Question

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Answer 1

Ok so
\(\LARGE(\frac{12xy^32^6}{4}x^5)(y)z^{12}=?\)

Answer 2

@jimthompson5910?

Answer 3

the fist one

Answer 4

first

Answer 5

We have \(\Large \frac{12xy^3z^6}{4x^5yz^{12}}\)

Let's focus on the red terms \(\Large \frac{12{\color{red}{x}}y^3z^6}{4{\color{red}{x^5}}yz^{12}}\)

If we pull out those red terms, then,
\(\large \frac{x}{x^5} = \frac{x}{x*x^4} = \frac{\cancel{x}}{\cancel{x}*x^4}= \frac{1}{x^4}\)

Do the same thing for the y terms, and the z terms as well. So you'll focus on each variable one at a time.

Answer 6

Also, you'll simplify the leading coefficients by saying \(\large \frac{12}{4} = \frac{4*3}{4} = \frac{\cancel{4}*3}{\cancel{4}} = \frac{3}{1} = 3\)

Answer 7

so the 3erd one

Answer 8

Choice C is correct

We could have the steps look like this
\(\Large \frac{12xy^3z^6}{4x^5yz^{12}}\)

\(\Large \frac{12}{4}*\frac{x}{x^5}*\frac{y^3}{y}*\frac{z^6}{z^{12}}\)

\(\large 3*x^{1-5}*y^{3-1}*z^{6-12}\)

\(\large 3*x^{-4}*y^{2}*z^{-6}\)

\(\Large \frac{3y^2}{x^4z^6}\)

Answer 9

In the third step, I used the rule \(\Large \frac{a^b}{a^c} =a^{b-c}\)