Check the picture below.
since we know the radius of the larger semicircle is 8, thus its diameter is 16, which is the length of one side of the equilateral triangle. We also know the smaller semicircle has a radius of 1/3, and thus a diameter of 2/3, namely the lenght of one side of the small equilateral triangle.
now, if we just can get the area of the larger figure and the area of the smaller one and subtract the smaller from the larger, we'll be in effect making a hole/gap in the larger and what's leftover is the shaded figure.
![\bf \stackrel{\textit{area of a semi-circle}}{A=\cfrac{1}{2}\pi r^2\qquad r=radius}~\hspace{10em}\stackrel{\textit{area of an equilateral triangle}}{A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\stackrel{side's}{length}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\left[ \stackrel{\textit{larger figure}}{\cfrac{1}{2}\pi 8^2~~+~~\cfrac{16^2\sqrt{3}}{4}} \right]\qquad -\qquad \left[ \cfrac{1}{2}\pi \left( \cfrac{1}{3} \right)^2 +\cfrac{\left( \frac{2}{3} \right)^2\sqrt{3}}{4}\right]}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20semi-circle%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20r%5E2%5Cqquad%20r%3Dradius%7D~%5Chspace%7B10em%7D%5Cstackrel%7B%5Ctextit%7Barea%20of%20an%20equilateral%20triangle%7D%7D%7BA%3D%5Ccfrac%7Bs%5E2%5Csqrt%7B3%7D%7D%7B4%7D%5Cqquad%20s%3D%5Cstackrel%7Bside%27s%7D%7Blength%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cleft%5B%20%5Cstackrel%7B%5Ctextit%7Blarger%20figure%7D%7D%7B%5Ccfrac%7B1%7D%7B2%7D%5Cpi%208%5E2~~%2B~~%5Ccfrac%7B16%5E2%5Csqrt%7B3%7D%7D%7B4%7D%7D%20%5Cright%5D%5Cqquad%20-%5Cqquad%20%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20%5Cleft%28%20%5Ccfrac%7B1%7D%7B3%7D%20%5Cright%29%5E2%20%2B%5Ccfrac%7B%5Cleft%28%20%5Cfrac%7B2%7D%7B3%7D%20%5Cright%29%5E2%5Csqrt%7B3%7D%7D%7B4%7D%5Cright%5D%7D)
![\bf \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\frac{4}{9}\sqrt{3}}{4} \right] \\\\\\ \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\sqrt{3}}{9} \right]~~\approx~~ 211.38 - 0.37~~\approx~~ 211.01](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%5B%2032%5Cpi%20%2B64%5Csqrt%7B3%7D%20%5Cright%5D%5Cqquad%20-%5Cqquad%20%5Cleft%5B%20%5Ccfrac%7B%5Cpi%20%7D%7B18%7D%2B%5Ccfrac%7B%5Cfrac%7B4%7D%7B9%7D%5Csqrt%7B3%7D%7D%7B4%7D%20%5Cright%5D%20%5C%5C%5C%5C%5C%5C%20%5Cleft%5B%2032%5Cpi%20%2B64%5Csqrt%7B3%7D%20%5Cright%5D%5Cqquad%20-%5Cqquad%20%5Cleft%5B%20%5Ccfrac%7B%5Cpi%20%7D%7B18%7D%2B%5Ccfrac%7B%5Csqrt%7B3%7D%7D%7B9%7D%20%5Cright%5D~~%5Capprox~~%20211.38%20-%200.37~~%5Capprox~~%20211.01)
Answer:
45
Step-by-step explanation:
Find the volume of a pyramid with a square base, where the perimeter of the base is 18.4\text{ cm}18.4 cm and the height of the pyramid is 20.6\text{ cm}20.6 cm. Round your answer to the nearest tenth of a cubic centimeter.
Answer:
The answer is that she would pay $65.56 in finance charges at the end of the month.
Step-by-step explanation:
Given: APR = 19.99%
Carry Over Balance: $398.97
The APR or Annual Percentage Rate, is calculated daily. You will need to get the daily periodic rate, or DPR, so divide the APR by 365:
19.99% = .1999
.1999 / 365 = .005477 (This is the Approximate DPR, rounded up to .005477)
To get the finance charge, multiply the average daily balance by the DPR and then by 30 days:
398.97 * .005477 * 30 = $65.56 finance charge for this carry over balance, at the end of the month. This assumes that the balance is the average daily balance.
Hope this helps!! Have a great day!
Allright So you know the answer cannot be A or D because Why.... It says in the text that ur finding the area and area is adding so it has nothing to do with multiplying ok
