Question

I don't need the answer, I just need someone to explain this to me

Answer 1

What would you like to discuss? Have you been able to get started on the Math (?) problem in question?

Answer 2

And this...

Answer 3

Reduce that 42/12.

Did you know that you can move that x^(-6) in the numerator to the denominator, writing it as x^6 in the denominator?

Answer 4

\[x ^{-a}=\frac{ 1 }{ x^a }\]

Answer 5

If so, please go through these 2 steps now. Your result?

Answer 6

Regarding reducing 42/12: What is the largest number by which you can divide both 42 and 12 without getting a remainder?

Answer 7

I'm sorry to see that you've just posted a question and logged off. Please, stick with your question for at least five minutes after posting it.

Answer 8

I got 7/2

Answer 9

@mathmale sorry, it's my first time using the app

Answer 10

(7/2) is fine. Next, you have x^(-8) in the numerator. How do you manipulate that so that you end up with a positive exponent? See our discussion, above.

Answer 11

Did you know that you can move that x^(-6) in the numerator to the denominator, writing it as x^6 in the denominator?

Answer 12

\[x ^{-a}=\frac{ 1 }{ x^a }\]

Answer 13

I add the two exponents together. Im then left with 7xy^1

Answer 14

Hold the (7/2).
Focus on:\[\frac{ x ^{-8} }{ x^5 }\]

Answer 15

Last, focus on simplifying\[\frac{ y^9 }{ y ^{12} }\]

Answer 16

"I add the two exponents together. Im then left with 7xy^1" ... but what happened to the (7/2)? Can't drop that '2.'

Looking at the two powers of x:\[\frac{ x ^{-8} }{ x^5 }=\frac{ 1 }{ x^5 }\frac{ x ^{-8} }{ }=?\]

Answer 17

Please review our earlier discussion to learn what to do with that x^(-8).

Answer 18

it would be 1/x^-8

Answer 19

No, it would be \[\frac{ 1 }{ x^8 }\]

You want to convert that exponent neg 8 to positve exp. 8.

Once again:\[\frac{ x ^{-8} }{ 1 }=\frac{ 1 }{ x^8 }\]

Answer 20

Notice how you now have +8 as the exponent of the variable x?

Answer 21

oh, oh. ok

Answer 22

So now you have\[\frac{ 7 }{ 2 }\frac{ 1 }{ x^5 x^8 }\frac{ y^9 }{ y ^{12} }\]

Answer 23

is it possible to cancel out the Ys?

Answer 24

Can you simplify this? what is the rule of exponents that applies to \[x^5 x^8?\]

Answer 25

Can't cancel out the y's. No.

Must use the rule of exponents\[\frac{ y^a }{ y^b }=y ^{a-b}\]

Answer 26

so for the Xs it would be 1/x^13

Answer 27

You would end up with x^13 in the DENOMINATOR.

Answer 28

\[\frac{ 7 }{ 2 }\frac{ 1 }{ x^{13} }\frac{ y^9 }{ y ^{12} }\]

Answer 29

Would I also have y^-3?

Answer 30

Please reduce the last factor (the one involving y).

Answer 31

Yes, y^(-3) is correct, BUT you want to write this with a positive, not a negative exponent. So, what are you going to do next?

Answer 32

y^3 and also put it in the numerator

Answer 33

Why put it into the numerator? You have \[y ^{-3}\] and must rewrite it as \[\frac{ 1 }{ y^? }\]

Answer 34

Example:\[\frac{ x ^{-8} }{ 1 }=\frac{ 1 }{ x^8 }\]

Answer 35

so it's \[\frac{ 1 }{ y^3 }\]

Answer 36

Yes. So, what is your final solution to this problem?

Answer 37

so the correct solution to this propblem is C.)

Answer 38

Yes, that's great. Thank youf or sticking with me throughout this discussion.

Answer 39

Thank YOU for helping me with this and explaining it. :)

Answer 40

;)