The answer is 22063 bill?????.
8/12 = 2/3
You should divide 8 and 12 by 4.
Working and answer in the photo above. Hope this helps
Answer:
B: (5, 12)
Step-by-step explanation:
Find the solution of the following system of linear equations:
3x- y=3
y=2x +2
Note that the 2nd equation has already been solved for y. Thus, subbing 2x + 2 for y in the 1st equation is the way to go:
3x - (2x + 2) = 3.
Simplifying, 3x - 2x - 2 = 3, and so x = 3 + 2 = 5.
Subbing 5 for x in the 2nd equation yields y = 2(5) + 2, or y = 12.
The solution is B: (5, 12)
Answer:
![2m\times 2m\times \frac{5}{4}m](https://tex.z-dn.net/?f=2m%5Ctimes%202m%5Ctimes%20%5Cfrac%7B5%7D%7B4%7Dm)
Step-by-step explanation:
We are given that a container with square bottom.
Let side of square=x
Height of container=h
Volume of container=![5m^3](https://tex.z-dn.net/?f=5m%5E3)
Cost of 1 square meter of material for the bottom=$10
Cost of 1 square meter of material for side=$8
We have to find the dimension of least expensive container.
Volume of container=![x^2h](https://tex.z-dn.net/?f=x%5E2h)
![x^2h=5](https://tex.z-dn.net/?f=x%5E2h%3D5)
![h=\frac{5}{x^2}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B5%7D%7Bx%5E2%7D)
Surface area of container=![x^2+2(x+x)h=x^2+4xh](https://tex.z-dn.net/?f=x%5E2%2B2%28x%2Bx%29h%3Dx%5E2%2B4xh)
Cost=![C(x)=10x^2+8(4xh)=10x^2+32x(\frac{5}{x^2})=10x^2+\frac{160}{x}](https://tex.z-dn.net/?f=C%28x%29%3D10x%5E2%2B8%284xh%29%3D10x%5E2%2B32x%28%5Cfrac%7B5%7D%7Bx%5E2%7D%29%3D10x%5E2%2B%5Cfrac%7B160%7D%7Bx%7D)
![C(x)=10x^2+\frac{160}{x}](https://tex.z-dn.net/?f=C%28x%29%3D10x%5E2%2B%5Cfrac%7B160%7D%7Bx%7D)
Differentiate w.r.t x
![\frac{dC}{dx}=20x-\frac{160}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7BdC%7D%7Bdx%7D%3D20x-%5Cfrac%7B160%7D%7Bx%5E2%7D)
![\frac{dC}{dx}=0](https://tex.z-dn.net/?f=%5Cfrac%7BdC%7D%7Bdx%7D%3D0)
![20x-\frac{160}{x^2}=0](https://tex.z-dn.net/?f=20x-%5Cfrac%7B160%7D%7Bx%5E2%7D%3D0)
![\frac{160}{x^2}=20x](https://tex.z-dn.net/?f=%5Cfrac%7B160%7D%7Bx%5E2%7D%3D20x)
![x^3=\frac{160}{20}=8](https://tex.z-dn.net/?f=x%5E3%3D%5Cfrac%7B160%7D%7B20%7D%3D8)
![x=\sqrt[3]{8}=2](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B8%7D%3D2)
Because side of container is always positive.
Again differentiate w.r.t x
![\frac{d^2C}{dx^2}=20+\frac{320}{x^3}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2C%7D%7Bdx%5E2%7D%3D20%2B%5Cfrac%7B320%7D%7Bx%5E3%7D)
Substitute x=2
![\frac{d^2C}{dx^2}=20+\frac{320}{2^3}=60>0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2C%7D%7Bdx%5E2%7D%3D20%2B%5Cfrac%7B320%7D%7B2%5E3%7D%3D60%3E0)
Hence, the cost is minimum at x=2
Substitute the value of x
m
Hence, the dimensions of the least expensive container are
![2m\times 2m\times \frac{5}{4}m](https://tex.z-dn.net/?f=2m%5Ctimes%202m%5Ctimes%20%5Cfrac%7B5%7D%7B4%7Dm)