Answer:
Step-by-step explanation:
The question isn’t clear, can you write it better so I can understand?
Step-by-step explanation:
area of rrctangle:
12×8= 96cm²
now if rectangle dissapears and then we "glue" what's left, we gonna get two circles
area of circle which gas diameter of 12cm( so radius will be 6cm)will be
πr²=36π≈113.04
another circles radius is 4cm,so area is:
16π≈50.25
sum of all areas: S≈96+113.04+50.25≈259.28cm2
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)