Question

PLease HELP HELP HELP HELP! (I SUCK AT MATH! NOT MY FAVE SUBJECT)

Answer 1

Answer:

y^2(y^4)=y^6

(2y)^2=4y^2

(y/4)^2=y^2/16

2(y^3)^3=2y^9

Step-by-step explanation:

Answer 2

Answers:

$y^2(y^4) = \boldsymbol{y^6}\\\\(2y)^2 = \boldsymbol{4y^2}\\\\(y/4)^2 = \boldsymbol{y^2/16}\\\\2(y^3)^3 = \boldsymbol{2y^9}$

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Explanation:

For the first expression, we use the rule that $a^b*a^c = a^{b+c}$. We add the exponents when multiplying exponentials of the same base. So, $y^2*y^4 = y^{2+4} = \boldsymbol{y^6}$

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In the second expression, we can expand out $(2y)^2$ into $(2y)*(2y)$. The coefficients 2 and 2 multiply to 2*2 = 4 as the final coefficient. The y terms multiply to $y*y = y^2$. So overall, that's how $(2y)^2 = \boldsymbol{4y^2}$

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We use the same idea for the third expression. We'll have two copies of y/4 multiplied together to end up with $\boldsymbol{y^2/16}$. We have $y^2$ in the numerator and 16 in the denominator.

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For the fourth expression, we can use the rule that $(a^b)^c = a^{b*c}$. In other words, if we raised $a^b$ to some other exponent c, then it's the same as writing $a^{b*c}$. We multiply the exponents.

The exponents in this case are 3 and 3 which multiply to 9. The 2 is then tacked up front to get $\boldsymbol{2y^9}$