Vanessa and William are stuck simplifying radical expressions. Vanessa has to simplify the quantity of x to the four thirds power, over x to the five sixths power. William has to simplify the sixteenth root of the quantity of x times x to the third times x to the fourth. 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. I just really need help putting stuff into sentences like explaining it.
@joyraheb ??
It'll be really helpful if you either drew or put your problem in equation mode. It's hard to read.
vanessas problem
Williams problem
wait! I sent vannessas twice
that is Williams.. does that help?
@AriPotta
@Sadworld
but not sure if that can even be more simplified..
how would I explain how to do that in sentences to them
@doulikepiecauseidont
Give me a few minutes.
Ok actually the first one is a lot easier than i thought
because they are like terms you can just subtract the fractions to get
\[x^\frac{ 4 }{ 3 }-x^\frac{ 5 }{ 6 }=x^\frac{ 1 }{ 2 }=\sqrt{x}\]
I think they're two separate expression since the question isn't to find the value of x.
ooh and the 2nd one would be \[^{16}\sqrt{x^8}=(x^\frac{ 1 }{ 16 })^8=\sqrt{x}\]
so they are the same equations...
they ask if they are the same, which they are.
I second @dtan5457 I got the same thing.
@Leahclaire
have to look up radical rules for this, can't remember anything..
im back
\[x ^{m}/x ^{n}=x ^{m-n}\]\[\sqrt[n]{x ^{m}} = x ^{\frac{ m }{ n }}\]
\[x ^{\frac{ 1 }{ 2 }}=\sqrt{x}\]
Whoops, had to go. you finished this? @Leahclaire
I guessed.. probably failed.
How?? I pretty much gave you the answers...which im not suppose to..
Here's a draft: "Vanessa and Williams will follow the laws of exponents (I don't know the names)..." so on and so on.
Oh. If your suppose to write an explanation. Easy, look at my answers, look at @Sadworld and the rules he supplied you with. It's not hard once you have both.
"Vanessa will subtract the exponents of both variables, etc. etc."
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