Acube with 2-inch sides is placed on a cube with 3-inch sides. then a cube with 1-inch sides is placed on the 2-inch cube. what is the surface area of the three cube tower? show your work.

Step-by-step explanation:

5 sides of the top cube is exposed.

so we get 1*1*5 = 5 in ^2

the second cube has 4 sides exposed also, so 2^2 * 4 = 16 in ^2, but also it has one side with the cube on top, so we have 4-1 = 3 on that side, so the overall is 19 in ^2

Then we have for the 3rd cube 5 sides exposed, so we have 3^2 * 5 = 45. We also have the area of the 2 in cube on it, so we get 3^2 - 2^2 = 9-4 =5.

So the overall is 45+5+19+5 = 55+19 =74

I hope im right sorry if im not!

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?