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tatuchka [14]
3 years ago
12

PLEASE HELPP FASSTTTTTT

Mathematics
1 answer:
lisabon 2012 [21]3 years ago
3 0

Answer:

sorry, i dont um, im setting, the acc, up and ,20

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%283a%5E2b%29%5E3%7D%7B9%28ab%29%5E4%7D" id="TexFormula1" title="\frac{(3a^2b)^3}{9(
VladimirAG [237]

You first multiply the outside exponents into the numbers in the parentheses.

When you have an exponent being multiplied directly to another exponent, you multiply the exponents together.

For example(because I am a bad explainer):

(x^{2} )^4= x^{2(4)} = x^8

(x^4)^3 = x^{4(3)} = x^{12}


When you divide an exponent by an exponent, you subtract the exponents

For example:

\frac{x^4}{x^1} =x^{4-1}=x^3


When you have a negative exponent, you move it to the other side of the fraction to make the exponent positive

For example:

x^{-3}=\frac{1}{x^3}

\frac{1}{y^{-5}}=\frac{y^5}{1}=y^5



\frac{(3a^2b)^3}{9(ab)^4}

You can think of it like this if you want:

\frac{(3^1a^2b^1)^3}{9(a^1b^1)^4}  Now multiply the outside exponents into the exponents in the parentheses

\frac{3^3a^6b^3}{9(a^4b^4)} =\frac{27a^6b^3}{9(a^4b^4)} Divide 27 and 9

\frac{3a^6b^3}{a^4b^4} =(3)(a^{6-4})(b^{3-4})=(3)(a^2)(b^{-1})=(3)(a^2)(\frac{1}{b^1})=\frac{3a^2}{b}



Your answer is \frac{3a^2}{b}



5 0
3 years ago
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