Answer:
i dont know.
Step-by-step explanation:
pls make it brainliest
<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>
<u><em>Step-by-step explanation:</em></u>
- Since the triangle is right use the tangent ratio to solve for x
- Multiply both sides by
x × 
( divide both sides by 
)
x = ≈ ( nearest tenth )
286 is the sum of the series below. You can plug this in or use long addition.
Either that or you make an average and that's your answer.
Answer:
Step-by-step explanation:
+
5
1
6
=
−
1
n+\frac{5}{16}=-1
n+165=−1
Solve
1
Subtract
5
1
6
\frac{5}{16}
165
from both sides of the equation
+
5
1
6
=
−
1
n+\frac{5}{16}=-1
n+165=−1
+
5
1
6
−
5
1
6
=
−
1
−
5
1
6
n+\frac{5}{16}{\color{#c92786}{-\frac{5}{16}}}=-1{\color{#c92786}{-\frac{5}{16}}}
n+165−165=−1−165
2
Simplify
Solution
=
−
2
1
