If x to the 12th power is divided by x to the 9th power:
Since they share same bases, and are being divided, it would be the same as x^{12 - 9), or x^{3}
Now, simply find the cube root of 125, which is 5.
Therefore, the number (x) must be equal to 5.
<em>Hope this helps! :)</em>
Area = b*h and in some cases * 2 so you need to do this, 64*52 + 40*28= 4448 yd2
Answer:
![=-\left(x-1\right)\left(x^4-6x^3-6x^2-14x-5\right)](https://tex.z-dn.net/?f=%3D-%5Cleft%28x-1%5Cright%29%5Cleft%28x%5E4-6x%5E3-6x%5E2-14x-5%5Cright%29)
Step-by-step explanation:
Factor out comon Term -1
![=-\left(x^5-7x^4-8x^2+9x+5\right)](https://tex.z-dn.net/?f=%3D-%5Cleft%28x%5E5-7x%5E4-8x%5E2%2B9x%2B5%5Cright%29)
Factor ![x^5-7x^4-8x^2+9x+5:\quad \left(x-1\right)\left(x^4-6x^3-6x^2-14x-5\right)](https://tex.z-dn.net/?f=x%5E5-7x%5E4-8x%5E2%2B9x%2B5%3A%5Cquad%20%5Cleft%28x-1%5Cright%29%5Cleft%28x%5E4-6x%5E3-6x%5E2-14x-5%5Cright%29)
![x^5-7x^4-8x^2+9x+5](https://tex.z-dn.net/?f=x%5E5-7x%5E4-8x%5E2%2B9x%2B5)
Use the rational root theorem
![a_0=5,\:\quad a_n=1](https://tex.z-dn.net/?f=a_0%3D5%2C%5C%3A%5Cquad%20a_n%3D1)
The dividers of
: 1, 5, The dividers of
: 1
Therefore, check the following rational numbers: ±![\frac{1,\:5}{1}](https://tex.z-dn.net/?f=%5Cfrac%7B1%2C%5C%3A5%7D%7B1%7D)
is a root of the expression, so factor out ![x-1](https://tex.z-dn.net/?f=x-1)
![=\left(x-1\right)\frac{x^5-7x^4-8x^2+9x+5}{x-1}](https://tex.z-dn.net/?f=%3D%5Cleft%28x-1%5Cright%29%5Cfrac%7Bx%5E5-7x%5E4-8x%5E2%2B9x%2B5%7D%7Bx-1%7D)
![\frac{x^5-7x^4-8x^2+9x+5}{x-1}=x^4-6x^3-6x^2-14x-5](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E5-7x%5E4-8x%5E2%2B9x%2B5%7D%7Bx-1%7D%3Dx%5E4-6x%5E3-6x%5E2-14x-5)
![=x^4-6x^3-6x^2-14x-5](https://tex.z-dn.net/?f=%3Dx%5E4-6x%5E3-6x%5E2-14x-5)
![=-\left(x-1\right)\left(x^4-6x^3-6x^2-14x-5\right)](https://tex.z-dn.net/?f=%3D-%5Cleft%28x-1%5Cright%29%5Cleft%28x%5E4-6x%5E3-6x%5E2-14x-5%5Cright%29)
It depends, because sometimes they are difficult and sometimes not ...
Have a good Saturday...