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Murljashka [212]
3 years ago
9

Determine whether the relation represents a function. Explain your reasoning. Your answer AY

Mathematics
1 answer:
tigry1 [53]3 years ago
6 0

A relation is a function if each element of the domain is paired with exactly one element of the range. ... If given a table, or a set of ordered pairs, you can look to see if any value of the domain has more than one corresponding value in the range.

tbh google

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Which expression is equivalent to StartFraction (3 m Superscript negative 2 Baseline n) Superscript negative 3 Baseline Over 6 m
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Option a: \frac{m^{5} }{162n} is the equivalent expression.

Explanation:

The expression is \frac{(3m^{-2} n)^{-3}}{6mn^{-2} } where m\neq 0, n\neq 0

Let us simplify the expression, to determine which expression is equivalent from the four options.

Multiplying the powers, we get,

\frac{3^{-3}m^{6} n^{-3}}{6mn^{-2} }

Cancelling the like terms, we have,

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This equation can also be written as,

\frac{m^{5}}{3^{3}6 n^{1} }

Multiplying the terms in denominator, we have,

\frac{m^{5} }{162n}

Thus, the expression which is equivalent to \frac{(3m^{-2} n)^{-3}}{6mn^{-2} } is \frac{m^{5} }{162n}

Hence, Option a is the correct answer.

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Step-by-step explanation:

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Complex Numbers
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