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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Your answer would be (b.) because -2+-3=-5
Answer:
the parabola can be written as:
f(x) = y = a*x^2 + b*x + c
first step.
find the vertex at:
x = -b/2a
the vertex will be the point (-b/2a, f(-b/2a))
now, if a is positive, then the arms of the parabola go up, if a is negative, the arms of the parabola go down.
The next step is to see if we have real roots by using the Bhaskara's equation:

Now, draw the vertex, after that draw the values of the roots in the x-axis, and now conect the points with the general draw of the parabola.
If you do not have any real roots, you can feed into the parabola some different values of x around the vertex
for example at:
x = (-b/2a) + 1 and x = (-b/2a) - 1
those two values should give the same value of y, and now you can connect the vertex with those two points.
If you want a more exact drawing, you can add more points (like x = (-b/2a) + 3 and x = (-b/2a) - 3) and connect them, as more points you add, the best sketch you will have.
Answer:
Monomial; Standard form: 18a^2b^2
Step-by-step explanation:
