The minimum surface area that such a box can have is 380 square
<h3>How to determine the minimum surface area such a box can have?</h3>
Represent the base length with x and the bwith h.
So, the volume is
V = x^2h
This gives
x^2h = 500
Make h the subject
h = 500/x^2
The surface area is
S = 2(x^2 + 2xh)
Expand
S = 2x^2 + 4xh
Substitute h = 500/x^2
S = 2x^2 + 4x * 500/x^2
Evaluate
S = 2x^2 + 2000/x
Differentiate
S' = 4x - 2000/x^2
Set the equation to 0
4x - 2000/x^2 = 0
Multiply through by x^2
4x^3 - 2000 = 0
This gives
4x^3= 2000
Divide by 4
x^3 = 500
Take the cube root
x = 7.94
Substitute x = 7.94 in S = 2x^2 + 2000/x
S = 2 * 7.94^2 + 2000/7.94
Evaluate
S = 380
Hence, the minimum surface area that such a box can have is 380 square
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Answer:
the slope is undefined
Step-by-step explanation:
x=-1
I'm assuming this is x^2 + 3x - 4 and x(x^2 + 3x - 2)
1.) First distribute x(x^2 + 3x - 2) to get x^3 + 3x^2 - 2x.
2.) Because you are subtracting all the terms from x^3 + 3x^2 - 2x, it's the same thing as distributing -1 to x^2 + 3x - 4 and then adding it to x^3 + 3x^2 - 2x.
3.) -1(x^2 + 3x - 4) = -x^2 - 3x + 4
4.) Add (x^3 + 3x^2 - 2x) + (-x^2 - 3x + 4)
5.) x^3 + 2x^2 - 5x + 4 is your final answer.
A. multiplication is your answer.
Answer:
The first term is 128
Step-by-step explanation:please see attachment for explanation
