Question

Help with Alg 2??

Answer 1

Vanessa and William are stuck simplifying radical expressions. Vanessa has to simplify\[\frac{ x\frac{ 4 }{ 3 } }{ x\frac{ 5 }{ 6 } } \] .

Answer 2

William has to simplify \[\sqrt[16]{x*x ^{3}*x ^{4}}\]

Answer 3

Using full sentences, describe how to fully simplify Vanessa and William's expressions. Describe if Vanessa and William started with equivalent expressions or if they started with expressions that are not equal.

Answer 4

@KamiBug @One098

Answer 5

The Quotient of Powers Property of Exponents says that when you divide two powers with the same base, you subtract the exponents.
The Product of Powers Property of Exponents says that to multiply two powers having the same base, add the exponents.
So for Vanessa's expression, subtract the exponents and for William's expression, add the exponents in the radical sign.
Also, remember \[\sqrt[a]{b^c} = b^\frac{ c }{ a }\]

Answer 6

the first expression:
\(\large \frac{x^{\frac{4}{3}}}{x^{\frac{5}{6}}}=x^{\frac{4}{3}-\frac{5}{6}}\)
exponent property

Answer 7

I'm using:
\(\Large \frac{a^n}{a^m}=a^{n-m}\) you should know this property

Answer 8

Is the Vanessa problem 1/2?

Answer 9

now it should be something with \(\large x^{?}\)
where the question mark you need to find what it is

Answer 10

Correct. :) Her simplified expression is x^ 1/2

Answer 11

Right. that's what I meant @KamiBug cx

Answer 12

oh yeah you are correct^_^
i didn't do the calculations

Answer 13

It's all good. What's the next part, with William.

Answer 14

use the radical property that @KamiBug showed you

Answer 15

to go from radical to exponents

Answer 16

you have \(\Large \sqrt[16]{x*x^3*x^4}=(x*x^3*x^4)^{\frac{1}{16}}\)

Answer 17

now use the power property and distribute 1/16

Answer 18

x^1/16*... I don't know?!

Answer 19

remember this \(\large (a*b^2)^3=a^3*(b^2)^3\)

Answer 20

you do the same thing

Answer 21

do you get it!?

Answer 22

I don't know. It's hard to understand.

Answer 23

you just distribute the power
here is how \((a*b)^m=a^m*b^m\)

Answer 24

if you have \((a*b*c*d)^m=a^m*b^m*c^m*d^m\) always works

Answer 25

let's pick a case similar to yours
\(\Huge(a^n*b^l)^{\frac{1}{r}}=(a^n)^{\frac{1}{r}}*(b^l)^{\frac{1}{r}}\)

Answer 26

good now?

Answer 27

OH!!! I got everything now!!

Answer 28

good^_^

Answer 29

But how does that relate to this problem?? Can you show me how to plug that into this??

Answer 30

simplified answer for vanessa