Part I
We have the size of the sheet of cardboard and we'll use the variable "x" to represent the length of the cuts. For any given cut, the available distance is reduced by twice the length of the cut. So we can create the following equations for length, width, and height.
width: w = 12 - 2x
length: l = 18 - 2x
height: h = x
Part II
v = l * w * h
v = (18 - 2x)(12 - 2x)x
v = (216 - 36x - 24x + 4x^2)x
v = (216 - 60x + 4x^2)x
v = 216x - 60x^2 + 4x^3
v = 4x^3 - 60x^2 + 216x
Part III
The length of the cut has to be greater than 0 and less than half the length of the smallest dimension of the cardboard (after all, there has to be something left over after cutting out the corners). So 0 < x < 6
Let's try to figure out an x that gives a volume of 224 in^3. Since this is high school math, it's unlikely that you've been taught how to handle cubic equations, so let's instead look at integer values of x. If we use a value of 1, we get a volume of:
v = 4x^3 - 60x^2 + 216x
v = 4*1^3 - 60*1^2 + 216*1
v = 4*1 - 60*1 + 216
v = 4 - 60 + 216
v = 160
Too small, so let's try 2.
v = 4x^3 - 60x^2 + 216x
v = 4*2^3 - 60*2^2 + 216*2
v = 4*8 - 60*4 + 216*2
v = 32 - 240 + 432
v = 224
And that's the desired volume.
So let's choose a value of x=2.
Reason?
It meets the inequality of 0 < x < 6 and it also gives the desired volume of 224 cubic inches.
Answer:
5⁴a²
Step-by-step explanation:
(5³a³)÷5a-¹×5-²a²
5³a³÷5a-¹×5-²a²
5³a³÷5¹-²×a-¹+²
5³a³÷5-¹a
5³a³/5-¹a
5³-(-¹)a³-¹
5⁴a²
To find the surface area of a cylinder, use the formula SA = 2(pi)(radius^2) + 2(pi)(radius)(height).
So now, plug in what you know.
SA = 2(3.14)(5^2) + 2(3.14)(5)(9)
SA = (6.28)(25) + (6.28)(45)
SA = 157 + 282.6
SA = 439.6 square cm
439.6 square cm is your surface area
Answer:
x=10 y=35 adjacent: AFE; CFD; BFC;BFD; AFB
Step-by-step explanation:
Answer: 16
Step-by-step explanation: 20/2 = 10
10x=160
10x/10=160/10
x=16
