Answers to Pretests

Solutions 1.6 Answers to Pretests

1.2 Algebra Pretest

1.2.1. Working with Real Numbers.

Answer.

$\displaystyle \frac{3}{2}-\frac{2}{3} = \frac{5}{6}$
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$\displaystyle \frac{3}{2}\cdot\frac{2}{3} = 1$
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$\displaystyle |3−6|+6|−3|=21$
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$\displaystyle \frac{3−6}{-3\times 6} = \frac{1}{6}$
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1.2.2. Working with Exponents.

Answer.

$\displaystyle \left(-\sqrt{2}\right)^2-\left(\sqrt{2}\right)^4+\left(\sqrt{2}\right)^0+\sqrt{\left(-2\right)^4} = 3$
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$\displaystyle \sqrt{(-3)^2+(-4)^2} = 5$
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$\displaystyle \left(\sqrt[3]{-8}\right)^{-1} = -\frac{1}{2}$
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$\displaystyle \left(\frac{4}{9}\right)^{3/2} = \frac{8}{27}$
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$\displaystyle 27^{-2/3}=\frac{1}{9}$
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$\displaystyle \frac{2^{12}}{2^{10}} = 4$
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1.2.3. Using Laws of Exponents.

Answer.

$\displaystyle \frac{x^5}{x^3} = x^2$
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$\displaystyle x^5\cdot x^3 = x^8$
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$\displaystyle \left(x^5 x^1\right)^3 = x^{18}$
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1.2.4. Converting to Exponential Form.

Answer.

\begin{equation*} \frac{3}{4x^5} - \sqrt{x^5} + \frac{1}{\sqrt[5]{x^2}} + \frac{2}{5 x^{-2}} = \frac{3}{4} x^5 - x^{5/2} + x^{-2/5} + \frac{2}{5} x^2 \end{equation*}

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1.2.5. Using Scientific Notation.

Answer.

$\displaystyle \left(1.5\times 10^3 \right)\times\left(4.0 \times 10^{-2}\right) = 6.0\times 10^{1} = 60$
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$\displaystyle \frac{6.5\times 10^3}{2.0 \times 10^{-2}} = 3.25 \times 10^{5} = 325000$
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$\displaystyle \frac{6.5 \times 10^{-2}}{2.0\times 10^3} = 3.25 \times 10^{-5} = 0.0000325$
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$\displaystyle \left(1.5\times 10^3 \right)+\left(4.0 \times 10^{-2}\right) = 1500.04 = 1.50004 \times 10^{3}$
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1.2.6. Reducing Radicals.

Answer.

$\displaystyle \sqrt{16 x^3 y^6 z^2} = 4|x y^3 z | \, \sqrt{x}$
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$\displaystyle \sqrt[3]{16 x^5 y^6 z^3} = 2 x y^2 z \, \sqrt[3]{2 x^2}$
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$\displaystyle \sqrt[4]{16 x^5 y^6 z^4} = 2 |x y z| \, \sqrt[4]{x y^2}$
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1.2.7. Operations with Polynomial Expressions.

Answer.

$\displaystyle -x^6+x^5+3 x^4-x^3-x^2-2 x$
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$\displaystyle 6 x^2-7 x-3$
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$\displaystyle 4 x y^2-2 y^3 -4 x^2+2 x y+y^2+3 x-2 y+1$
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1.2.8. Operations with Rational Expressions.

Answer.

$\displaystyle \displaystyle \frac{7-x^2}{x^2-2 x-3} \frac{x(7-x)}{(x-3) (x+1)}$
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$\displaystyle \displaystyle \frac{x+1}{2x-6}$
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$\displaystyle \displaystyle \frac{x (x+3)}{x+1} = \frac{x^2+3x}{x+1}$
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1.2.9. Factoring Trinomials.

Answer.

$\displaystyle x^2+2 x-15 = (x-3) (x+5) $
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$\displaystyle x^2-25x = x(x-25)$
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$\displaystyle x^2-25 = (x-5)(x+5)$
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$\displaystyle x^2-10 x+25 = (x-5)^2$
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$\displaystyle 2 x^2-5 x-3 = (2x+1)(x-3)$
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$\displaystyle 2 x^2+5 x-3 = (2x-1)(x+3)$
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$\displaystyle x^2+2 x-15 = (x-3)(x+5)$
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$\displaystyle 9x^4-y^2 = (3x^2-y)(3x^2+y)$
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$\displaystyle a^2-b^2 = (a-b)(a+b)$
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$\displaystyle a^2+2ab+b^2 = (a+b)^2$
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$\displaystyle a^2-2ab+b^2 = (a-b)^2$
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1.2.10. Finding the Domain.

Answer.

$\displaystyle (-\infty,2]$
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$\displaystyle (-\infty,2)\cup(2,\infty)$
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$\displaystyle (-\infty,2)$
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$\displaystyle [0,\infty)$
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$\displaystyle (-\infty,\infty)$
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1.2.11. Solving Equations.

Answer.

$\displaystyle x = 4$
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$\displaystyle x=-4,3$
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$\displaystyle x =0,-\frac{1}{4}$
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$\displaystyle x = \pm 3$
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$\displaystyle x=0,6$
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$\displaystyle x=-2,0,2$
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$\displaystyle x=1,\frac{5}{3}$
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$\displaystyle x=\frac{7}{6}$
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$\displaystyle x=0$
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$\displaystyle x=\pm \sqrt{7}$
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1.3 Functions Pretest

1.3.1. Cartesian Plane.

Answer.

It’s a circle centered at $(3,1)$ with radius $r = 4\text{.}$ This is described by the equation $(x-3)^2+(y-1)^2 = 4^2\text{.}$ This meets the $x$-axis at two points: $(3-\sqrt{15},0)$ and $(3+\sqrt{15},0)\text{.}$

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1.3.2. Linear Equations.

Answer.

$\displaystyle 2x-6y = 3$
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$\displaystyle y = -3x+2$
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$\displaystyle y = -4$
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$x = 0$ is the $y$-axis
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1.3.3. Point-Slope Form.

Answer.

Point-Slope Form: $y-y_0 = m(x-x_0)\text{.}$ In this case, $m=-\frac{7}{8}$ and the line is given by

\begin{equation*} y-3 = -\frac{7}{8}\left(x+2\right) \quad \Rightarrow \quad y = -\frac{7}{8} x + \frac{10}{8} \end{equation*}

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1.3.4. Linear Models.

Answer.

Rate of change: $m = \frac{-2}{10}$ thousands of dollars per year.

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\begin{equation*} V(t) = -\frac{2}{10}(t-15)+19 \end{equation*}

Purchase price: $V(0) = 22$ thousands of dollars.

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1.3.5. Evaluating Functions.

Answer.

$\displaystyle f(-2)=-16$
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$\displaystyle f(-t)= -3 t^2-t-2$
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$\displaystyle f(t+2)=-3 t^2-11 t-12$
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$\displaystyle f(t)+2 = t-3 t^2$
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1.3.6. Piecewise-Defined Functions.

Answer.

$\displaystyle g(-2) = -2$
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$\displaystyle g(0) = 3$
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$\displaystyle g(2) = 3$
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$g(4)$ is undefined!
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$\displaystyle g(6) = 0$
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1.3.7. Graphs of Parent Functions.

Answer.

You may quickly compare your results with those in Section 7.3 or use a graphing tool.

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1.3.8. Transforming Parent Functions.

Answer.

1) Shift the graph left $3$ units; 2) reflect over the $x$-axis; 3) shift the graph up $6$ units.

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Zeros at $x=-3\pm\sqrt{6}\text{.}$

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1.3.9. Composition of Functions.

Answer.

$\displaystyle (f\circ g)(2) = 0$
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$\displaystyle (g\circ f)(8) = 4$
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$\displaystyle (g\circ f)(0) = -4$
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$(f\circ f)(8)$ is undefined!
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$\displaystyle (g\circ g)(x) = x^4$
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$\displaystyle (g\circ f)(x) = x-4$
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1.3.10. Inverse Functions.

Answer.

$\displaystyle f^{-1}(1) = 3$
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$\displaystyle (f^{-1})(-1) = 0$
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$\displaystyle (f\circ f^{-1})(3) = 3$
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$\displaystyle (f^{-1}\circ f)(1) = 1$
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1.3.11. Finding an Inverse Function.

Answer.

$f(x) = 4-3x\text{,}$ $f^{-1}(x) = \frac{4-x}{3}$
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$g(x) = \sqrt[3]{4-3x}\text{,}$ $g^{-1}(x) = \frac{1}{3} \left(4-x^3\right)$
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$h(x) = \frac{4}{3x}\text{,}$ $h^{-1}(x) = \frac{4}{3x}$
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$k(x) = \frac{4-3x}{1+x}\text{,}$ $k^{-1}(x) = \frac{4-x}{x+3}$
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1.4 Exponential and Logarithms Pretest

1.4.1. Exponential and Logarithmic Functions.

Answer.

See Section 8.4.

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1.4.2. Transforming Graphs of Exponential Functions.

Answer.

The graph is 1) reflected over the $y$-axis; 2) scaled vertically by a factor of $3\text{,}$ then shifted up $4$ units.

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The $y$-intercept is at $(0,7)\text{.}$ There is a horizontal asymptote at $y=3\text{.}$ The domain is still $(-\infty,\infty)\text{,}$ but the range is now $(3,\infty)\text{.}$

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1.4.3. Working with Logarithmic Expressions.

Answer.

$\log_2 (4) = 2\text{,}$ because $2^2 = 4\text{.}$
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$\log_4 (2) = 1/2\text{,}$ because $4^{1/2} = 2\text{.}$
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$\log_2 (0)$ is undefined because $2^x \neq 0$ is never zero.
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$\log_{1/2} (8) = -3$ because $(1/2)^{-3} = 8\text{.}$
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$\ln \left(\frac{1}{\sqrt[3]{e}}\right) = - \frac{1}{3}$ because $e^{-1/3} = \frac{1}{\sqrt[3]{e}}\text{.}$
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$\ln(1) = 0$ because $e^0 = 1\text{.}$
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1.4.4. Using Properties of Logarithms 1.

Answer.

\begin{equation*} \log_b\left(\frac{X Y^2}{b^3 \, \sqrt{Z}}\right) = -6 \end{equation*}

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1.4.6. Using Properties of Logarithms 3.

Answer.

\begin{equation*} 3^{2\log_3(5)} = 25 \end{equation*}

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1.4.7. Solving Exponential and Logarithmic Equations.

Answer.

$\displaystyle e^{-3x}=7 \quad \Rightarrow \quad x = - \frac{1}{3}\ln(7)$
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$\displaystyle \frac{1}{2-e^x}=2 \quad \Rightarrow \quad x = \ln(3/2)$
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1.4.8. Exponential Models.

Answer.

Initially, $T(0) = 140+60 = 200$ degrees.
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Solve $T(t) = 100$ to find $t \approx 6.26$ minutes.
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Room temperature is about 60 degrees. Why?
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1.5 Trigonometry Pretest

1.5.1. Radian Measure.

Answer.

Draw a circle of any radius. Measure an arc along the circle twice the radius in length. The corresponding angle is two radians. Since the circumference of a circle is $2\pi r\text{,}$ an angle of $\pi$ radians corresponds to an arc half the circumference. So $\pi$ radians is $180^\circ\text{.}$

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1.5.2. Arc-Length.

Answer.

\begin{equation*} s = (\text{$4$ inches}) \times 120^\circ \times \frac{\pi}{180^\circ} = \text{$8.38$ inches} \end{equation*}

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1.5.3. Right Triangle Trigonometry.

Answer.

The hypotenuse is $8$ inches. The side adjacent to $\theta$ is $\sqrt{60} = 2\sqrt{15}\text{.}$

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1.5.4. Trigonometric Functions.

Answer.

Using $x=-3\text{,}$ $y=1\text{,}$ and $r=\sqrt{10}$ we have $\sin(\theta) = 1/\sqrt{10}\text{,}$ $\cos(\theta) = -3/\sqrt{10} \text{,}$ $\tan(\theta) = -1/3\text{,}$ $\csc(\theta) = \sqrt{10}\text{,}$ $\sec(\theta) = -\sqrt{10}/3\text{,}$ and $\cot(\theta) = -3\text{.}$

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1.5.5. Trigonometric Functions.

Answer.

Using $x=1\text{,}$ $y=-4\text{,}$ and $r=\sqrt{17}$ we have $\sin(\theta) = -4/\sqrt{17}\text{,}$ $\cos(\theta) = 1/\sqrt{17} \text{,}$ $\tan(\theta) = -4\text{,}$ $\csc(\theta) = -\sqrt{17}/4\text{,}$ $\sec(\theta) = -sqrt{17}\text{,}$ and $\cot(\theta) = -1/4\text{.}$

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