Answer:
169π
Step-by-step explanation:
circle eq: (x-a)²+(y-b)²=r², where (a,b) is center and r is radius
r²=169
r = ±13
since a radius length must be positive, r = +13, not -13
A of circle: πr²
r = 13
169π
given that the circle eq has r², you could've noticed that you can take the constant in the circle eq and multiply that by π to get the same answer
Explanation:
Addition of fractions can be accomplished using the formula ...
a/b + c/d = (ad +bc)/(bd)
Usually, you are asked to find the common denominator and rewrite the fractions using that denominator. It is not necessary, but it can save a step in the reduction of the final result. Here, we'll use the formula, then reduce the result to lowest terms.
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13. 5/6 +9/11 = (5·11 +6·9)/(6·11) = 109/66 = 1 43/66
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14. 7/20 -5/8 = (7·8 -20·5)/(20·8) = -44/160 = -11/40
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15. 1/5 -1/12 = (1·12 -5·1)/(5·12) = 7/60
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Dividing fractions can be accomplished different ways. I was taught to multiply by the inverse of the divisor. ("Invert and multiply.") Here, that means the problem (2/7) / (1/13) can be rewritten as ...
(2/7) × (13/1) . . . . . where 13/1 is the inverse of 1/13.
You can also express the fractions over a common denominator. In that case, the quotient is the ratio of the numerators. Perhaps a little less obvious is that you can express the fractions using a common numerator. Then the quotient is the inverse of the ratio of the denominators: (2/7) / (2/26) = 26/7. (You can see how this works if you "invert and multiply" the fractions with common numerators. Those numerators cancel.)
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16. (2/7)/(1/13) = 2/7·13/1 = 26/7 = 3 5/7
so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check
so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.
![\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cstackrel%7Bf%28x%29%7D%7B2x%5E3-x%5E2-5x%7D~~%20-%20~~%5B%5Cstackrel%7Bg%28x%29%7D%7B-x%5E2%2B3x%7D%5D%5Cimplies%202x%5E3-x%5E2-5x%2Bx%5E2-3x%20%5C%5C%5C%5C%5C%5C%202x%5E3-8x%5Cimplies%202%28x%5E3-4x%29%5Cimplies%20%5Cdisplaystyle%202%5Cint%5Climits_%7B-2%7D%5E%7B0%7D%20%28x%5E3-4x%29dx%20%5Cimplies%202%5Cleft%5B%20%5Ccfrac%7Bx%5E4%7D%7B4%7D-2x%5E2%20%5Cright%5D_%7B-2%7D%5E%7B0%7D%5Cimplies%20%5Cboxed%7B8%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}](https://tex.z-dn.net/?f=%5Cstackrel%7Bg%28x%29%7D%7B-x%5E2%2B3x%7D~~%20-%20~~%5B%5Cstackrel%7Bf%28x%29%7D%7B2x%5E3-x%5E2-5x%7D%5D%5Cimplies%20-x%5E2%2B3x-2x%5E3%2Bx%5E2%2B5x%20%5C%5C%5C%5C%5C%5C%20-2x%5E3%2B8x%5Cimplies%202%28-x%5E3%2B4x%29%20%5C%5C%5C%5C%5C%5C%20%5Cdisplaystyle%202%5Cint%5Climits_%7B0%7D%5E%7B2%7D%20%28-x%5E3%2B4x%29dx%20%5Cimplies%202%5Cleft%5B%20-%5Ccfrac%7Bx%5E4%7D%7B4%7D%2B2x%5E2%20%5Cright%5D_%7B0%7D%5E%7B2%7D%5Cimplies%20%5Cboxed%7B8%7D%20~%5Chfill%20%5Cboxed%7B%5Cstackrel%7B%5Ctextit%7Btotal%20area%7D%7D%7B8~~%20%2B%20~~8~~%20%3D%20~~16%7D%7D)
Answer:
1st option will be the answer.
