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Mathematics 18 Online
OpenStudy (anonymous):

(30x^3y^-5 over 150x^-4y^2)^2 raise the quantity in the parentheses to indicated exponent, then simplify--use positive exponents

OpenStudy (anonymous):

cud u write it as an equation. its not clear

OpenStudy (anonymous):

((30x^3y^-5)/(150x^-4y^2))^2

Parth (parthkohli):

Hey, Alice! Welcome to Openstudy! Is this the problem? \( \color{Black}{\Rightarrow \Large \left ({30x^3y^{-5} \over 150x^{-4}y^{2}} \right )^2 }\)

OpenStudy (anonymous):

ohh okk

OpenStudy (anonymous):

yes

Parth (parthkohli):

Okay. So you have to compute these first: \( \color{Black}{\Rightarrow \Large {30 \over 150}}\) \( \color{Black}{\Rightarrow \Large {x^3 \over x^{-4}}}\) \( \color{Black}{\Rightarrow \Large {y^{-5} \over y^2}}\)

OpenStudy (anonymous):

x^a / x^b = x ^ (a-b)

Parth (parthkohli):

Do you know the properties of exponents? The ones you gotta use here are: \( \color{Black}{\Rightarrow x^a \div x^b = x^{a - b} }\) \( \color{Black}{\Rightarrow (x^a)^b = x^{a\times b} }\)

Parth (parthkohli):

For example: \( \color{Black}{\Rightarrow \Large {x^3 \over x^{-4} }\normalsize = x^{3 - (-4)} = x^7 }\)

Parth (parthkohli):

Did you get it? @Ladyalice25

OpenStudy (anonymous):

so the equation so far will look like x^7/5y^7 is that correct

Parth (parthkohli):

No!

Parth (parthkohli):

\( \color{Black}{\Rightarrow \Large {30 \over 150} = {3 \over 15} = {1 \over 5} = \normalsize 0.2}\) \( \color{Black}{\Rightarrow \Large x^3 \div x^{-4} = x^7}\) \( \color{Black}{\Rightarrow \Large {y^{-5} \div y^2} = y^{-5 - 2}=y^{-7}}\)

Parth (parthkohli):

Combine 'em all \( \color{Black}{\Rightarrow\left ( 0.2x^7y^{-7} \right) ^2 }\)

Parth (parthkohli):

The final answer will then look like \( \color{Black}{\Rightarrow 0.2^2 \times x^{7 \times 2} \times y^{-7 \times 2} }\) \( \color{Black}{\Rightarrow 0.04 \times x^{14} \times y^{-14} }\)

OpenStudy (anonymous):

k thank you know i understand it

Parth (parthkohli):

You're welcome!

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