The diameter of a spherical balloon that is being filled with air is increasing at the rate of 3 inches more than the time, t. W
1 answer:
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Answer:
CosC =
Step-by-step explanation:
Cos C =
<em>Where Adjacent = 24 , Hyp = 40</em>
Cos C =
<u><em>In simplest form</em></u>
CosC =
Check the picture below.
so the lunchbox is really a semi-circle with a radius of 3, sitting on top of a square of 6x6, with a lenght of 10.
the volume of the lunchbox then will be the area of the semi-circle and square times the length.
![\bf \left[ \stackrel{\textit{semi-circle's area}}{\frac{\pi 3^2}{2}}+\stackrel{\textit{square's area}}{(6\cdot 6)} \right]\stackrel{length}{10}\implies \left( \frac{9\pi }{2}+36\right)(10)](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%5B%20%5Cstackrel%7B%5Ctextit%7Bsemi-circle%27s%20area%7D%7D%7B%5Cfrac%7B%5Cpi%203%5E2%7D%7B2%7D%7D%2B%5Cstackrel%7B%5Ctextit%7Bsquare%27s%20area%7D%7D%7B%286%5Ccdot%206%29%7D%20%5Cright%5D%5Cstackrel%7Blength%7D%7B10%7D%5Cimplies%20%5Cleft%28%20%20%5Cfrac%7B9%5Cpi%20%7D%7B2%7D%2B36%5Cright%29%2810%29)
so the lunchbox is really a semi-circle with a radius of 3, sitting on top of a square of 6x6, with a lenght of 10.
the volume of the lunchbox then will be the area of the semi-circle and square times the length.
Answer:
Step-by-step explanation: see attachment for solution
