ANSWER ASAP GIVING BRAILIEST AND STUFF! Please answer the following essay question:
Suppose each dimension of a cube is increased. What happens to the surface area when each dimension is doubled, tripled or, quadrupled?

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24s^2, 54s^2, 96s^2

Step-by-step explanation:

Let s represent the initial side length of the cube.  Then the area of each face of the cube is A = 6s^2 (recalling that the area of a square of side length s is s^2).

a) Now suppose we double the side length.  The total area of the 6 faces of the cube will now be A = 6(2s)^2, or 24s^2 (a 24 times larger surface area),

b) tripled:  A = 6(3s)^2 = 54x^2

c) quadrupled?  A = 6(4s)^2 = 96s^2

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .

1. minimum

2. 6400

3. eight

hope this !