You use the quadratic formula:
2x {}^{2} - 3x = 5
2x {}^{2} - 3x - 5 = 0
x = \frac{3 + - \sqrt{9 -4(2)( - 5)}}{2 \times 2}
x = \frac{3 + - \sqrt{49} }{4}
x = \frac{3 + 7}{4} \: and \: x = \frac{3 - 7}{4}
x = \frac{5}{2} \: and \: x = - 1
Answer:
The distance is 3.
Step-by-step explanation:
Since the x coordinates are the same, we take (-1)-(-4) = 3.
5+(-4)= 1
14+(-13)= 1
So the answer is B. =
I hope this helps
Answer:
and 
Step-by-step explanation:
Given

Required
Find x and y
In the second equation. Assume that:

Substitute
in the first equation


Collect like terms


Multiply through by 5


Solve for x

Substitute this value of x in 





if the diameter is 20, the its radius must be half that or 10.
![\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ A=5\pi \\ r=10 \end{cases}\implies \begin{array}{llll} 5\pi =\cfrac{\theta \pi (10)^2}{360}\implies 5\pi =\cfrac{5\pi \theta }{18} \\\\\\ \cfrac{5\pi }{5\pi }=\cfrac{\theta }{18}\implies 1=\cfrac{\theta }{18}\implies 18=\theta \end{array}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20sector%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B%5Ctheta%20%5Cpi%20r%5E2%7D%7B360%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20%5Ctheta%20%3D%5Cstackrel%7Bdegrees%7D%7Bangle%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20A%3D5%5Cpi%20%5C%5C%20r%3D10%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%205%5Cpi%20%3D%5Ccfrac%7B%5Ctheta%20%5Cpi%20%2810%29%5E2%7D%7B360%7D%5Cimplies%205%5Cpi%20%3D%5Ccfrac%7B5%5Cpi%20%5Ctheta%20%7D%7B18%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B5%5Cpi%20%7D%7B5%5Cpi%20%7D%3D%5Ccfrac%7B%5Ctheta%20%7D%7B18%7D%5Cimplies%201%3D%5Ccfrac%7B%5Ctheta%20%7D%7B18%7D%5Cimplies%2018%3D%5Ctheta%20%5Cend%7Barray%7D)