volume occupied by the balls is [Image] in^3. The volume of the empty space in the can is [Image] in^3.)
- 20.86 in^3 (To answer this question, subtract the volume of the three balls (spheres) from the can’s volume. The volume of a can is [Image]. The volume of this can is [Image] in^3. The formula for the volume of a sphere is [Image]. The volume of each ball is [Image] in^3. Since there are 3 balls, the total volume occupied by the balls is [Image] in^3. The volume of the empty space in the can is [Image] in^3.)
A: 8.94 in^3

Q: Determine whether the pair of solids are similar, congruent, or neither. Figures are not necessarily drawn to scale.

[Image] [Image]
- similar (Since these are both cubes, they have the same dimensions all the way around. Therefore, all corresponding dimensions between the two will have the same ratio, 8 to 5. That makes them similar solids.)
- congruent (Since these are both cubes, they have the same dimensions all the way around. Therefore, all corresponding dimensions between the two will have the same ratio, 8 to 5. That makes them similar solids.)
- neither (Since these are both cubes, they have the same dimensions all the way around. Therefore, all corresponding dimensions between the two will have the same ratio, 8 to 5. That makes them similar solids.)
- cannot be determined (Since these are both cubes, they have the same dimensions all the way around. Therefore, all corresponding dimensions between the two will have the same ratio, 8 to 5. That makes them similar solids.)
A: similar

Q: Determine whether the pair of solids are similar, congruent, or neither. Figures are not necessarily drawn to scale.

[Image] [Image]
- similar (The ratio of the heights is 5:3 and ratio of the diameters is also 5:3. Therefore, all corresponding dimensions between the two have the same ratio. That makes them similar solids.)
- congruent (The ratio of the heights is 5:3 and ratio of the diameters is also 5:3. Therefore, all corresponding dimensions between the two have the same ratio. That makes them similar solids.)
- neither (The ratio of the heights is 5:3 and ratio of the diameters is also 5:3. Therefore, all corresponding dimensions between the two have the same ratio. That makes them similar solids.)
- cannot be determined (The ratio of the heights is 5:3 and ratio of the diameters is also 5:3. Therefore, all corresponding dimensions between the two have the same ratio. That makes them similar solids.)
A: similar

Q: Determine whether the pair of solids are similar, congruent, or neither. Figures are not necessarily drawn to scale.

[Image] [Image]
- similar (The ratio of the heights is 4:1 and the other dimensions have ratios of 2:1. Since all corresponding dimensions do not have the same ratio, these prisms are neither similar nor congruent solids.)
- congruent (The ratio of the heights is 4:1 and the other dimensions have ratios of 2:1. Since all corresponding dimensions do not have the same ratio, these prisms are neither similar nor congruent solids.)
- neither (The ratio of the heights is 4:1 and the other dimensions have ratios of 2:1. Since all corresponding dimensions do not have the same ratio, these prisms are neither similar nor congruent solids.)
- cannot be determined (The ratio of the heights is 4:1 and the other dimensions have ratios of 2:1. Since all corresponding dimensions do not have the same ratio, these prisms are neither similar nor congruent solids.)
A: neither

Q: Determine whether the pair of solids are similar, congruent, or neither. Figures are not necessarily drawn to scale.

[Image] [Image]
- similar (The given dimensions have ratios of 2:1. However, since there are no measures given for the third dimension, it is not possible to determine the relationship between these prisms.)
- congruent (The given dimensions have ratios of 2:1. However, since there are no measures given for the third dimension, it is not possible to determine the relationship between these prisms.)
- neither (The given dimensions have ratios of 2:1. However, since there are no measures given for the third dimension, it is not possible to determine the relationship between these prisms.)
- cannot be determined (The given dimensions have ratios of 2:1. However, since there are no measures given for the third dimension, it is not possible to determine the relationship between these prisms.)
A: cannot be determined

Q: Determine whether the pair of solids are similar, congruent, or neither. Figures are not necessarily drawn to scale.

[Image] [Image]
- similar (The ratio of the diameters is 4:3 and the ratio of the heights is 5:3. Therefore, these cylinders are neither similar nor congruent.)
- congruent (The ratio of the diameters is 4:3 and the ratio of the heights is 5:3. Therefore, these cylinders are neither similar nor congruent.)
- neither (The ratio of the diameters is 4:3 and the ratio of the heights is 5:3. Therefore, these cylinders are neither similar nor congruent.)
- cannot be determined (The ratio of the diameters is 4:3 and the ratio of the heights is 5:3. Therefore, these cylinders are neither similar nor congruent.)
A: neither

Q: Determine whether the pair of solids are similar, congruent, or neither. Figures are not necessarily drawn to scale.

[Image] [Image]
- similar (The radius of the cylinder on the left is 7, so its diameter is 14. That is the same as the cylinder on the right. They also have the same height. Therefore, these cylinders are congruent.)
- congruent (The radius of the cylinder on the left is 7, so its diameter is 14. That is the same as the cylinder on the right. They also have the same height. Therefore, these cylinders are congruent.)
- neither (The radius of the cylinder on the left is 7, so its diameter is 14. That is the same as the cylinder on the right. They also have the same height. Therefore, these cylinders are congruent.)
- cannot be determined (The radius of the cylinder on the left is 7, so its diameter is 14. That is the same as the cylinder on the right. They also have the same height. Therefore, these cylinders are congruent.)
A: congruent

Q: Determine whether the pair of solids are similar, congruent, or neither. Figures are not necessarily drawn to scale.

[Image] [Image]
- similar (All three corresponding measures of these two pyramids have the same ratio: 1.2:1. It is shown that [Image]. Therefore, these two pyramids are similar.)
- congruent (All three corresponding measures of these two pyramids have the same ratio: 1.2:1. It is shown that [Image]. Therefore, these two pyramids are similar.)
- neither (All three corresponding measures of these two pyramids have the same ratio: 1.2:1. It is shown that [Image]. Therefore, these two pyramids are similar.)
- cannot be determined (All three corresponding measures of these two pyramids have the same ratio: 1.2:1. It is shown that [Image]. Therefore, these two pyramids are similar.)
A: similar

Q: Determine whether the pair of solids are similar, congruent, or neither. Figures are not necessarily drawn to scale.

[Image] [Image]
- similar (All three corresponding measures of these two pyramids do not have the same ratio. It is true that [Image], but [Image]. Therefore, these two pyramids are not similar. They are also not congruent since their corresponding parts are not the same measure.)
- congruent (All three corresponding measures of these two pyramids do not have the same ratio. It is true that [Image], but [Image]. Therefore, these two pyramids are not similar. They are also not congruent since their corresponding parts are not the same measure.)
- neither (All three corresponding measures of these two pyramids do not have the same ratio. It is true that [Image], but [Image]. Therefore, these two pyramids are not similar. They are also not congruent since their corresponding parts are not the same measure.)
- cannot be determined (All three corresponding measures of these two pyramids do not have the same ratio. It is true that [Image], but [Image]. Therefore, these two pyramids are not similar. They are also not congruent since their corresponding parts are not the same measure.)
A: neither

Q: Determine whether the pair of solids are similar, congruent, or neither. Figures are not necessarily drawn to scale.

[Image] [Image]
- similar (The diameter of each sphere is 26 centimeters since the one on the right has a radius of 13, which is 26 when doubled. That means the two spheres are the exact same size, therefore they are congruent.)
- congruent (The diameter of each sphere is 26 centimeters since the one on the right has a radius of 13, which is 26 when doubled. That means the two spheres are the exact same size, therefore they are congruent.)
- neither (The diameter of each sphere is 26 centimeters since the one on the right has a radius of 13, which is 26 when doubled. That means the two spheres are the exact same size, therefore they are congruent.)
- cannot be determined (The diameter of each sphere is 26 centimeters since the one on the right has a radius of 13, which is 26 when doubled. That means the two spheres are the exact same size, therefore they are congruent.)
A: congruent

Q: Determine whether the pair of solids are similar, congruent, or neither. Figures are not necessarily drawn to scale.

[Image] [Image]
- similar (Any two spheres that are not congruent are similar since they have only one defining measure. So these two spheres are similar.)
- congruent (Any two spheres that are not congruent are similar since they have only one defining measure. So these two spheres are similar.)
- neither (Any two spheres that are not congruent are similar since they have only one defining measure. So these two spheres are similar.)
- cannot be determined (Any two spheres that are not congruent are similar since they have only one defining measure. So these two spheres are similar.)
A: similar

Q: Determine whether each statement is sometimes, always, or never true.

Two spheres are congruent.
- sometimes (Two spheres can be congruent only if their radii are equal. So this statement is sometimes true.)
- always (Two spheres can be congruent only if their radii are equal. So this statement is sometimes true.)
- never (Two spheres can be congruent only if their radii are equal. So this statement is sometimes true.)
A: sometimes

Q: Determine whether each statement is sometimes, always, or never true.

Congruent spheres have equal surface areas.
- sometimes (Two spheres that are congruent are the same size and shape. Therefore, they would have the same surface area. So this statement is always true.)
- always (Two spheres that are congruent are the same size and shape. Therefore, they would have the same surface area. So this statement is always true.)
- never (Two spheres that are congruent are the same size and shape. Therefore, they would have the same surface area. So this statement is always true.)
A: always

Q: Determine whether each statement is sometimes, always, or never true.

Two solids with equal volumes are congruent.
- sometimes (Two solids with the same volume may be congruent but do not have to be. So this statement is sometimes true.)
- always (Two solids with the same volume may be congruent but do not have to be. So this statement is sometimes true.)
- never (Two solids with the same volume may be congruent but do not have to be. So this statement is sometimes true.)
A: sometimes

Q: Determine whether each statement is sometimes, always, or never true.

A cylinder is similar to a prism.
- sometimes (A cylinder has a circular base and a prism has a polygon for a base. Therefore, they cannot be similar. They are distinctly different shapes. So this statement is never true.)
- always (A cylinder has a circular base and a prism has a polygon for a base. Therefore, they cannot be similar. They are distinctly different shapes. So this statement is never true.)
- never (A cylinder has a circular base and a prism has a polygon for a base. Therefore, they cannot be similar. They are distinctly different shapes. So this statement is never true.)
A: never

Q: Determine whether each statement is sometimes, always, or never true.

A pyramid and a cone with the same height are similar.
- sometimes (A cone has a circular base and a pyramid has a polygon for a base. Therefore, they cannot be similar. They are distinctly different shapes. So this statement is never true.)
- always (A cone has a circular base and a pyramid has a polygon for a base. Therefore, they cannot be similar. They are distinctly different shapes. So this statement is never true.)
- never (A cone has a circular base and a pyramid has a polygon for a base. Therefore, they cannot be similar. They are distinctly different shapes. So this statement is never true.)
A: never

Q: Determine whether each statement is sometimes, always, or never true.

Non-similar pyramids have different volumes.
- sometimes (If two pyramids are not similar, they can still have the same volume. Lack of similarity has no effect on the relationship between their volumes. So this statement is sometimes true.)
- always (If two pyramids are not similar, they can still have the same volume. Lack of similarity has no effect on the relationship between their volumes. So this statement is sometimes true.)
- never (If two pyramids are not similar, they can still have the same volume. Lack of similarity has no effect on the relationship between their volumes. So this statement is sometimes true.)
A: sometimes

Q: Determine whether each statement is sometimes, always, or never true.

If the volumes of two triangular prisms are the same, the prisms are congruent.
- sometimes (If two triangular prisms have the same volume, they may be congruent, but not necessarily. So this statement is sometimes true.)
- always (If two triangular prisms have the same volume, they may be congruent, but not necessarily. So this statement is sometimes true.)
- never (If two triangular prisms have the same volume, they may be congruent, but not necessarily. So this statement is sometimes true.)
A: sometimes

Q: Determine whether each statement is sometimes, always, or never true.

If the scale factor of two similar solids is a:b, then the ratio of their volumes is a^3:b^3.
- sometimes (This is a theorem for similar solids. So this statement is always true.)
- always (This is a theorem for similar solids. So this statement is always true.)
- never (This is a theorem for similar solids. So this statement is always true.)
A: always

Q: Determine whether each statement is sometimes, always, or never true.

If the scale factor of two similar solids is a:b, then the ratio of their surface areas is a:b.
- sometimes (The theorem for similar solids states the ratio of their surface areas must be a^2:b^2. So this statement is never true.)
- always (The theorem for similar solids states the ratio of their surface areas must be a^2:b^2. So this statement is never true.)
- never (The theorem for similar solids states the ratio of their surface areas must be a^2:b^2. So this statement is never true.)
A: never

Q: Determine whether each statement is sometimes, always, or never true.

Congruent solids are also similar to one another.
- sometimes (The scale factor for all congruent solids is 1. So this statement is always true.)
- always (The scale factor for all congruent solids is 1. So this statement is always true.)
- never (The scale factor for all congruent solids is 1. So this statement is always true.)
A: always

Q: Graph a rectangular solid that contains the given point and the origin as vertices.

A(–2, 4, 1)
- [Image] (To graph this solid, begin by going backwards on the x-axis, since the x-coordinate is negative. Recall that the x-axis is the one that “sticks out” of the paper. Since the y- and z-coordinates are both positive, then go y units to the right and z units up. This point is the back right upper point of the solid and the origin is the lower left front point. Draw segments parallel to the axes to connect these into a rectangular solid.)
- [Image] (To graph this solid, begin by going backwards on the x-axis, since the x-coordinate is negative. Recall that the x-axis is the one that “sticks out” of the paper. Since the y- and z-coordinates are both positive, then go y units to the right and z units up. This point is the back right upper point of the solid and the origin is the lower left front point. Draw segments parallel to the axes to connect these into a rectangular solid.)
- [Image] (To graph this solid, begin by going backwards on the x-axis, since the x-coordinate is negative. Recall that the x-axis is the one that “sticks out” of the paper. Since the y- and z-coordinates are both positive, then go y units to the right and z units up. This point is the back right upper point of the solid and the origin is the lower left front point. Draw segments parallel to the axes to connect these into a rectangular solid.)
- [Image] (To graph this solid, begin by going backwards on the x-axis, since the x-coordinate is negative. Recall that the x-axis is the one that “sticks out” of the paper. Since the y- and z-coordinates are both positive, then go y units to the right and z units up. This point is the back right upper point of the solid and the origin is the lower left front point. Draw segments parallel to the axes to connect these into a rectangular solid.)
A: [Image]

Q: Graph a rectangular solid that contains the given point and the origin as vertices.

B(1, –3, 3)
- [Image] (To graph this solid, begin by going forward on the x-axis, since the x-coordinate is positive. Recall that the x-axis is the one that “sticks out” of the paper. Since the y-coordinate is negative and z-coordinate is positive, then go y units to the left and z units up. This point is the front left upper point of the solid and the origin is the lower right back point. Draw segments parallel to the axes to connect these into a rectangular solid.)
- [Image] (To graph this solid, begin by going forward on the x-axis, since the x-coordinate is positive. Recall that the x-axis is the one that “sticks out” of the paper. Since the y-coordinate is negative and z-coordinate is positive, then go y units to the left and z units up. This point is the front left upper point of the solid and the origin is the lower right back point. Draw segments parallel to the axes to connect these into a rectangular solid.)
- [Image] (To graph this solid, begin by going forward on the x-axis, since the x-coordinate is positive. Recall that the x-axis is the one that “sticks out” of the paper. Since the y-coordinate is negative and z-coordinate is positive, then go y units to the left and z units up. This point is the front left upper point of the solid and the origin is the lower right back point. Draw segments parallel to the axes to connect these into a rectangular solid.)
- [Image] (To graph this solid, begin by going forward on the x-axis, since the x-coordinate is positive. Recall that the x-axis is the one that “sticks out” of the paper. Since the y-coordinate is negative and z-coordinate is positive, then go y units to the left and z units up. This point is the front left upper point of the solid and the origin is the lower right back point. Draw segments parallel to the axes to connect these into a rectangular solid.)
A: [Image]

Q: Determine the distance between the points. Round to the nearest tenth.

D(–1, –4, 5) and E(0, –2, –5)
- 3.6 (For two points D(x_1, y_1, z_1) and E(x_2, y_2, z_2) the distance formula is [Image]. For example, if the given points are D(2, –1, 5) and E(–3, 4, –1), then the distance between them would be [Image] = [Image], rounding to the nearest tenth.)
- 1.7 (For two points D(x_1, y_1, z_1) and E(x_2, y_2, z_2) the distance formula is [Image]. For example, if the given points are D(2, –1, 5) and E(–3, 4, –1), then the distance between them would be [Image] = [Image], rounding to the nearest tenth.)
- 2.2 (For two points D(x_1, y_1, z_1) and E(x_2, y_2, z_2) the distance formula is [Image]. For example, if the given points are D(2, –1, 5) and E(–3, 4, –1), then the distance between them would be [Image] = [Image], rounding to the nearest tenth.)
- 10.2 (For two points D(x_1, y_1, z_1) and E(x_2, y_2, z_2) the distance formula is [Image]. For example, if the given points are D(2, –1, 5) and E(–3, 4, –1), then the distance between them would be [Image] = [Image], rounding to the nearest tenth.)
A: 10.2

Q: Determine the distance between the points. Round to the nearest tenth.

B(0, 2, 2) and C [Image], [Image]
- 1.4 (For two points B(x_1, y_1, z_1) and C(x_2, y_2, z_2) the distance formula is [Image]. For example, if the given points are B(0, 4, 4) and C [Image], then the distance between them would be [Image] = 9.7, rounding to the nearest tenth.)
- 2.8 (For two points B(x_1, y_1, z_1) and C(x_2, y_2, z_2) the distance formula is [Image]. For example, if the given points are B(0, 4, 4) and C [Image], then the distance between them would be [Image] = 9.7, rounding to the nearest tenth.)
- 6.2 (For two points B(x_1, y_1, z_1) and C(x_2, y_2, z_2) the distance formula is [Image]. For example, if the given points are B(0, 4, 4) and C [Image], then the distance between them would be [Image] = 9.7, rounding to the nearest tenth.)
- 1.6 (For two points B(x_1, y_1, z_1) and C(x_2, y_2, z_2) the distance formula is [Image]. For example, if the given points are B(0, 4, 4) and C [Image], then the distance between them would be [Image] = 9.7, rounding to the nearest tenth.)
A: 6.2

Q: Determine the coordinates of the midpoint M of the segment joining the pair of points.

S [Image] and T [Image]
- [Image] (For two points S(x_1, y_1, z_1) and T(x_2, y_2, z_2) the midpoint formula is [Image]. For example, if the given points are S(2, –1, 5) and T(–3, 4, –1), then the midpoint of the segment joining the points would be [Image].)
- [Image] (For two points S(x_1, y_1, z_1) and T(x_2, y_2, z_2) the midpoint formula is [Image]. For example, if the given points are S(2, –1, 5) and T(–3, 4, –1), then the midpoint of the segment joining the points would be [Image].)
- [Image] (For two points S(x_1, y_1, z_1) and T(x_2, y_2, z_2) the midpoint formula is [Image]. For example, if the given points are S(2, –1, 5) and T(–3, 4, –1), then the midpoint of the segment joining the points would be [Image].)
- [Image] (For two points S(x_1, y_1, z_1) and T(x_2, y_2, z_2) the midpoint formula is [Image]. For example, if the given points are S(2, –1, 5) and T(–3, 4, –1), then the midpoint of the segment joining the points would be [Image].)
A: [Image]

Q: Determine the coordinates of the midpoint M of the segment joining the pair of points.

F [Image], 4, –2 [Image] and G [Image], 3, –4 [Image]
- [Image] (For two points F(x_1, y_1, z_1) and G(x_2, y_2, z_2) the midpoint formula is [Image]. For example, if the given points are F [Image], –5, 2 [Image] and G [Image], 3, 4 [Image], then the midpoint of the segment joining the points would be [Image].)
- [Image] (For two points F(x_1, y_1, z_1) and G(x_2, y_2, z_2) the midpoint formula is [Image]. For example, if the given points are F [Image], –5, 2 [Image] and G [Image], 3, 4 [Image], then the midpoint of the segment joining the points would be [Image].)
- [Image] (For two points F(x_1, y_1, z_1) and G(x_2, y_2, z_2) the midpoint formula is [Image]. For example, if the given points are F [Image], –5, 2 [Image] and G [Image], 3, 4 [Image], then the midpoint of the segment joining the points would be [Image].)
- [Image] (For two points F(x_1, y_1, z_1) and G(x_2, y_2, z_2) the midpoint formula is [Image]. For example, if the given points are F [Image], –5, 2 [Image] and G [Image], 3, 4 [Image], then the midpoint of the segment joining the points would be [Image].)
A: [Image]

A food manufacturer packages vegetables in the cylindrical can shown in the diagram at left for sale to restaurants. They want to sell the vegetables to consumers in grocery stores as well, and need to provide them in smaller cans. They have decided to halve the height of the can.
[Image] [Image]
a.      Find the surface area of each can. Use 3.14 for p. Explain the effect of halving the height on the amount of material used to produce the can.
b.      Find the volume of each can. Use 3.14 for p. Explain the effect of halving the height on the amount of product that can fit inside.

Architecture students were assigned the task of building models of structures of various shapes. One student built a model of a barn and a grain silo. The barn consisted of a rectangular solid with a pyramid on top. The dimensions of the rectangular solid are shown. The height of the pyramid is 5 centimeters. The silo is a cylinder with dimensions as shown.
[Image]
What is the combined volume of the barn and the silo to the nearest cubic centimeter? Show your work. Use 3.14 for p and round to the nearest hundredth.