Answer:
9/4<x≤4 or all real numbers
Step-by-step explanation:
I'm not sure what you are asking
1. Let a and b be coefficients such that

Combining the fractions on the right gives



so that

2. a. The given ODE is separable as

Using the result of part (1), integrating both sides gives

Given that y = 1 when x = 1, we find

so the particular solution to the ODE is

We can solve this explicitly for y :


![\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)
2nd point distance-1st point distant
Using the left or right side of triangle, you can conclude that the midsegment will divide the triangle in both midlines of the sides. This means the length of the line would be half of the line below it. The equation would be:
47/4x+2= 94/ 4x+44
4x+44/ 4x+2 = 94/47
4x+ 44 / 4x+2 =2
4x+44 = 2(4x+2)
4x+44 = 8x+4
4x-8x= 4-44
-4x= -40
x= 10
Then the length of midline would be:
4x+2= 4(10)+2= 42
34 books would be the answer.