Ashipping 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 the 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

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

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

hello from mrbilldoesmath!

answer:   x^2 + y^2 = 1024

discussion:

the equation of a circle with center (h,k) and radius "r" is

(x-h)^2 + (y-k)^2 = r^2.

in our case

(x-0)^2 + (y-0) ^2 = 32^2 = 1024

you,

mrb