c = 2(3.14)(radius)
c = 2(3.14)(3)
<em>multiply all those numbers together to get your circumference.</em>
c = 18.84.
Answer:
3
Step-by-step explanation:
To find the circumference of this circle, you need to use the formula C=2(3.14)r
which means circumference=2 times pi times the radius so, you would multiply 2.8(the radius of your circle) times 2. Then you would multiply that times 3.14(pi rounded to the nearest hundredth). And there u go. U have ur answer.
C=2(2.8)(3.14)
C= 5.6(3.14)
C=17.584
C=17.58 feet
Answer:
it would be the bottom left graph
Step-by-step explanation:
Rise over Run
1. Let a and b be coefficients such that
![\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}](https://tex.z-dn.net/?f=%5Cdfrac1%7Bx%282x%2B3%29%7D%20%3D%20%5Cdfrac%20ax%20%2B%20%5Cdfrac%20b%7B2x%2B3%7D)
Combining the fractions on the right gives
![\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}](https://tex.z-dn.net/?f=%5Cdfrac1%7Bx%282x%2B3%29%7D%20%3D%20%5Cdfrac%7Ba%282x%2B3%29%20%2B%20bx%7D%7Bx%282x%2B3%29%7D)
![\implies 1 = (2a+b)x + 3a](https://tex.z-dn.net/?f=%5Cimplies%201%20%3D%20%282a%2Bb%29x%20%2B%203a)
![\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23](https://tex.z-dn.net/?f=%5Cimplies%20%5Cbegin%7Bcases%7D3a%3D1%20%5C%5C%202a%2Bb%3D0%5Cend%7Bcases%7D%20%5Cimplies%20a%3D%5Cdfrac13%2C%20b%20%3D%20-%5Cdfrac23)
so that
![\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}](https://tex.z-dn.net/?f=%5Cdfrac1%7Bx%282x%2B3%29%7D%20%3D%20%5Cboxed%7B%5Cdfrac13%20%5Cleft%28%5Cdfrac1x%20-%20%5Cdfrac2%7B2x%2B3%7D%5Cright%29%7D)
2. a. The given ODE is separable as
![x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}](https://tex.z-dn.net/?f=x%282x%2B3%29%20%5Cdfrac%7Bdy%7Ddx%7D%20%3D%20y%20%5Cimplies%20%5Cdfrac%7Bdy%7Dy%20%3D%20%5Cdfrac%7Bdx%7D%7Bx%282x%2B3%29%7D)
Using the result of part (1), integrating both sides gives
![\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C](https://tex.z-dn.net/?f=%5Cln%7Cy%7C%20%3D%20%5Cdfrac13%20%5Cleft%28%5Cln%7Cx%7C%20-%20%5Cln%7C2x%2B3%7C%5Cright%29%20%2B%20C)
Given that y = 1 when x = 1, we find
![\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)](https://tex.z-dn.net/?f=%5Cln%7C1%7C%20%3D%20%5Cdfrac13%20%5Cleft%28%5Cln%7C1%7C%20-%20%5Cln%7C5%7C%5Cright%29%20%2B%20C%20%5Cimplies%20C%20%3D%20%5Cdfrac13%5Cln%285%29)
so the particular solution to the ODE is
![\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)](https://tex.z-dn.net/?f=%5Cln%7Cy%7C%20%3D%20%5Cdfrac13%20%5Cleft%28%5Cln%7Cx%7C%20-%20%5Cln%7C2x%2B3%7C%5Cright%29%20%2B%20%5Cdfrac13%5Cln%285%29)
We can solve this explicitly for y :
![\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)](https://tex.z-dn.net/?f=%5Cln%7Cy%7C%20%3D%20%5Cdfrac13%20%5Cleft%28%5Cln%7Cx%7C%20-%20%5Cln%7C2x%2B3%7C%20%2B%20%5Cln%285%29%5Cright%29)
![\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|](https://tex.z-dn.net/?f=%5Cln%7Cy%7C%20%3D%20%5Cdfrac13%20%5Cln%5Cleft%7C%5Cdfrac%7B5x%7D%7B2x%2B3%7D%5Cright%7C)
![\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|](https://tex.z-dn.net/?f=%5Cln%7Cy%7C%20%3D%20%5Cln%5Cleft%7C%5Csqrt%5B3%5D%7B%5Cdfrac%7B5x%7D%7B2x%2B3%7D%7D%5Cright%7C)
![\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}](https://tex.z-dn.net/?f=%5Cboxed%7By%20%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B5x%7D%7B2x%2B3%7D%7D%7D)
2. b. When x = 9, we get
![y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}](https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B45%7D%7B21%7D%7D%20%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B15%7D7%7D%20%5Capprox%20%5Cboxed%7B1.29%7D)