Question

When 27 times x^2 times z all over negative 3 times x^2 z to the fourth power is completely simplified, the exponent on the variable z is _____

Answer 1

Hold on gimme a sec

Answer 2

Ok

Answer 3

-9/z^5 I think

Answer 4

So what would z be?

Answer 5

Hmm not totally sure

Answer 6

I need Z lol

Answer 7

Hmm I know you can ask @texaschic101 to help

Answer 8

\[\large \frac{27x^2z}{-3x^2z^4}\]

We just want the exponent on variable 'z' so focus on that

\[\large \frac{z}{z^4} = z^{1 - 4} = z^{?}\]

Answer 9

-3

Answer 10

Right...now I'm not sure what they mean by completely simplified...

we can have \(\huge \frac{-9}{z^{3}}\) or we can have \(\large -9z^{-3}\)

Answer 11

Im so confused

Answer 12

\[\large \frac{27\cancel{x^2}z}{-3\cancel{x^2}z^4}\]

so all we really have is
\[\large \frac{27z}{-3z^4}\]

If we divide the 27/-3 we have -9...so now we have

\[\large -9 \times \frac{z}{z^4}\]

Answer 13

So it would be 4?

Answer 14

Now we just need to remember that

\[\large \frac{n^a}{n^b} = n^{a - b}\]

so

\[\large \frac{z}{z^4} = z^{1 - 4} = z^{-3}\]

Answer 15

So altogether we have \(\large -9z^{-3}\)

The exponent that goes with the 'z' would be -3

\[\large -9z^{\color \red{-3}}\]

Answer 16

Yes but I put -3 last time and it was wrong

Answer 17

Well then maybe they want to keep this as

\[\large \frac{\cancel{z^1}}{z^{\cancel{4} \rightarrow 3}} = \frac{1}{z^3}\]

Answer 18

So it would be positive 3

Answer 19

Yeah if they don't want -3 then they didnt want that final simplification step...so yeah 3 would be correct then

Answer 20

Ok thanks

Answer 21

Yes it was positive 3

Answer 22

Alright that's what they wanted ....great! and anytime :)