I think it might be this answer <span> 20501.6 </span>
Answer:
D,0,2,-2
Step-by-step explanation:
2x^5-3x^3-20x=0
x(2x^4-3x^2-20)=0
x=0
or 2x^4-3x^2-20=0
put x²=t
2t²-3t-20=0
-20×2=-40
8-5=3
8×-5=-40
2t²-(8-5)t-20=0
2t²-8t+5t-20=0
2t(t-4)+5(t-4)=0
(t-4)(2t+5)=0
t=4
x²=4
x=2,-2
t=-5/2
x²=-5/2
it gives imaginary root. so real rational roots are 0,2,-2
The quickest way to tell if a line is perpendicular is to find the slope, so you'll need to get the slope for both lines
PQ = ((-8) - (-9))/(2 - 0) = 1/2
and, for reference, a perpendicular slope is the negative reciprocal of the original (so we're looking for -2)
RS = (3 - 4)/(3 - 1) = -1/2
so, no, these two lines aren't perpendicular because the slope of RS is only the negative of PQ, not the negative <em>reciprocal</em>.
