Answer:
6m+36
Step-by-step explanation:
Distributive property: 6(m)+6(6)
6m+36
Answer:
a) h = 123/x^2
b) S = x^2 +492/x
c) x ≈ 6.27
d) S'' = 6; area is a minimum (Y)
e) Amin ≈ 117.78 m²
Step-by-step explanation:
a) The volume is given by ...
V = Bh
where B is the area of the base, x^2, and h is the height. Filling in the given volume, and solving for the height, we get:
123 = x^2·h
h = 123/x^2
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b) The surface area is the sum of the area of the base (x^2) and the lateral area, which is the product of the height and the perimeter of the base.

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c) The derivative of the area with respect to x is ...

When this is zero, area is at an extreme.
![0=2x -\dfrac{492}{x^2}\\\\0=x^3-246\\\\x=\sqrt[3]{246}\approx 6.26583](https://tex.z-dn.net/?f=0%3D2x%20-%5Cdfrac%7B492%7D%7Bx%5E2%7D%5C%5C%5C%5C0%3Dx%5E3-246%5C%5C%5C%5Cx%3D%5Csqrt%5B3%5D%7B246%7D%5Capprox%206.26583)
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d) The second derivative is ...

This is positive, so the value of x found represents a minimum of the area function.
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e) The minimum area is ...

The minimum area of metal used is about 117.78 m².
Answer:
3
Step-by-step explanation:
We can use the formula [ y2-y1/x2-x1 ] to solve.
7-1/2-0
6/2
3
Best of Luck!
Answer:
12 possibilities
Step-by-step explanation:
In the first urn, we have 4 balls, and all of them are different, as they have different labels, so the group of two red balls r1 and r2 is different from the group of red balls r2 and r3.
The same thing occurs in the second urn, as all balls have different labels.
The problem is a combination problem (the group r1 and r2 is the same group r2 and r1).
For the first urn, we have a combination of 4 choose 2:
C(4,2) = 4!/2!*2! = 4*3*2/2*2 = 2*3 = 6 possibilities
For the second urn, we also have a combination of 4 choose 2, so 6 possibilities.
In total we have 6 + 6 = 12 possibilities.