If the side lengths of B are 2x as long as the side lengths of A then the volume of B should be the volume of A^2
216^2=46,656
46,656-216=46,440

This can be more easily displayed with smaller numbers

If cube X were 2*2*2 then the volume would be 8
If cube Y were 4*4*4 then the volume would be 64
8(volume of cube X)^2=64(volume of cube Y)

7 0

3 years ago

Write 6 2/3 as a improper fraction

kirill115 [55]

Hi, i think the answer 20/3 hope this helped :D

3 0

3 years ago

Read 2 more answers

Please solve this thank you :)

snow_lady [41]

Answer:

Scale= 3 x=9 y=15

Step-by-step explanation:

8 0

3 years ago

Find the length and width of a rectangle that has the given area and a minimum perimeter. Area: 162 square feet

Gnom [1K]

Answer:

The width and length of rectangle is 12.728 m

Step-by-step explanation:

Let the length of the rectangle = L

let the width of the rectangle = W

The subjective function is given by;

F(p) = 2(L + W)

F = 2L + 2W

Area of the rectangle is given by;

A = LW

LW = 162 ft²

L = 162 / W

Substitute in the value of L into subjective function;

$f = 2l + 2w\\\\f = 2(\frac{162}{w} )+2w\\\\f = \frac{324}{w} + 2w\\\\\frac{df}{dw} = \frac{-324}{w^2} +2\\\\$

Take the second derivative of the function, to check if it will given a minimum perimeter

$\frac{d^2f}{dw^2}= \frac{648}{w^3} \\\\Thus, \frac{d^2f}{dw^2}>0, \ since,\frac{648}{w^3} >0 \ (minimum \ function \ verified)$

Determine the critical points of the first derivative;

df/dw = 0

$\frac{-324}{w^2} +2 = 0\\\\-324 + 2w^2=0\\\\2w^2 = 324\\\\w^2 = \frac{324}{2} \\\\w^2 = 162\\\\w= \sqrt{162}\\\\w = 12.728 \ m$

L = 162 / 12.728

L = 12.728 m

Therefore, the width and length of rectangle is 12.728 m

3 0

3 years ago