Answer:
I would cost $4.75
Step-by-step explanation:
9.50 / 2
4.75
Hope it´s helpfull :)
<span>Let p, np be the roots of the given QE.So p+np = -b/a, and np^2 = c/aOr (n+1)p = -b/a or p = -b/a(n+1)So n[-b/a(n+1)]2 = c/aor nb2/a(n+1)2 = cor nb2 = ac(n+1)2
Which will give can^2 + (2ac-b^2)n + ac = 0, which is the required condition.</span>
All you have to do is multiply it then you get the answer
Answer:
a 106
b 92
Step-by-step explanation:
Hope this helps!
You can do it! :) I believe in you