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2 years ago

Can some one help me with this question its hard :( down below ill give brainliest

Mathematics

1 answer:

Alborosie2 years ago

6 0

Exponent law:
m^x * m^y = m^(x + y)
In other terms, when multiplying same bases, the exponents add
m^(2 + 2) = m^4

Solution: A. m^4

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3 years ago

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Step-by-step explanation:

Hi, to answer this question we have to write an equation:

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Feel free to ask for more if needed or if you did not understand something

4 0

3 years ago

Divide. (18x^3 + 12x^2 - 3x) ÷ 6x^2

nlexa [21]

$\bold{[ \ Answer \ ]}$

$\boxed{\bold{\frac{x^3\left(6x^2+4x-1\right)}{2}}}$

$\bold{[ \ Explanation \ ]}$

$\bold{Divide: \ \left(18x^3\:+\:12x^2\:-\:3x\right)\:\div \:6x^2}$

$\bold{-------------------}$

$\bold{Rewrite}$

$\bold{18x^3+12x^2-3x \ = \ x^2\frac{x\left(6x^2+4x-1\right)}{2}}$

$\bold{Rewrite}$

$\bold{x^2\frac{x\left(6x^2+4x-1\right)}{2}}$

$\bold{Multiply \ Fractions \ (a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c})}$

$\bold{\frac{x\left(6x^2+4x-1\right)x^2}{2}}$

$\bold{Rewrite}$

$\bold{x\left(6x^2+4x-1\right)x^2 \ = \ x^3\left(6x^2+4x-1\right)}$

$\bold{Simplify}$

$\bold{\frac{x^3\left(6x^2+4x-1\right)}{2}}$

$\boxed{\bold{[] \ Eclipsed \ []}}$

3 0

3 years ago

Read 2 more answers