Your answer should be 33 m ^2
A) Pine Road and Oak Street form a right angle, so we can extract the relation
![\tan30^\circ=\dfrac x{11\sqrt3}\implies\dfrac1{\sqrt3}=\dfrac x{11\sqrt3}\implies x=11](https://tex.z-dn.net/?f=%5Ctan30%5E%5Ccirc%3D%5Cdfrac%20x%7B11%5Csqrt3%7D%5Cimplies%5Cdfrac1%7B%5Csqrt3%7D%3D%5Cdfrac%20x%7B11%5Csqrt3%7D%5Cimplies%20x%3D11)
where
![x](https://tex.z-dn.net/?f=x)
is the distance we want to find (bottom side of the rectangle).
Alternatively, we can use the other given angle by solving for
![x](https://tex.z-dn.net/?f=x)
in
![\tan60^\circ=\dfrac{11\sqrt3}x](https://tex.z-dn.net/?f=%5Ctan60%5E%5Ccirc%3D%5Cdfrac%7B11%5Csqrt3%7Dx)
but we'll find the same solution either way.
b) Pine Road and Oak Street form a right triangle, with Main Street as its hypotenuse. We can use the Pythagorean theorem to find how long it is.
![(\text{Pine})^2+(\text{Oak})^2=(\text{Main})^2](https://tex.z-dn.net/?f=%28%5Ctext%7BPine%7D%29%5E2%2B%28%5Ctext%7BOak%7D%29%5E2%3D%28%5Ctext%7BMain%7D%29%5E2)
Let
![y](https://tex.z-dn.net/?f=y)
be the length of Main Street. Then
![(11\sqrt3)^2+11^2=x^2\implies x^2=484\implies x=\pm22](https://tex.z-dn.net/?f=%2811%5Csqrt3%29%5E2%2B11%5E2%3Dx%5E2%5Cimplies%20x%5E2%3D484%5Cimplies%20x%3D%5Cpm22)
but of course the distance has to be positive, so
![x=22](https://tex.z-dn.net/?f=x%3D22)
.
Good evening
The absolute value has the following property:
![|x|\leq a\iff -a\leq x\leq a](https://tex.z-dn.net/?f=%7Cx%7C%5Cleq%20a%5Ciff%20-a%5Cleq%20x%5Cleq%20a)
Apllying:
The average temperature range is between 45 and 65 degrees.
If you are solving for x then here’s how:
(3x2 + 4x - 12) = (x + 5)
Remove the x on the left by remove one x from both of the sides.
(3x2 + 3x - 12) = 5
Add 12 onto both sides
(3x2 + 3x) = 17
Multiply
(6 + 3x) = 17
Take away 6 from both sides
3x = 11
Divide by 3 on both sides
x = 3.666 recurring
Answer:
hope this helps you.........