She bought x number of $7 cd and used and $10 gift card so
7x-10=32
7x=42
x=6 cds
![x^ \frac{m}{n}= \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%20%5Cfrac%7Bm%7D%7Bn%7D%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D%20%20)
pemdas, so exponent first before multiply
4(x^1/2)=4x^2
this is different from
(4x)^1/2
so
![x^ \frac{1}{2}= \sqrt[2]{x^1}](https://tex.z-dn.net/?f=x%5E%20%5Cfrac%7B1%7D%7B2%7D%3D%20%5Csqrt%5B2%5D%7Bx%5E1%7D%20%20)
times that by 4
4√x
You can make it easier by replacing x^n with another variable, factoring, then putting x^n back in the end.
Using exponent and algebra rules, rewrite x^2n - 2x^n + 1 as
(x^n)^2 - (2 x x^2) + 1
Then, let x^n = m.
m^2 - 2m + 1
Now factor that: (m - 1)^2
And now put x^n back: (x^n - 1)^2
Answer:
x-- 3
Step-by-step explanation:
The solutions for ‘x’ are 2 and -5
<u>Step-by-step explanation:</u>
Given equation:

Since the base on both sides as ‘12’ are the same, we can write it as



Often, the value of x is easiest to solve by
by factoring a square factor, setting each factor to zero, and then isolating each factor. Whereas sometimes the equation is too awkward or doesn't matter at all, or you just don't feel like factoring.
<u>The Quadratic Formula:</u> For
, the values of x which are the solutions of the equation are given by:

Where, a = 1, b = 3 and c = -10



So, the solutions for ‘x’ are


The solutions for ‘x’ are 2 and -5