<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>
Answer:
.
Step-by-step explanation:
All coterminal angles of an angle
are defined as
or 
where, n is an integer.
The given angle is

So, all coterminal angles of an angle
are

For n=1,




Since,
between 0 and 2π, therefore, the required coterminal angle is
.
Answer:
10^2+2×2b×10+2b^2 = 100+40b+2b^2
Answer:
the answer is the first one
Step-by-step explanation:
39.99 rounded up is 40.00
19.99 rounded up is 20.00
40.00*0.05 = 2, 40-2 = 38
20.00*0.15 = 3, 20-3 = 17*.05 = 0.85, 17-0.85 = 16.15
38+16.15 = 54.15
$2.16 per 0.9 pounds of granola