How does the volume of a square pyramid change if the base area is quadrupled and the height is reduced to 1/9 of its original size

The volume is reduced to 4/9 of its original size

Step-by-step explanation:

In this question, we are concerned with how the volume of a square pyramid changes given that its base area is quadrupled and the height is reduced to 1/9 of its original size

The easiest way to do this is to choose arbitrary values for the original base area and the original height, then we can calculate the base area of the new height and the new base area .

Mathematically, the volume of a square pyramid can be calculated using the formula

V = Ah/3

where A is the base area and h is the height of the pyramid

Now let’s say that the original area is 16 units^2 and the height is 27 units

Now the volume at this point would be

V = (27 * 16)/3 = 144 units^3

Now for the new pyramid, we have the base area quadrupled, meaning it is multiplied by 4, the new base area is thus 16 * 4 = 64 square units

The height reduced to 1/9 and thus becomes 1/9 * 27 = 3 units

The new volume is thus

V = Ah/3 = (64 * 3)/3 = 64 units^3

so how do they compare?

144 to 64 = 9/4

This means that the volume of the square pyramid is reduced to 4/9 of its original size

(4/9 of 144 is 64)

if you explain the the question more correctly i might be able to answer this better

step-by-step explanation:

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,