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Quiz 2
1. Simplify $\left( \dfrac{x}{3y} \right)^2 \left( \dfrac{2y}{x} \right)^3 x^{-2}$.
Solution: Calculate using the rules of exponents:
$$\begin{array}{ll}
\left( \dfrac{x}{3y} \right)^2 \left( \dfrac{2y}{x} \right)^3 x^{-2} &= \left( \dfrac{x^2}{9y^2} \right) \left( \dfrac{8y^3}{x^3} \right) \dfrac{1}{x^2} \\
&=\dfrac{8}{9} x^{2-3-2} y^{3-2} \\
&=\dfrac{8}{9} x^{-3} y^{1} \\
&= \dfrac{8y}{9x^3}.
\end{array}$$
2. What is...
a.) $\sqrt{25}$
b.) $\sqrt[3]{27}$
c.) $\sqrt[5]{-32}$
Solution: For a.), the answer is $5$. For b.), the answer is $3$. For c.), the answer is $-2$.
3. Collect like terms:
$$5 \{ 4x^3 + 2[3x-(2-x)] \}.$$
Solution: Compute
$$\begin{array}{ll}
5\{4x^3+2[3x-(2-x)] \} &= 5\{ 4x^3 + 6x -2(2-x)\} \\
&= 5\{4x^3 +6x -4 + 2x \} \\
&= 20x^3 + 30x -20 + 10x \\
&=20x^3 +40x-20.
\end{array}$$
4. Write as an algebraic expression and simplify: "the total amount of money (in cents) of $x$ nickels, $2x+3$ dimes, and $x+2$ quarters"
Solution: Since nickels are worth $5$ cents, dimes are worth $10$ cents, and quarters are worth $25$ cents, we write
$$5x+10(2x+3)+25(x+2).$$
Now simplify this expression:
$$\begin{array}{ll}
5x+10(2x+3)+25(x+2) &= 5x + 20x+30+25x+50 \\
&=50x+80
\end{array}$$
5. Multiply $(x+3)(x-1)$.
Solution: Using the distributive law $a(b+c)=ab+ac$ we get
$$\begin{array}{ll}
(x+3)(x-1) &= (x+3)x + (x+3)(-1) \\
&= x^2+3x -x-3\\
&= x^2+2x-3.
\end{array}$$
