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Oliga [24]
3 years ago
9

2. Guelmy played 4 different video games in 2 hours. At this same rate, how many video games would he play in 5 hours?​

Mathematics
1 answer:
Inga [223]3 years ago
4 0

Answer:

10 video games. Since he is playing 2 video games every hour.

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Which expressions are equivalent to \dfrac{4^{-3}}{4^{-1}} 4 −1 4 −3 ​ start fraction, 4, start superscript, minus, 3, end super
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Answer:

\dfrac{4^{-3}}{4^{-1}} = \dfrac{4^{1}}{4^{3}}

\dfrac{4^{-3}}{4^{-1}} = \dfrac{1}{4^{2}}

Step-by-step explanation:

Given

\dfrac{4^{-3}}{4^{-1}}

Required

Choose equivalent expressions

Choosing the first answer:

\dfrac{4^{-3}}{4^{-1}}

Split expressions

4^{-3} * \frac{1}{4^{-1}}

Apply laws of indices: (a^{-b} = \frac{1}{a^b})

\frac{1}{4^3} * \frac{1}{4^{-1}}

Apply laws of indices: (a^{-b} = \frac{1}{a^b})

\frac{1}{4^3} * \frac{1}{1/4}

\frac{1}{4^3} * \frac{4^1}{1}

\frac{4^1}{4^3}

Hence:

\dfrac{4^{-3}}{4^{-1}} = \dfrac{4^{1}}{4^{3}}

Choosing the second:

\dfrac{4^{-3}}{4^{-1}}

Apply law of indices: (\frac{a^m}{a^n} = a^{m-n})

So,

\dfrac{4^{-3}}{4^{-1}} = 4^{-3-(-1)}

\dfrac{4^{-3}}{4^{-1}} = 4^{-3+1)}

\dfrac{4^{-3}}{4^{-1}} = 4^{-2}

Apply law of indices: (a^{-b} = \frac{1}{a^b})

So:

\dfrac{4^{-3}}{4^{-1}} = \dfrac{1}{4^{2}}

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