LxWxH = 1018464
the other shape's volume is = 143 mm^3 this equals 2924207
to find the difference you would take them away from each other
1905743
hope this helped
HeyItsEihab
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)
5*9=45
3*9=27
45-27=8
9*2=18
Therfeore the answer is A.
Answer:
range = 51
Step-by-step explanation:
Range = biggest - smallest
62 - 11 = 51
Answer:
Since students need at least a C for the grade to transferdes that are less than C into one category instead of grouping D's and F's separately. How many ...