20 edges
F=12, V=10
V -E+F=2
10-E+12=2
-E+22=2
-E=-20
E=20
Given:
The given quadratic equation is:
To find:
The solution for the given quadratic equation in simplest form by using the quadratic formula.
Solution:
Consider a quadratic equation is defined as 
, then the quadratic formula is:
The given quadratic equation is:
We have, 
. Using the quadratic formula, we get
![[\because \sqrt{-a}=i\sqrt{a}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Csqrt%7B-a%7D%3Di%5Csqrt%7Ba%7D%5D)
The solutions of the given equations are 
and 
.
Hence, the correct option is B.
By comparing the right triangle ABC with the triangle XYZ, we will see that:
<h3>
How to find the given segments?</h3>
So we know that the largest triangle is a dilation of scale factor 2 of the smaller one.
Also, remember that for right triangle we have the relation:
Tan(θ) = (opposite cathetus)/(adjacent cathetus).
Now, the problem tells us that:
tan(x) = 5/2.5
For the angle x, the adjacent cathetus is XY, and the opposite cathetus is YZ.
Then we have:
YZ = 5
XY = 2.5
Knowing that the correspondent sides in the larger triangle will be 2 times the above ones, we can see that:
AC = 2*XY = 2*2.5 = 5
CB = 2*YZ = 2*5 = 10
If you want to learn more about right triangles, you can read:
brainly.com/question/2217700
keeping in mind that standard form for a linear equation means
• all coefficients must be integers, no fractions
• only the constant on the right-hand-side
• all variables on the left-hand-side, sorted
• "x" must not have a negative coefficient
![\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-3})~\hspace{10em} slope = m\implies 6 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-3)=6[x-(-1)]\implies y+3=6(x+1) \\\\\\ y+3=6x+6\implies y=6x+3\implies -6x+y=3\implies 6x-y=-3](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B-3%7D%29~%5Chspace%7B10em%7D%20slope%20%3D%20m%5Cimplies%206%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-%28-3%29%3D6%5Bx-%28-1%29%5D%5Cimplies%20y%2B3%3D6%28x%2B1%29%20%5C%5C%5C%5C%5C%5C%20y%2B3%3D6x%2B6%5Cimplies%20y%3D6x%2B3%5Cimplies%20-6x%2By%3D3%5Cimplies%206x-y%3D-3)
Answer:
P'( 4, -4)
Step-by-step explanation:
The double reflection across the x-axis does nothing, so the net effect is a single reflection across the y-axis. That effect is to negate the x-coordinate:
reflection across y-axis: (x, y) ⇒ (-x, y)
Then P(-4, -4) ⇒ P'(4, -4)
