- Simplify: \( 18-(-5)+2\cdot(-6) \)
- If \( \frac{3}{5}+\frac{1}{6}-\frac{1}{2} \) is simplified and reduced to the lowest terms, what is the denominator of the resulting fraction?
- \( (\frac{3}{2}\div\frac{3}{5})-(\frac{5}{6}\cdot\frac{1}{3}) =?\)
- Half of a class earned A's on a test, and of these student's \( \frac{3}{5} \) also have an A in the class. What percent of the class earned A's on the test and have A's in the class?
- A man buys \( 3.45 \) pounds of apples, \( 2.74 \) pounds of bananas, and \( \frac{11}{5} \) pounds of oranges. How many total pounds of fruit did he buy?
- Jessica earns \( $22.50 \) per hour, and earns \( 1.5 \) times that for any overtime hours she works. This week Jessica worked \( 40 \) regular hours and \( 15 \) overtime hours. How many dollars did she earn?
- Convert the product \( 3,500 \cdot 10,000 \) into scientific notation.
- Convert \( 4 < \sqrt{x} < 7 \) into an equivalent expression without the radical.
- What value of \( x \) solves the proportion \( \frac{10}{x}=\frac{5}{4} \) ?
- It takes a man \( 27 \) minutes to jog \( x \) miles. At the same speed the man jogs \( 7 \) miles in \( 63 \) minutes. Find \( x \).
- What is \( 25\% \) of \( 80\% \) of \( 200 \) ?
- A group of \( 12 \) sit down at a restaurant; \( 4 \) people order pizza, \( 5 \) order hamburgers, and \( 3 \) order hot dogs. What percentage of the group ordered pizza or hamburgers?
- The mean grade point average (GPA) in a class of \( 24 \) students is \( 3.10 \). By how much will the average GPA of the class increase if one student with a GPA of \( 3.60 \) joins the class?
- A data set consists of the numbers \( 8 \), \( 13 \), \( 9 \), and \( x \). If the average is \( 10 \), what is the median?
- If \( x=-2 \), what is the value of \( (x^2-1)(x-3) \) ?
- A person's maximum heart rate (\( MHR \)) is equal to \( 220 \) minus their age in years. Different people have different resting heart rates (\( RHR \)), which depend on many factors. Weekly exercise that elevates your heart rate to is recommended to maintain cardiovascular health. One suggested training heart rate (\( THR \)) is given by the formula: $$ THR = RHR + 0.75(MHR - RHR) $$ Based on this formula, at what \( THR \) should a \( 36 \) year old with a resting heart rate of \( 60 \) beats per minute exercise?
- Your expected outcome ( /( E /)) from a bet is represented by the equation: $$ E = p_w \cdot w - p_l \cdot l $$ Where \( p_w \) is the probability of winning, \( p_l \) is the probability of losing, \( w \) is the payout you receive for a win, and \( l \) is the amount you pay for a loss. You enter a bet where you must pay \( $50 \) if you lose, but you get \( $100 \) if you win. The probability of winning is \( 25\% \), and the probability of losing is \( 75\% \). What is your expected outcome from this bet?
- A rocket travelled for \( 2 \) hours at a speed of \( 1,200 \) miles per hour, and then for \( x \) hours at a speed of \( 900 \) miles per hour. If the average speed of the flight was \( 1,140 \) miles per hour, then what is \( x \) ?
- Expand the factored polynomial: \( (2a+b)(a-3b) \)
- Add and combine like terms: \( 3x^{2}y^{2} + x^{2}y - 4xy^{2} - 2xy + 2(x^{2}y^{2} - 3x^{2}y - xy^{2} + 4xy) \)
- Factor the polynomial: \( x^{2} - 14x + 49 \)
- If \( x-5=-3(1-x) \), then \( x= \) ?
- Factor the polynomial: \( x(x+3)+2 \)
- Solve for \( x \): \( \frac{5}{3} + \frac{1}{2} = x + \frac{5}{6} \)
- Assume \( a \), \( b \), \( c \) are non-zero. Simplify the expression: $$ \frac{(2a)^{3}b^{7}c^{5}}{(2a^{4}bc^{2})^{2}} $$
- For non-zero values of \( x \), \( y \), \( z \), simplify the expression: $$ \sqrt[3]{27x^6y^2z^3} $$
- For \( x>0 \) and \( y>0 \), rationalize the denominator of the expression: $$ \frac{4\sqrt{x}}{3\sqrt{x}+3\sqrt{y}} $$
- For \( x \neq 4 \), simplify the expression: $$ \frac{x^2-10x+24}{x-4} $$
- What is the slope of the line perpendicular to \( 2x-3y+4=0 \) ?
- What is the equation of the line passing through points \( (0,1) \) and \( (2,7) \) ?
Solutions
- Solution: $$
\begin{align*}
& 18-(-5)+2\cdot(-6)= \\[5pt]
& 18+5-12= \\[5pt]
& 11
\end{align*}
$$
- Solution: $$
\begin{align*}
& \frac{3}{5} + \frac{1}{6} - \frac{1}{2} = \\[5pt]
& \frac{3 \cdot 6}{5 \cdot 6} + \frac{1 \cdot 5}{6 \cdot 5} - \frac{1 \cdot 15}{2 \cdot 15} = \\[5pt]
& \frac{18}{30} + \frac{5}{30} - \frac{15}{30} = \\[5pt]
& \frac{8}{30}=\frac{4 \cdot 2}{15 \cdot 2}=\frac{4}{15} \\[5pt]
& \text{The denominator is 15}
\end{align*}
$$
- Solution: $$
\begin{align*}
& (\frac{3}{2} \div \frac{3}{5}) - (\frac{5}{6} \cdot \frac{1}{3}) = \\[5pt]
& (\frac{3}{2} \cdot \frac{5}{3}) - (\frac{5}{6} \cdot \frac{1}{3}) = \\[5pt]
& \frac{15}{6} - \frac{5}{18} = \frac{45}{18} - \frac{5}{18} = \\[5pt]
& \frac{40}{18} = \frac{20}{9} \\[5pt]
\end{align*}
$$
- \( \frac{1}{2} \cdot \frac{3}{5} = \frac{3}{10} = 30\% \)
- \( 3.45+2.74+\frac{11}{5} = 3.45+2.74+2.2 = 8.39 \text{ pounds} \)
- \( 40(22.50)+(15 \cdot 1.5)(22.50) = 62.5(22.50) = $1,406.25 \)
- \( 3,500 \cdot 10,000 = 35,000,000 = 3.5 \cdot 10^7 \)
- Squaring each expression gives the solution: \( 16 < x < 49 \)
- \( \frac{10}{x} = \frac{5}{4} \rightarrow 40 = 5x \rightarrow x = 8 \)
- From the text you can derive the proportion: $$ \frac{x}{27} = \frac{7}{63} \rightarrow x = (27)\frac{7}{63} = 3$$
- \( 0.25(0.80)(200)=40 \)
- \( \frac{4+5}{12} = \frac{9}{12} = 75\% \)
- Since the mean GPA of the initial \( 24 \) students is \( 3.10 \), the sum of all their GPA's must be: \( (24)(3.10) \). Adding another student with a GPA of \( 3.60 \) would make the class average: $$ \frac{24(3.10)+3.60}{24+1} = 3.12 $$ The question asked for the difference between the old and new GPA averages, which is: $$ 3.12 - 3.10 = 0.02 $$
- We must first calculate \( x \): $$
\begin{align*}
& \frac{8+9+13+x}{4} = 10 \rightarrow \\[5pt]
& 30 + x =40 \rightarrow \\[5pt]
& x = 10
\end{align*}
$$
Since the data set has an even number of points, the medican is the mean of the middle two points:
$$ (9+10)/2 = 9.5 $$
- Solution: $$
\begin{align*}
& ((-2)^2-1)((-2)-3)= \\[5pt]
& (4-1)(-5)= \\[5pt]
& -15
\end{align*}
$$
- First calculate: \( MHR=220-36=184 \), then substitute \( MHR \) and \( RHR \) into the equation to solve for \( THR \): $$ THR = 60 + 0.75(184-60) = 153 $$
- Plug all the values inot the equation to get: $$ E = (0.25)(100)-(0.75)(50)=-12.5 $$ Which means, on average, you would expect to lose \( $12.50 \) from this bet.
- Solution: $$
\begin{align*}
& \frac{2(1200)+x(900)}{2+x} = 1140 \rightarrow \\[5pt]
& 2400 + 900x = 2280 + 1140x \rightarrow \\[5pt]
& 120 = 240x \rightarrow \\[5pt]
& x = 0.5 \text{ hours}
\end{align*}
$$
- Solution: $$
\begin{align*}
& (2a+b)(a-3b) = \\[5pt]
& 2a^2-6ab+ab-3b^2 = \\[5pt]
& 2a^2-5ab-3b^2
\end{align*}
$$
- Solution: $$
\begin{align*}
& 3x^2y^2+x^2y-4xy^2-2xy+2(x^2y^2-3x^2y-xy^2+4xy) \rightarrow \\[5pt]
& 3x^2y^2+x^2y-4xy^2-2xy+(2x^2y^2-6x^2y-2xy^2+8xy) \rightarrow \\[5pt]
& (3+2)x^2y^2+(1-6)x^2y+(-4-2)xy^2+(-2+8)xy \rightarrow \\[5pt]
& 5x^2y^2-5x^2y-6xy^2+6xy
\end{align*}
$$
- Solution: $$
\begin{align*}
& x^2-14x+49 = \\[5pt]
& (x-7)(x-7) = \\[5pt]
& (x-7)^2
\end{align*}
$$
- Solution: $$
\begin{align*}
& x-5 = -3(1-x) \rightarrow \\[5pt]
& x-5 = 3x-3 \rightarrow \\[5pt]
& 2x = -2 \rightarrow \\[5pt]
& x = -1
\end{align*}
$$
- Solution: $$
\begin{align*}
& x(x+3)+2 = \\[5pt]
& x^2+3x+2 = \\[5pt]
& (x+1)(x+2)
\end{align*}
$$
- Solution: $$
\begin{align*}
& \frac{5}{3} + \frac{1}{2} = x + \frac{5}{6} \rightarrow \\[5pt]
& \frac{10}{6} + \frac{3}{6} = x + \frac{5}{6} \rightarrow \\[5pt]
& x = \frac{10+3-5}{6} \rightarrow \\[5pt]
& x = \frac{8}{6} = \frac{4}{3}
\end{align*}
$$
- Solution: $$
\begin{align*}
& \frac{(2a)^3b^7c^5}{(2a^4bc^2)^2}= \\[5pt]
& \frac{8a^3b^7c^5}{4a^8b^2c^4} = \\[5pt]
& \frac{2b^5c}{a^5}
\end{align*}
$$
- \( \sqrt[3]{27x^6y^2z^3} = 3x^2y^{\frac{2}{3}}z \)
- Solution: $$
\begin{align*}
& \frac{4\sqrt{x}}{3\sqrt{x}+3\sqrt{y}} \cdot \frac{3\sqrt{x}-3\sqrt{y}}{3\sqrt{x}-3\sqrt{y}} = \\[5pt]
& \frac{12x-12\sqrt{xy}}{9x-9y} = \\[5pt]
& (\frac{4}{3})(\frac{x-\sqrt{xy}}{x-y})
\end{align*}
$$
- Solution: $$
\begin{align*}
& \frac{x^2-10x+24}{x-4} = \\[5pt]
& \frac{(x-4)(x-6)}{x-4} = \\[5pt]
& x-6
\end{align*}
$$
- Perpendicular lines have slopes that are inverses with opposite signs. We first convert the given equation into slope intercept form: $$ 2x-3y+4=0 \rightarrow y=\frac{2}{3}x+\frac{4}{3} $$ From this, we see the slope of the perpendicular line will be: $$ -\frac{3}{2} $$
- We must first calculate the slope: $$ \frac{7-1}{2-0} = \frac{6}{2} = 3 $$ The first point, \( (0,1) \), is the y-intercept. Hence, in slope intercept form the equation for the line is: $$ y = 3x + 1 $$