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)
Y=2/-8x-1.5 ( pretty much add 2x to both sides, then divide all parts by negative 8. So it'd be 2 over -8 and then 12 divided by -8 which equals -1.5)
C for change it is kinda obvious because change starts with a c and apples start with an a and bananas start with a b
Answer:
625000
Step-by-step explanation:
x+12%= 700000
x*1.12=700000
x=700000/1.12
x=625000