Question

6a exam - exponents #1
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question 1 of 9
multiply.
$9v(-3v^8)$
simplify your answer as much as possible.

question 2 of 9
simplify.
$(2x^3y^2)^3(x^3yz^2)$

question 3 of 9
simplify.
$\frac{xy^6}{x^3y^5}$

question 4 of 9
evaluate the expressions.
$-3\left(\frac{2}{3}\
ight)^0 =$
$-(5)^0 =$

Explanation:

Question 1

Step1: Multiply coefficients

$9 \times (-3) = -27$

Step2: Multiply variable terms

$v \times v^8 = v^{1+8} = v^9$

Step3: Combine results

$-27 \times v^9$

Question 2

Step1: Expand the power term

$(2x^3y^2)^3 = 2^3 \times (x^3)^3 \times (y^2)^3 = 8x^9y^6$

Step2: Multiply by the second term

$8x^9y^6 \times x^3yz^2 = 8 \times x^{9+3} \times y^{6+1} \times z^2$

Step3: Simplify exponents

$8x^{12}y^7z^2$

Question 3

Step1: Simplify $x$ terms

$\frac{x}{x^3} = x^{1-3} = x^{-2} = \frac{1}{x^2}$

Step2: Simplify $y$ terms

$\frac{y^6}{y^5} = y^{6-5} = y^1 = y$

Step3: Combine results

$\frac{1}{x^2} \times y = \frac{y}{x^2}$

Question 4

Step1: Evaluate first expression

Any non-zero number to 0 power is 1: $-3 \times 1 = -3$

Step2: Evaluate second expression

Any non-zero number to 0 power is 1: $-1 \times 1 = -1$

Answer:

1. $\boldsymbol{-27v^9}$
2. $\boldsymbol{8x^{12}y^7z^2}$
3. $\boldsymbol{\frac{y}{x^2}}$
4. $\boldsymbol{-3}$; $\boldsymbol{-1}$