The first step for solving this equation is to determine the defined range.

, x ≠ 1
Remember that when the denominators of both fractions are the same,, you need to set the numerators equal. This will look like the following:

= 5
Take the root of both sides of the equation and remember to use both positive and negative roots.
x +/-
![\sqrt[4]{5}](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B5%7D)
Separate the solutions.
x =
![\sqrt[4]{5}](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B5%7D)
, x ≠ 1
x = -
Check if the solution is in the defined range.
x =
x = -
This means that the final solution to your question are the following:
x =
x = -
Let me know if you have any further questions.
:)
The area of a hexagon with a radius of 5 inches is 86.603.
6--3=9
15-6=9,
so
24+9=33,
33+9=42,
42+9=51
33,42,51
For this case we must indicate which of the equations shown can be solved using the quadratic formula.
By definition, the quadratic formula is applied to equations of the second degree, of the form:

Option A:

Rewriting we have:

This equation can be solved using the quadratic formula
Option B:

Rewriting we have:

It can not be solved with the quadratic formula.
Option C:

Rewriting we have:

This equation can be solved using the quadratic formula
Option D:

Rewriting we have:

It can not be solved with the quadratic formula.
Answer:
A and C
Answer:
PQ:80
QR:160
PR:120
Step-by-step explanation:
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×
=
×
∠
×

×
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=
×
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<em>hope it helps </em>
<em>have a great day!!</em>