Question

leveled practice in 7 and 8, evaluate the cube root or square root.

1. relate the volume of the cube to the length of each edge.

v = 8 cm³
edge length edge length edge length
□ cm × □ cm × □ cm
∛8 = □

1. relate the area of the square to the length of each side.

a = 16 cm²
side length side length
□ cm × □ cm
√16 = □

1. would you classify the number 169 as a perfect square, a perfect cube, both, or neither? explain.
2. the volume of a cube is 512 cubic inches. what is the length of each side of the cube?
3. a square technology chip has an area of 25 square centimeters. how long is each side of the chip?
4. would you classify the number 200 as a perfect square, a perfect cube, both, or neither? explain.
5. a company is making building blocks. what is the length of each side of the block?

v = 1 ft³

Explanation:

Step1: Recall volume formula for cube

The volume formula for a cube is $V = s^3$, where $s$ is the length of an edge. For a square, the area formula is $A=s^2$, where $s$ is the side - length.

Step2: Solve problem 7

Given $V = 8\ cm^3$, using $V=s^3$, we need to find $s$ such that $s^3=8$. Since $2\times2\times2 = 8$, $\sqrt[3]{8}=2$. So the edge - lengths are $2\ cm\times2\ cm\times2\ cm$.

Step3: Solve problem 8

Given $A = 16\ cm^2$, using $A = s^2$, we need to find $s$ such that $s^2=16$. Since $4\times4 = 16$, $\sqrt{16}=4$. So the side - lengths are $4\ cm\times4\ cm$.

Step4: Solve problem 9

For a perfect square, if $n = m^2$ for some integer $m$. For a perfect cube, if $n = k^3$ for some integer $k$.
We know that $13\times13=169$, so $169$ is a perfect square. But there is no integer $k$ such that $k^3 = 169$. So $169$ is a perfect square only.

Step5: Solve problem 10

Given $V = 512$ cubic inches. Using $V=s^3$, we need to find $s$ such that $s^3=512$. Since $8\times8\times8=512$, $s = 8$ inches.

Step6: Solve problem 11

Given $A = 25$ square centimeters. Using $A=s^2$, we need to find $s$ such that $s^2=25$. Since $5\times5 = 25$, $s = 5$ centimeters.

Step7: Solve problem 12

For a perfect square, we check if $200=m^2$ for an integer $m$. $\sqrt{200}\approx14.14$, not an integer.
For a perfect cube, we check if $200 = k^3$ for an integer $k$. $\sqrt[3]{200}\approx5.85$, not an integer. So $200$ is neither a perfect square nor a perfect cube.

Step8: Solve problem 13

Given $V = 1\ ft^3$. Using $V=s^3$, we need to find $s$ such that $s^3=1$. Since $1\times1\times1 = 1$, $s = 1$ foot.

Answer:

1. $2\ cm\times2\ cm\times2\ cm$, $\sqrt[3]{8}=2$
2. $4\ cm\times4\ cm$, $\sqrt{16}=4$
3. A perfect square only. Because $13\times13 = 169$ and there is no integer $k$ such that $k^3=169$.
4. 8 inches
5. 5 centimeters
6. Neither. $\sqrt{200}$ and $\sqrt[3]{200}$ are not integers.
7. 1 foot