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3 years ago

7

Solve the following NO BOTS OR PEOPLE WANTING POINTS

1 answer:

elena-s [515]3 years ago

8 0

Answer:

-1

Step-by-step explanation:

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1/2 would not be a useful benchmark to compare these 2 fractions because both of them are greater than half. The benchmark fraction 1/2 is most useful when one fraction is less than half and the other more than 1/2. 5/8 is more than 4/8(1/2), and 9/10 is more than 5/10(1/2).

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ASAP someone please simply two exponents

zhannawk [14.2K]

Answer:

1.) $m^{15}$

2.) $=\frac{1}{y^{15}}$

Give me a comment if you want the explanation.

1.) $\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$\left(m^3\right)^5=m^{3\cdot \:5}$

$\mathrm{Multiply\:the\:numbers:}\:3\cdot \:5=15$

$=m^{15}$

2.) $\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$\left(y^{-3}\right)^5=y^{-3\cdot \:5}$

$=y^{-3\cdot \:5}$

$=y^{-15}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}$

$=\frac{1}{y^{15}}$

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3 years ago

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The answer is A.

Hope I helped.

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