Fernando has two equally-sized containers of clay which he is using to build a pyramid-like structure. The finished structure will consist of four square-based, two-inch thick slabs of clay, with each slab two inches less wide than the one beneath it. Fernando has one-half of a container of clay left after forming the first two slabs. Does he have enough clay to finish the structure?

Fernando does not have enough clay to finish the structure since, the expressions for the volume left are not the same.

Step-by-step explanation:

Let w be the length of the longest slab. Since each slab is 2 inches less than the one beneath it, we have the width of the other 3 slabs as w - 2, w - 2 - 2 = w - 4 and w - 4 - 2 = w - 6 respectively.

Since the base of each slab is a square and has thickness 2 inches, the volume of each slab from largest to smallest is thus 2w², 2(w - 2)², 2(w - 4)² and 2(w - 6)² respectively.

We have that we have half of the container of clay left after forming the first two slabs(which are the bottom-most slabs). Let V be the volume of each clay container. Since we have half left, that is V/2, we have used 2V - V/2 = 3V/2 to make the two bottom-most slabs.

So, 3V/2 = 2w² + 2(w - 2)²

= 2w² + 2(w² - 4w + 4)

= 2w² + 2w² - 8w + 8

= 4w² - 8w + 8

= 4(w² - 2w + 2)   (1)

Also we want to know if V/2 is equal to the volume of the two top-most slabs.

So, V/2 = 2(w - 4)² + 2(w - 6)²

= 2(w² - 8w + 16) + 2(w² - 12w + 36)

= 2w² - 16w + 32 + 2w² - 24w + 72

= 4w² - 40w + 104

= 4(w² - 10w + 26)

From (1) V/2 = 4(w² - 2w + 2)/3 ≠ 4(w² - 10w + 26)

Since the expressions for the remaining volume are not the same, Fernando does not have enough clay to finish the structure.

i think ur answer is   (3,-1)

80π/3 or about 83.775

step-by-step explanation:

since a cone is 1/3 of a cylinder, we need to find the base's area. since area = πr^2, we get 16π. then, multiplying this by 5, we get 80π. dividing by 3, since the cone's volume is 1/3 of that of a cylinder's, we get 80π/3.

hope this !