18pie cubed because it's such a small section
The decimal form would be 3.6
Step-by-step explanation:
1) the length of the side is:
![\frac{4-\sqrt{2} }{\sqrt{2}}=2\sqrt{2}-1.](https://tex.z-dn.net/?f=%5Cfrac%7B4-%5Csqrt%7B2%7D%20%7D%7B%5Csqrt%7B2%7D%7D%3D2%5Csqrt%7B2%7D-1.)
2) the required perimeter is:
![P=4(2\sqrt{2} -1)=8\sqrt{2} -4.](https://tex.z-dn.net/?f=P%3D4%282%5Csqrt%7B2%7D%20-1%29%3D8%5Csqrt%7B2%7D%20-4.)
3) the required area is:
![A=(2\sqrt{2} -1)^2=9-4\sqrt{2}.](https://tex.z-dn.net/?f=A%3D%282%5Csqrt%7B2%7D%20-1%29%5E2%3D9-4%5Csqrt%7B2%7D.)
Answer: The last option.
Step-by-step explanation:
1. Calculate the area of the rectangle as following:
![A_r=6m*8m=48m^{2}](https://tex.z-dn.net/?f=A_r%3D6m%2A8m%3D48m%5E%7B2%7D)
2. Find the base of the triangle applying the Pythagorean Theorem. Then:
![b=\sqrt{(10m)^{2}-(8m)^{2}}=6m](https://tex.z-dn.net/?f=b%3D%5Csqrt%7B%2810m%29%5E%7B2%7D-%288m%29%5E%7B2%7D%7D%3D6m)
3. Then, the area of the right triangle is:
![A_t=\frac{6m*8m}{2}=24m^{2}](https://tex.z-dn.net/?f=A_t%3D%5Cfrac%7B6m%2A8m%7D%7B2%7D%3D24m%5E%7B2%7D)
4. The area of the trapezoid is the sum of the area of the triangle and the area of the rectangle. Then:
![A_{trapezoid}=24m^{2}+48m^{2}=72m^{2}](https://tex.z-dn.net/?f=A_%7Btrapezoid%7D%3D24m%5E%7B2%7D%2B48m%5E%7B2%7D%3D72m%5E%7B2%7D)
The first sentence is not true in general. Consider the equation
![|x|=-1](https://tex.z-dn.net/?f=%7Cx%7C%3D-1)
. There are no solutions. Now consider
![|x|=0](https://tex.z-dn.net/?f=%7Cx%7C%3D0)
. There is only one solution,
![x=0](https://tex.z-dn.net/?f=x%3D0)
.
But whatever. You're asked to demonstrate that
![|x|=9](https://tex.z-dn.net/?f=%7Cx%7C%3D9)
has two solutions (which is true; the right hand side must be a positive integer in order to have two solutions). This follows immediately from the definition of absolute value, which says
![|x|=\begin{cases}x&\text{for }x\ge0\\-x&\text{for }x](https://tex.z-dn.net/?f=%7Cx%7C%3D%5Cbegin%7Bcases%7Dx%26%5Ctext%7Bfor%20%7Dx%5Cge0%5C%5C-x%26%5Ctext%7Bfor%20%7Dx%3C0%5Cend%7Bcases%7D)
So suppose
![x\ge0](https://tex.z-dn.net/?f=x%5Cge0)
. Then
![|x|=9\implies x=9](https://tex.z-dn.net/?f=%7Cx%7C%3D9%5Cimplies%20x%3D9)
Now suppose
![x](https://tex.z-dn.net/?f=x%3C0)
. Then
![|x|=9\implies -x=9\implies x=-9](https://tex.z-dn.net/?f=%7Cx%7C%3D9%5Cimplies%20-x%3D9%5Cimplies%20x%3D-9)
So two solutions to
![|x|=9](https://tex.z-dn.net/?f=%7Cx%7C%3D9)
are
![x=\pm9](https://tex.z-dn.net/?f=x%3D%5Cpm9)
.