Answer:
1. Find the difference between the areas.
Area of the small rectangle: [tex](x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14[/tex]
Area of the big rectangle: [tex](x+9)(x+11)=x^2+11x+9x+99=x^2+20x+99[/tex]
The difference is: [tex]11x+85[/tex]
[tex]( x^2+20x+99)- (x^2+9x+14)=x^2+20x+99-x^2-9x-14=11x+85[/tex]
2.
You can solve this question just by looking at the graph.
a) The height is 4 meters.
[tex]f(d)=h=-2d^2+7d+4[/tex]
To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.
[tex]f(0)=h=-2(0)^2+7(0)+4[/tex]
[tex]h=4[/tex]
The height is 4 meters.
b) 9 meters.
For [tex]d=1[/tex]
[tex]f(1)=h=-2(1)^2+7(1)+4[/tex]
[tex]f(1)=h=-2+7+4[/tex]
[tex]h=9[/tex]
b) The ball travels 4 meters.
But to calculate it, it is when [tex]h=0[/tex]
[tex]0=-2d^2+7d+4[/tex]
Using the quadratic formula:
[tex]$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$[/tex]
[tex]$d=\frac{-7 \pm \sqrt{7^2-4\left(-2\right)4}}{2\left(-2\right)}$[/tex]
[tex]$d=\frac{-7\pm\sqrt{81}}{-4}$[/tex]
[tex]$d=\frac{-7\pm9}{-4}$[/tex]
It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.
[tex]$d=-\frac{1}{2} \text{ or }d=4$[/tex]
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:
[tex]V=\pi r^2h[/tex]
[tex]V=\pi (2.5)^2 \cdot 5[/tex]
[tex]V=31.25 \pi[/tex]
[tex]V_{c} \approx 98.17 \text{ m}^3[/tex]
The volume of the sphere:
[tex]$V=\frac{4}{3} \pi r^2$[/tex]
[tex]V_{s} \approx 65.4 \text{ m}^3[/tex]
The volume of the capsule is approximately [tex]163.57 \text{ m}^3[/tex]
