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Mathematics 7 Online
OpenStudy (stryder246):

PLEASE HELP WILL EARN MEDAL When 27 times x squared times z all over negative 3 times x squared times z to the sixth power is completely simplified, the exponent on the variable z is _____.

OpenStudy (briellshaw):

if you say they will earn a medal some people with give you any answer using the main words so ask for an explanation behind the answer

OpenStudy (stryder246):

here is the equation

OpenStudy (stryder246):

and ok briellshaw

OpenStudy (stryder246):

Could you please help me Mehek14

Mehek (mehek14):

What is 27/-3 = ?

OpenStudy (stryder246):

-9

Mehek (mehek14):

Ok so -9 is the numerator

OpenStudy (stryder246):

yes

Mehek (mehek14):

\(\dfrac{x^2}{x^2}=?\)

OpenStudy (stryder246):

this is the part that i get stuck on but i think that it is x^4

OpenStudy (stryder246):

or just x with no exponent

Mehek (mehek14):

well when we divide, you subtract the exponents

OpenStudy (stryder246):

ok so it will be x with no exponent

Mehek (mehek14):

there would be no x

Mehek (mehek14):

Now \(\dfrac{z}{z^5}\)

OpenStudy (stryder246):

i thought that it is z/x^6

OpenStudy (stryder246):

z/z^6 my bad

OpenStudy (stryder246):

\[z/z^6\]

OpenStudy (stryder246):

are you still there

Mehek (mehek14):

Yes

OpenStudy (stryder246):

ok wouldn't it be \[^{z^-5}\]

Mehek (mehek14):

no

OpenStudy (stryder246):

i think that the answer is -5

Mehek (mehek14):

Nope

Mehek (mehek14):

\(z=z^1\\z^5\) Subtract the exponents

OpenStudy (stryder246):

how did you get z^5

Mehek (mehek14):

From the question you posted?

OpenStudy (stryder246):

it says z^6

Mehek (mehek14):

Oh my bad cx \(\dfrac{z}{z^6}\)

OpenStudy (stryder246):

its ok and 1-6=-5

Mehek (mehek14):

it should look like \(\dfrac{9}{z^5}\) Your answer wants the exponent of z which is `5`

OpenStudy (stryder246):

i thought that the exponent is -5

Mehek (mehek14):

positive exponent.

OpenStudy (stryder246):

ok so the answer is 5

OpenStudy (stryder246):

Ok thanks for using you time to help me solve my problem! Thanks Again!

Mehek (mehek14):

You're welcome :) and yes 5

OpenStudy (stryder246):

Ok Sweet!

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