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Bingel [31]
3 years ago
13

Look at this expression.

Mathematics
1 answer:
k0ka [10]3 years ago
4 0
\bf \cfrac{14ab^3}{7a^{-2}b^{-1}}\\\\
-----------------------------\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\\ \quad \\
%  negative exponential denominator
a^{{ n}} \implies \cfrac{1}{a^{- n}}
\qquad \qquad 
\cfrac{1}{a^{- n}}\implies \cfrac{1}{\frac{1}{a^{ n}}}\implies a^{{ n}} \\\\
-----------------------------\\\\


\bf \cfrac{14}{7}\cdot \cfrac{a}{1}\cdot \cfrac{1}{a^{-2}}\cdot \cfrac{b^3}{1}\cdot \cfrac{1}{b^{-1}}\implies  2\cdot a\cdot a^2\cdot b^3\cdot b^1\implies 2a^{1+2}b^{3+1}
\\\\\\
2a^3b^4
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Help! I don't know how to do this! I will mark Brainliest!
gogolik [260]

Answer:

- 3y² - 7y + 6

Step-by-step explanation:

Given

(- 3y + 2)(y + 3)

Each term in the second factor is multiplied by each term in the first factor, that is

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= - 3y² - 9y + 2y + 6 ← collect like terms

= - 3y² - 7y + 6

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3 years ago
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