Answer:
1. Find the difference between the areas.
<u>Area of the small rectangle</u>: ![(x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14](https://tex.z-dn.net/?f=%28x%2B2%29%28x%2B7%29%3Dx%5E2%2B7x%2B2x%2B14%3Dx%5E2%2B9x%2B14)
<u>Area of the big rectangle</u>: ![(x+9)(x+11)=x^2+11x+9x+99=x^2+20x+99](https://tex.z-dn.net/?f=%28x%2B9%29%28x%2B11%29%3Dx%5E2%2B11x%2B9x%2B99%3Dx%5E2%2B20x%2B99)
The difference is: ![11x+85](https://tex.z-dn.net/?f=11x%2B85)
![( x^2+20x+99)- (x^2+9x+14)=x^2+20x+99-x^2-9x-14=11x+85](https://tex.z-dn.net/?f=%28%20x%5E2%2B20x%2B99%29-%20%28x%5E2%2B9x%2B14%29%3Dx%5E2%2B20x%2B99-x%5E2-9x-14%3D11x%2B85)
2.
You can solve this question just by looking at the graph.
a) The height is 4 meters.
![f(d)=h=-2d^2+7d+4](https://tex.z-dn.net/?f=f%28d%29%3Dh%3D-2d%5E2%2B7d%2B4)
To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.
![f(0)=h=-2(0)^2+7(0)+4](https://tex.z-dn.net/?f=f%280%29%3Dh%3D-2%280%29%5E2%2B7%280%29%2B4)
![h=4](https://tex.z-dn.net/?f=h%3D4)
The height is 4 meters.
b) 9 meters.
For ![d=1](https://tex.z-dn.net/?f=d%3D1)
![f(1)=h=-2(1)^2+7(1)+4](https://tex.z-dn.net/?f=f%281%29%3Dh%3D-2%281%29%5E2%2B7%281%29%2B4)
![f(1)=h=-2+7+4](https://tex.z-dn.net/?f=f%281%29%3Dh%3D-2%2B7%2B4)
![h=9](https://tex.z-dn.net/?f=h%3D9)
b) The ball travels 4 meters.
But to calculate it, it is when ![h=0](https://tex.z-dn.net/?f=h%3D0)
![0=-2d^2+7d+4](https://tex.z-dn.net/?f=0%3D-2d%5E2%2B7d%2B4)
Using the quadratic formula:
![$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$](https://tex.z-dn.net/?f=%24d%3D%5Cfrac%7B-b%5Cpm%20%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D%24)
![$d=\frac{-7 \pm \sqrt{7^2-4\left(-2\right)4}}{2\left(-2\right)}$](https://tex.z-dn.net/?f=%24d%3D%5Cfrac%7B-7%20%5Cpm%20%5Csqrt%7B7%5E2-4%5Cleft%28-2%5Cright%294%7D%7D%7B2%5Cleft%28-2%5Cright%29%7D%24)
![$d=\frac{-7\pm\sqrt{81}}{-4}$](https://tex.z-dn.net/?f=%24d%3D%5Cfrac%7B-7%5Cpm%5Csqrt%7B81%7D%7D%7B-4%7D%24)
![$d=\frac{-7\pm9}{-4}$](https://tex.z-dn.net/?f=%24d%3D%5Cfrac%7B-7%5Cpm9%7D%7B-4%7D%24)
It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.
![$d=-\frac{1}{2} \text{ or }d=4$](https://tex.z-dn.net/?f=%24d%3D-%5Cfrac%7B1%7D%7B2%7D%20%5Ctext%7B%20or%20%7Dd%3D4%24)
3.
In this question, we have to find the area of the cylinder and the sphere.
From the information given, we have
a = 5mm and d = 5mm, therefore the radius is 2.5 mm.
The volume of a cylinder:
![V=\pi r^2h](https://tex.z-dn.net/?f=V%3D%5Cpi%20r%5E2h)
![V=\pi (2.5)^2 \cdot 5](https://tex.z-dn.net/?f=V%3D%5Cpi%20%282.5%29%5E2%20%5Ccdot%205)
![V=31.25 \pi](https://tex.z-dn.net/?f=V%3D31.25%20%5Cpi)
![V_{c} \approx 98.17 \text{ m}^3](https://tex.z-dn.net/?f=V_%7Bc%7D%20%5Capprox%2098.17%20%5Ctext%7B%20m%7D%5E3)
The volume of the sphere:
![$V=\frac{4}{3} \pi r^2$](https://tex.z-dn.net/?f=%24V%3D%5Cfrac%7B4%7D%7B3%7D%20%20%5Cpi%20r%5E2%24)
![V_{s} \approx 65.4 \text{ m}^3](https://tex.z-dn.net/?f=V_%7Bs%7D%20%5Capprox%2065.4%20%5Ctext%7B%20m%7D%5E3)
The volume of the capsule is approximately ![163.57 \text{ m}^3](https://tex.z-dn.net/?f=163.57%20%20%5Ctext%7B%20m%7D%5E3)