Answer:
5 to the power of 3 is 125, so since the denominator is 3 you need a numerator that will simplify to 3.
9 divided by 3 equals 3, so your numerator will be 9.
Hope it helps!
Step-by-step explanation:
<span><span><span><span>I think is this 9<span>x8</span></span>+<span>x7</span></span>−x</span>+<span>13</span></span>
Answer:
62398
Step-by-step explanation:
Hope this helps :)
Answer: well, the answer can really be any two numbers, as long as y is one more than x. EXAMPLE: 16 = 15 + 1, 7 = 6 + 1, -54 = -55 + 1, etc
Step-by-step explanation:
The root is
-sqr root of 5.
First, we put these roots in the forn of
![(x - a)](https://tex.z-dn.net/?f=%28x%20-%20a%29)
where a is the root
So we have
![(x - ( - 2))(x - \sqrt{5} )(x - \frac{10}{3} )](https://tex.z-dn.net/?f=%28x%20-%20%28%20-%202%29%29%28x%20-%20%20%5Csqrt%7B5%7D%20%29%28x%20-%20%20%5Cfrac%7B10%7D%7B3%7D%20%29)
![(x + 2)(x - \sqrt{5} )(3x - 10)](https://tex.z-dn.net/?f=%28x%20%2B%202%29%28x%20-%20%20%5Csqrt%7B5%7D%20%29%283x%20-%2010%29)
![(3 {x}^{2} - 4x - 20)(x - \sqrt{5} )](https://tex.z-dn.net/?f=%283%20%7Bx%7D%5E%7B2%7D%20%20-%204x%20-%2020%29%28x%20-%20%20%5Csqrt%7B5%7D%20%29)
To get rid of that square root, let have another root that js the conjugate posive root of 5.
![(3 {x}^{2} - 4x - 20)(x - \sqrt{5} )(x + \sqrt{5} )](https://tex.z-dn.net/?f=%283%20%7Bx%7D%5E%7B2%7D%20%20-%204x%20-%2020%29%28x%20-%20%20%5Csqrt%7B5%7D%20%29%28x%20%2B%20%20%5Csqrt%7B5%7D%20%29)
![(3 {x}^{2} - 4x - 20)(x {}^{2} + 5)](https://tex.z-dn.net/?f=%283%20%7Bx%7D%5E%7B2%7D%20%20-%204x%20-%2020%29%28x%20%7B%7D%5E%7B2%7D%20%20%2B%205%29)
Which will gives us a rational coeffeicent of degree 4.
Why we didn't do
![(x - \sqrt{5} )](https://tex.z-dn.net/?f=%28x%20-%20%20%5Csqrt%7B5%7D%20%29)
?
Because
![(x - \sqrt{5} ) {}^{2} = {x}^{2} - 2 \sqrt{5} + 5](https://tex.z-dn.net/?f=%28x%20-%20%20%5Csqrt%7B5%7D%20%29%20%7B%7D%5E%7B2%7D%20%20%3D%20%20%7Bx%7D%5E%7B2%7D%20%20-%202%20%5Csqrt%7B5%7D%20%20%2B%205)
If we foiled out we will still have a irrational coeffceint.