an equation is undefined, when its denominator is 0, or turns to 0, because we can't divide anything by 0.
so let's simply set the denominator of this one = 0.
![\bf x^2-7x+10=0\implies (x-2)(x-5)=0\implies x= \begin{cases} 2\\ 5 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(2)^2-9}{(2)^2-7(2)+10}\implies \cfrac{-5}{-10+10}\implies \cfrac{-5}{0}\leftarrow und efined \\\\\\ \cfrac{(5)^2-9}{(5)^2-7(5)+10}\implies \cfrac{16}{-10+10}\implies \cfrac{16}{0}\leftarrow und efined](https://tex.z-dn.net/?f=%5Cbf%20x%5E2-7x%2B10%3D0%5Cimplies%20%28x-2%29%28x-5%29%3D0%5Cimplies%20x%3D%20%5Cbegin%7Bcases%7D%202%5C%5C%205%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B%282%29%5E2-9%7D%7B%282%29%5E2-7%282%29%2B10%7D%5Cimplies%20%5Ccfrac%7B-5%7D%7B-10%2B10%7D%5Cimplies%20%5Ccfrac%7B-5%7D%7B0%7D%5Cleftarrow%20und%20efined%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B%285%29%5E2-9%7D%7B%285%29%5E2-7%285%29%2B10%7D%5Cimplies%20%5Ccfrac%7B16%7D%7B-10%2B10%7D%5Cimplies%20%5Ccfrac%7B16%7D%7B0%7D%5Cleftarrow%20und%20efined)
Y=-1x+1 is the correct answer because you take the two points and go y2-y1/x2-x1, then one you figure that out you take y=mx+b and input your values (you can choose either point) so your work should look like: -4-4/5-3=-8/8=-1. -4=-1 (5)+b
-4= -5+b
+5. +5
1=b
y=-1x+1
The answer is D.) Neither ordered pair is a solution.
The reason being:
If we look at the ordered pair (3,-8) and plug it into the equation like so:
3(3)-(-8)4=11
9-32=11
-23=11
FALSE
Now let's look at the ordered pair (4,4):
3(4)-4(4)=11
12-16=11
-4=11
FALSE
We see that neither ordered pair works, so the answer is D.) neither is a solution.
Hope this helped :)
$4.30
Do ten times $3.57 and subtract it from $40
