A shipping box in the shape of a cube has an edge length of 8 inches. It can hold 8 mugs with no extra space between them. If th

Question

3 years ago

5

e dimensions of the shipping box are all doubled, what is the maximum number of mugs that the larger box can hold? 16 24 48 64

2 answers:

andriy [413] · Answer 1 · 2022-06-11T07:49:12+03:00

4 0

Answer:

64

Step-by-step explanation:

Volume of a cube is:

V = s³

When s = 8 in:

V = (8 in)³ = 512 in³

When s is doubled (s = 16 in):

V = (16 in)³ = 4096 in³

Now we can write a proportion:

8 mugs / 512 in³ = x mugs / 4096 in³

x = 64

blagie [28] · Answer 2 · 2022-06-11T09:03:02+03:00

<h2>Answer:</h2>

The maximum number of mugs that the larger box can hold is:

64

<h2>Step-by-step explanation:</h2>

The edge length of box is: 8 inches.

This means that the volume of box is: 8×8×8=512 cubic inches.

Also, the number of mugs a box of edge length 8 in. can hold is: 8

Hence, volume of 1 mug= Volume of box/Number of mugs

Volume of 1 mug= 512/8=64 cubic inches

Now, if the dimension of the box is doubled.

This means that the Volume of new box= 8×Volume of old box

Volume of new box=8×512=4096 cubic inches

Let n be the number of mugs that the new box can hold.

Volume of new box=n×Volume of 1 mug

⇒ 4096=n×64

⇒ n=4096/64

⇒ n=64

Hence, the answer is:

64