tis always a good idea when factoring to start off with a quick prime factoring.
![\begin{array}{llll} 52&=&2\cdot 2\cdot 13\\\\ 77&=&\boxed{7}\cdot 11\\\\ 91&=&\boxed{7}\cdot 13\\\\ 124&=&2\cdot 2\cdot 31\\\\ 217&=&\boxed{7}\cdot 31 \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bllll%7D%2052%26%3D%262%5Ccdot%202%5Ccdot%2013%5C%5C%5C%5C%2077%26%3D%26%5Cboxed%7B7%7D%5Ccdot%2011%5C%5C%5C%5C%2091%26%3D%26%5Cboxed%7B7%7D%5Ccdot%2013%5C%5C%5C%5C%20124%26%3D%262%5Ccdot%202%5Ccdot%2031%5C%5C%5C%5C%20217%26%3D%26%5Cboxed%7B7%7D%5Ccdot%2031%20%5Cend%7Barray%7D)
Answer:
A(-6,-5) B(4,1)
Step-by-step explanation:
am not that sure
X/4-9=5
x/4=14
x=14/4
x=7/2
x=3.5
Separate the variables:
![y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx](https://tex.z-dn.net/?f=y%27%20%3D%20%5Cdfrac%7Bdy%7D%7Bdx%7D%20%3D%20%28y%2B1%29%28y-2%29%20%5Cimplies%20%5Cdfrac1%7B%28y%2B1%29%28y-2%29%7D%20%5C%2C%20dy%20%3D%20dx)
Separate the left side into partial fractions. We want coefficients a and b such that
![\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%28y%2B1%29%28y-2%29%7D%20%3D%20%5Cdfrac%20a%7By%2B1%7D%20%2B%20%5Cdfrac%20b%7By-2%7D)
![\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdfrac1%7B%28y%2B1%29%28y-2%29%7D%20%3D%20%5Cdfrac%7Ba%28y-2%29%2Bb%28y%2B1%29%7D%7B%28y%2B1%29%28y-2%29%7D)
![\implies 1 = a(y-2)+b(y+1)](https://tex.z-dn.net/?f=%5Cimplies%201%20%3D%20a%28y-2%29%2Bb%28y%2B1%29)
![\implies 1 = (a+b)y - 2a+b](https://tex.z-dn.net/?f=%5Cimplies%201%20%3D%20%28a%2Bb%29y%20-%202a%2Bb)
![\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13](https://tex.z-dn.net/?f=%5Cimplies%20%5Cbegin%7Bcases%7Da%2Bb%3D0%5C%5C-2a%2Bb%3D1%5Cend%7Bcases%7D%20%5Cimplies%20a%20%3D%20-%5Cdfrac13%20%5Ctext%7B%20and%20%7D%20b%20%3D%20%5Cdfrac13)
So we have
![\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx](https://tex.z-dn.net/?f=%5Cdfrac13%20%5Cleft%28%5Cdfrac1%7By-2%7D%20-%20%5Cdfrac1%7By%2B1%7D%5Cright%29%20%5C%2C%20dy%20%3D%20dx)
Integrating both sides yields
![\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cdfrac13%20%5Cleft%28%5Cdfrac1%7By-2%7D%20-%20%5Cdfrac1%7By%2B1%7D%5Cright%29%20%5C%2C%20dy%20%3D%20%5Cint%20dx)
![\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C](https://tex.z-dn.net/?f=%5Cdfrac13%20%5Cleft%28%5Cln%7Cy-2%7C%20-%20%5Cln%7Cy%2B1%7C%5Cright%29%20%3D%20x%20%2B%20C)
![\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C](https://tex.z-dn.net/?f=%5Cdfrac13%20%5Cln%5Cleft%7C%5Cdfrac%7By-2%7D%7By%2B1%7D%5Cright%7C%20%3D%20x%20%2B%20C)
![\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C](https://tex.z-dn.net/?f=%5Cln%5Cleft%7C%5Cdfrac%7By-2%7D%7By%2B1%7D%5Cright%7C%20%3D%203x%20%2B%20C)
![\dfrac{y-2}{y+1} = e^{3x + C}](https://tex.z-dn.net/?f=%5Cdfrac%7By-2%7D%7By%2B1%7D%20%3D%20e%5E%7B3x%20%2B%20C%7D)
![\dfrac{y-2}{y+1} = Ce^{3x}](https://tex.z-dn.net/?f=%5Cdfrac%7By-2%7D%7By%2B1%7D%20%3D%20Ce%5E%7B3x%7D)
With the initial condition y(0) = 1, we find
![\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12](https://tex.z-dn.net/?f=%5Cdfrac%7B1-2%7D%7B1%2B1%7D%20%3D%20Ce%5E%7B0%7D%20%5Cimplies%20C%20%3D%20-%5Cdfrac12)
so that the particular solution is
![\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cdfrac%7By-2%7D%7By%2B1%7D%20%3D%20-%5Cdfrac12%20e%5E%7B3x%7D%7D)
It's not too hard to solve explicitly for y; notice that
![\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}](https://tex.z-dn.net/?f=%5Cdfrac%7By-2%7D%7By%2B1%7D%20%3D%20%5Cdfrac%7B%28y%2B1%29-3%7D%7By%2B1%7D%20%3D%201-%5Cdfrac3%7By%2B1%7D)
Then
![1 - \dfrac3{y+1} = -\dfrac12 e^{3x}](https://tex.z-dn.net/?f=1%20-%20%5Cdfrac3%7By%2B1%7D%20%3D%20-%5Cdfrac12%20e%5E%7B3x%7D)
![\dfrac3{y+1} = 1 + \dfrac12 e^{3x}](https://tex.z-dn.net/?f=%5Cdfrac3%7By%2B1%7D%20%3D%201%20%2B%20%5Cdfrac12%20e%5E%7B3x%7D)
![\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}](https://tex.z-dn.net/?f=%5Cdfrac%7By%2B1%7D3%20%3D%20%5Cdfrac1%7B1%2B%5Cfrac12%20e%5E%7B3x%7D%7D%20%3D%20%5Cdfrac2%7B2%2Be%5E%7B3x%7D%7D)
![y+1 = \dfrac6{2+e^{3x}}](https://tex.z-dn.net/?f=y%2B1%20%3D%20%5Cdfrac6%7B2%2Be%5E%7B3x%7D%7D)
![y = \dfrac6{2+e^{3x}} - 1](https://tex.z-dn.net/?f=y%20%3D%20%5Cdfrac6%7B2%2Be%5E%7B3x%7D%7D%20-%201)
![\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}](https://tex.z-dn.net/?f=%5Cboxed%7By%20%3D%20%5Cdfrac%7B4-e%5E%7B3x%7D%7D%7B2%2Be%5E%7B3x%7D%7D%7D)
The blank would be 50% because there are only 2 options it can land on and they are both even