Question

plz help ss below

Answer 1

@snowflake0531

Answer 2

\(\Large\dfrac {\frac{a}{b}}{\frac{c}{d}} = \dfrac{a}{b} \div \dfrac{c}{d}\)

\(\large = \dfrac{a}{b} \times \dfrac{d}{c}\)

Answer 3

so
(3r^2t-6rt^2/6r^2t^3)*(6r^3t^4/8t^3r-4t^2r^2)
what should i do next?

Answer 4

(6r^3*t^4)/(6r^2*t^3)
can be simplified to
6r*t?

Answer 5

but im a bit stuck on how to do the other part

Answer 6

\( \dfrac{3r^2t-6rt^2}{6r^2t^3} \times \dfrac{6r^3t^4}{8t^3r-4t^2r^2}\)

And rearranging it a bit

\( = \dfrac{6r^3t^4}{6r^2t^3} \times \dfrac{3r^2t-6rt^2}{8t^3r-4t^2r^2}\)

Answer 7

oh yea
so rt
then i would have to cancel out the top, can you help me with that?

Answer 8

That's what I'm here to help you do!

So it would just be 'rt'


\( \dfrac{\cancel{6}\cancel{r^3t^4}rt}{\cancel{6}\cancel{r^2t^3}} \times \dfrac{3r^2t-6rt^2}{8t^3r-4t^2r^2}\)


so we have

\(rt \times \dfrac{3r^2t-6rt^2}{8t^3r-4t^2r^2}\)

So let's simplify that fraction

Answer 9

we need to factor

\( 3r^2t - 6rt^2\)

and also

\( 8t^3r - 4t^2 r^2\)

Answer 10

ok so
i kinda know that if the base is the same u minus the exponeys
then it be
3/8*(r*t^-2)? for the first then
3/2*(r^-1*t^0)

Answer 11

so rt*(3/8*(r*t^-2)-3/2*(r^-1*1))??

Answer 12

I'm somewhat confused how you got all that

Answer 13

We want to factor the numerator and the denominator so that way we can see if there's anything in common to simplify it more

Answer 14

oh ok yea i prob got something wrong, how would i simplfy?

Answer 15

what can you factor out of \( 3r^2t−6rt^2\)

so there's 3 and 6
the GCF of that is 3

and then there's r^2 and r
what's the GCF of that?

and t and t^2

and so then we can find the GCF between 3r^2t and 6rt^2
and then factor that out

Answer 16

so
3*r*t?

Answer 17

then
r-2t?

Answer 18

Exactly! So now we have

\(rt \times \dfrac{3rt(r-2t)}{8t^3r-4t^2r^2}\)


What about \( 8t^3 r - 4t^2r^2\)
What is the GCF of those two terms?

Answer 19

4*t^2*r(2*t-r^2)

Answer 20

4t^2r is correct but check again what you get when you factor it out

what is (4t^2r^2) / (4t^2r)

Answer 21

4*t^2*r(2*t-r) my bad

Answer 22

Good good!

and what happens if you factor out a negative 1 from it as well?

hint: we would get the same thing in parenthesis that the numerator has

Answer 23

-4*t^2*r(2*t+r)??

Answer 24

almost but not quite, it would be

\(-4t^2r(-2t+r)\)

and then that could be re-written as \(-4t^2r(r-2t)\)

Do you see that now?

Answer 25

ohh yes
3rt/-4t^2r cus u can factor out the r-2t in both

Answer 26

3rt/-4t^2r*rt=3rt^2/-4t^2r??

Answer 27

You're on the right track but I think maybe because you're typing it all out, the letters and stuff get mashed up and it's hard to see so you made an error. I'll write it in \(\LaTeX\) so that way you can see it clearly


\(rt \times \dfrac{3rt\cancel{(r-2t)}}{-4t^2r\cancel{(r-2t)}}\)


\(=rt \times \dfrac{3rt}{-4t^2r}\)

now can you simplify it? we have rt in the numerator and t^2 * r in the denominator

Answer 28

so 3/-4t*rt=3rt/-4t

Answer 29

good and now we have

\(-\dfrac{3rt}{4t}\)

do you want to do something about that 't' in the numerator and denominator?

Answer 30

oh -3r/4

Answer 31

and you're all done!

Answer 32

thanks!

Answer 33

You're welcome!!