Answer:
33.06
Step-by-step explanation:
you would do 28.5*0.16 and you would get 4.56 ss you would add that to the cost and you would get 33.06
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:
60.6528 kg
Step-by-step explanation:
Assuming the solid ball of iron is a sphere with a radius of 18 cm (or 0.18 m), we can compute the volume of the sphere:

Therefore, the volume of the solid ball is 0.007776 m³
Since we know that density is mass/volume, then we can find the mass of the iron:

Thus, the mass of the solid ball of iron is 60.6528 kg
Answer:
The average number of customers per day during the sale was 150.
Step-by-step explanation:
This question can be solved using a rule of three.
Before the sale, the average number of customers per day was 120, which is 100% = 1.
Now, there are x customers, and since it is an increase of 25%, it is 100+25 = 125% = 1.25. So
120 - 1
x - 1.25
x = 1.25*120 = 150
The average number of customers per day during the sale was 150.