Question

More rational expression help

Answer 1

oh GOSH!, lol bring it on dtan!

Answer 2

Using only positive exponents, rewrite and simplify
\[\frac{ 6p^{-2}q^4 }{ 2^{-3}p^5q^{-1} }\]

Answer 3

if anyone has a lot of spare time can they walk me through this..

Answer 4

Oh, change every negative exponent to a positive one first.

Answer 5

yes numerator i had 1/6p^2

Answer 6

then 2^-3 i can move up to=8 right?

Answer 7

same with q?

Answer 8

Ok, let's start with only the numerators first.

Answer 9

sure

Answer 10

\[\frac{ 6p^{-2}q^4 }{ 2^{-3}p^5q^{-1} }\]\[6p^{-2} \implies \frac{1}{p^2}\]Therefore \(6p^{-2}q^4 \implies \dfrac{6q^4}{p^2}\)

Answer 11

so basically 6p^-2=6 and 1/p^2

Answer 12

All the positive variables stay on top, ad all the negative ones get pushed to the denominator.

Answer 13

i see.

Answer 14

ooh. now i can move the 2^-3 to the top

Answer 15

and q^-1

Answer 16

just multiply p^2 and p^5?

Answer 17

to get
\[\frac{48q^5 }{ p^7 }\]

Answer 18

correct??

Answer 19

Let's check: \[2^{-3}p^5q^{-1} = \frac{8q}{p^5}\]

Answer 20

\[\frac{ 6p^{-2}q^4 }{ 2^{-3}p^5q^{-1} } = \frac{6q^4\cdot 8 \cdot q}{p^5 \cdot p^2 } = \frac{ 48q^5}{p^7}~ \checkmark\]

Answer 21

Good job.

Answer 22

i think my main problem derived from

thinking \[6p^{-2}=6\frac{ 1 }{ p^2 }\]

Answer 23

that there will be 2 deniminators

Answer 24

so it becomes a complex fraction

Answer 25

Yeah that's the same thing as writing \[\frac{6}{p^2}\]

Answer 26

instead of moving the p^2 to just the regular denominator, i made a new one above the regular denominator

Answer 27

if that makes sense..

Answer 28

\[6\cdot \frac{1}{p^2} = \frac{6}{1}\cdot \frac{1}{p^2} = \frac{6}{p^2}\]q

Answer 29

No, it doesn't really make sense.

Answer 30

let me try to write in the equations

Answer 31

\[\frac{ \frac{1 }{ p^2 } }{ 2^{-3}p^5q^{-1} }\]

Answer 32

that's what i did at first..

Answer 33

O_o I see

Answer 34

instead of putting it with the rest of the denominators..?

Answer 35

im assuming that's NOT the way to do it lol

Answer 36

Oh, I think you're making it a little bit more difficult than it needs to be, but that's just my opinion.

Answer 37

For problems like these i just switch the negative exponents with the positive ones.

i'll be like "oh, \(p^{-2}\) s a negative exponent, let me just drop it in the denominator"

Or like " \(2^{-3}\) is a negative exponent, and is in the denominator,let me just bring it into the numerator"

Answer 38

I'm lagging really really bad right now, sorry for the delay.

Answer 39

it's fine. these questions are starting to become clearer and clearer thanks to people like you.

Answer 40

Aww thank you! :')