Answer:
50x105m
Step-by-step explanation:
310 (length of field) - 65-65= 200/4=50
Start with the perimeter subtract 65 twice because the length is on two sides. Now the length and width will be equal so divide by 4. Add 65 back because you subtracted it earlier.
At 08:00 there are 18 bags
At 08:10 there are an extra 10 produced = 28 but 1/2 are consumed leaving 14
At 08:20 there are an extra 10 produced = 24 but 1/2 are consumed leaving 12
At 08:30 there are an extra 10 produced = 22 but 1/2 are consumed leaving 11
At 08:36 another 6 have been produced = 17 bags
The initial velocity is 19 ft/s.
This is in the form f(t) = -16t²+v₀t+h₀, where v₀ is the initial velocity. 19 is in the place of v₀, so it is the initial velocity.
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.
![\frac{dA}{dw}=\frac{d}{dw}[w(100-2w)]](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdw%7D%3D%5Cfrac%7Bd%7D%7Bdw%7D%5Bw%28100-2w%29%5D)

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.