7. Match each expression in the left column with an equivalent expression in the right column. A. ?! + m + N + M! 1. (r1 - 36).4
1 answer:
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Answer:
Note: The expression "h =300- pi r /pi r" would equate to 299, as written. I find h = (300 - π )/πr to be the equation for height.
Step-by-step explanation:
The total surface area of a sylinder is 2πr(h+r) where h is the height and r is the radius. That consists of top and bottom disks, each with an area of π, for a total of 2π
. The side is a continous sheet that has an area of 2πrh, where h is the height. 2πr is the circumference of the can, so the surface area of the side is 2πrh.
The total surace area of a closed cylinder is therefore 2π + 2πrh.
2π + 2πrh = 600
π + πrh = 300
πrh = 300 - π
h = (300 - π )/πr
Answer:
For maximum area of the rectangular exercise run dimensions will be 50ft by 25ft.
Step-by-step explanation:
Let the length of the rectangular exercise run = l ft
and width of the run = w ft
Sinoman has to cover a rectangular exercise run from three sides with the fencing material,
So length of the material = (l + 2w) ft
l + 2w = 100
l = 100 - 2w --------(1)
Area of the rectangular area covered = Length × width
A = lw
A = w(100 - 2w) [(l = 100 - 2w)from equation (1)
For maximum area we find the derivative of area and equate it to zero.
A' = 100 - 4w
For A' = 0
100 - 4w = 0
4w = 100
w = 25 ft
From equation (1)
l = 100 - 2w
l = 100 - 2×(25)
l = 50 ft
Therefore, for maximum area of the rectangular exercise run dimensions will be 50ft by 25ft.
