A1^(r)^(n-1)
a6= 1000^(-2)^5
1000^-16
1/1000^16
a6= 0
Answer= 0
1. is line FG
2. is line IH
3. is angle J
4. is angle H
5. is 12 cm.
6. is 4 cm.
7. is 135
8. is 150
9. sorry can't help you on this one.
10. is 150+90+90=330
<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>