Question

1. mrs. drew wants to build a square sandbox with an area of 121 square feet. what is the total length of wood drew needs to make the sides of the sandbox? 15. communicate and justify diego says that if you cube the number 4 and then take the cube - root of the result, you end up with 8. is diego correct? explain. 16. higher order thinking tala is packing a moving box. she has a square - framed poster with an area of 9 square feet. the cube - shaped box has a volume of 30 cubic feet. will the poster fit flat in the box? explain. the volume of the box is 30 cubic feet. assessment practice 17. which expression has the greatest value? a. (sqrt{81}-2) b. (sqrt{49}-sqrt{16}) c. (sqrt{35}+sqrt{16}) d. (sqrt{25}-3) 18. a toy has various shaped objects that a child can push through matching holes. the area of the square hole is 8 square centimeters. the volume of a cube - shaped block is 64 cubic centimeters. part a part b which edge length can you find? explain. will the block fit in the square hole? explain.

Explanation:

Step1: Solve problem 14

The area of a square is $A = s^{2}$, where $s$ is the side - length. Given $A = 121$ square feet. We need to find $s$. So, $s=\sqrt{A}$.
$s=\sqrt{121}=11$ feet.

Step2: Solve problem 16

The area of the poster is $A = 9$ square feet, so the side - length of the square poster $s_{1}=\sqrt{9}=3$ feet. The volume of the cube - shaped box is $V = 30$ cubic feet. The side - length of the cube $s_{2}=\sqrt[3]{30}\approx3.11$ feet. Since $s_{2}>s_{1}$, the poster will fit in the box.

Step3: Solve problem 17

• For option A: $\sqrt{81}-2=9 - 2=7$.
• For option B: $\sqrt{49}-\sqrt{16}=7 - 4 = 3$.
• For option C: $\sqrt{36}+\sqrt{16}=6 + 4=10$.
• For option D: $\sqrt{25}-3=5 - 3=2$. So, option C has the greatest value.

Step4: Solve problem 18 Part A

The area of the square hole is $A = 8$ square centimeters, so the side - length of the square hole $s_{h}=\sqrt{8}\approx2.83$ centimeters. The volume of the cube - shaped block is $V = 64$ cubic centimeters, so the side - length of the cube $s_{b}=\sqrt[3]{64}=4$ centimeters. We can find the side - length of the cube from its volume.

Step5: Solve problem 18 Part B

Since $s_{b}=4$ centimeters and $s_{h}\approx2.83$ centimeters, $s_{b}>s_{h}$, the block will not fit in the square hole.

Answer:

1. 11 feet
2. Yes, because the side - length of the poster is 3 feet and the side - length of the box is approximately 3.11 feet.
3. C. $\sqrt{36}+\sqrt{16}$
4. Part A: We can find the side - length of the cube from its volume, which is 4 centimeters.

Part B: No, because the side - length of the block (4 centimeters) is greater than the side - length of the hole (approximately 2.83 centimeters).