Answer:2n +10
Step-by-step explanation:
2*n= 2n
2*5= 10
Answer:
The area is changing by the rate of 44.62 meters per sec.
Step-by-step explanation:
Let x be the side of the square and r be the radius of the circle,
Then, the area outside the circle but inside the square is,
V = Area of square - area of circle,
∵ Area of a square = side² and area of a circle =
(radius)²,
Thus,
![V=x^2-\pi(r)^2](https://tex.z-dn.net/?f=V%3Dx%5E2-%5Cpi%28r%29%5E2)
Differentiating with respect to t ( time )
![\frac{dV}{dt}=2x\frac{dx}{dt} -2\pi r\frac{dr}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D2x%5Cfrac%7Bdx%7D%7Bdt%7D%20-2%5Cpi%20r%5Cfrac%7Bdr%7D%7Bdt%7D)
We have,
x = 20 meters, r = 3 meters,
![\frac{dr}{dt}=4\text{ meters per sec}](https://tex.z-dn.net/?f=%5Cfrac%7Bdr%7D%7Bdt%7D%3D4%5Ctext%7B%20meters%20per%20sec%7D)
![\implies \frac{dV}{dt}=2(20)(3)-2\pi(3)(4)](https://tex.z-dn.net/?f=%5Cimplies%20%5Cfrac%7BdV%7D%7Bdt%7D%3D2%2820%29%283%29-2%5Cpi%283%29%284%29)
![=120-24\pi](https://tex.z-dn.net/?f=%3D120-24%5Cpi)
![=44.6017763138](https://tex.z-dn.net/?f=%3D44.6017763138)
![\approx 44.62\text{ meter per sec}](https://tex.z-dn.net/?f=%5Capprox%2044.62%5Ctext%7B%20meter%20per%20sec%7D)
Answer:
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Step-by-step explanation:
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Answer:
A = 981.3 ft^2
Step-by-step explanation:
Here we're calculating the area of a semicircle of radius 25 ft.
The area of a circle is A = (pi)(r)^2, and that of a semicircle of the same radius is A = (1/2)(pi)(r)^2.
Here pi = 3.14 and r = 25 ft, and so the area/space in question is
A = (1/2)(3.14)(25)^2
A = 981.3 ft^2