Answer:
2 RootIndex 4 StartRoot 4 EndRoot
Step-by-step explanation:
we have
Decompose the number 64 in prime factors
substitute
![64^{\frac{1}{4}}=(2^{4}2^{2})^{\frac{1}{4}}=2^{\frac{4}{4}}2^{\frac{2}{4}}=2\sqrt[4]{4}](https://tex.z-dn.net/?f=64%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%3D%282%5E%7B4%7D2%5E%7B2%7D%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%3D2%5E%7B%5Cfrac%7B4%7D%7B4%7D%7D2%5E%7B%5Cfrac%7B2%7D%7B4%7D%7D%3D2%5Csqrt%5B4%5D%7B4%7D)
Answer: D) No; If x is 5, the expression on the left simplifies to 8, making the inequality false.
In other words, if x = 5, then x+3 becomes 5+3 which ultimately becomes 8, but this is not greater than 8 on the right side.
Your steps could look like this
x+3 > 8
5+3 > 8 ... replace x with 5
8 > 8 ... this is a false inequality
All I can say to help you with this, is to rewrite the division as a fraction,
x^3y / y^4x
Sorry, if it's not much help.
use long division or synthetic division
I will use long
y-3 times y=y^3-3y
subtract that from y^2-y-11
we are left with
2y-11
multiply y-3 by 2 and get 2y-6
subtract that from 2y-11
get 5 as a remainder
answer is y+2+5/(y-3)
