Chapter 3 Review Exercises Algebra 2 Cc 2015 Hwk Ch 3 Rev Solve Quad

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ii.1 The Rectangular Coordinate Systems and Graphs

1.

$x$	$y=\frac{\mathrm{one}}{\mathrm{ii}}\mathrm{10}+2$	$\left(x,y\right)$
$\mathrm{-2}$	$y=\frac{\mathrm{i}}{2}\left(\mathrm{-two}\right)+\mathrm{ii}=\mathrm{ane}$	$\left(\mathrm{-2},1\right)$
$\mathrm{-i}$	$y=\frac{1}{2}\left(\mathrm{-1}\right)+\mathrm{two}=\frac{3}{\mathrm{ii}}$	$\left(-\mathrm{i},\frac{3}{2}\right)$
$0$	$y=\frac{1}{2}\left(0\right)+2=2$	$\left(0,2\right)$
$1$	$y=\frac{1}{2}\left(\mathrm{ane}\right)+\mathrm{ii}=\frac{\mathrm{v}}{\mathrm{ii}}$	$\left(1,\frac{5}{2}\right)$
$\mathrm{two}$	$y=\frac{1}{2}\left(2\right)+2=\mathrm{three}$	$\left(2,\mathrm{three}\right)$

two.

x-intercept is $\left(4,0\right);$ y-intercept is $\left(0,3\right).$

4.

$\left(-5,\frac{5}{\mathrm{two}}\right)$

2.ii Linear Equations in Ane Variable

v.

$x=-\frac{7}{17}.$ Excluded values are $\mathrm{10}=-\frac{1}{2}$ and $\mathrm{ten}=-\frac{\mathrm{one}}{\mathrm{three}}.$

x.

Horizontal line: $y=2$

11.

Parallel lines: equations are written in slope-intercept class.

two.3 Models and Applications

2.

$C=\mathrm{two.five}x+3,650$

iv.

$L=37$ cm, $\mathrm{Due\; west}=18$ cm

two.4 Complex Numbers

1.

$\sqrt{\mathrm{-24}}=0+2i\sqrt{\mathrm{six}}$

3.

$(3\mathrm{-iv}i)-(2+\mathrm{v}i)=\mathrm{ane}\mathrm{-nine}i$

2.5 Quadratic Equations

1.

$\left(\mathrm{10}-6\right)\left(x+\mathrm{i}\right)=0;x=\mathrm{six},x=-1$

2.

$\left(x\mathrm{-vii}\right)\left(x+\mathrm{iii}\right)=0,$ $x=7,$ $x=\mathrm{-3.}$

three.

$\left(x+5\right)\left(x\mathrm{-5}\right)=0,$ $x=\mathrm{-five},$ $x=5.$

4.

$\left(3\mathrm{ten}+\mathrm{two}\right)\left(\mathrm{iv}\mathrm{ten}+1\right)=0,$ $x=-\frac{\mathrm{ii}}{\mathrm{three}},$ $\mathrm{10}=-\frac{\mathrm{one}}{4}$

5.

$\mathrm{10}=0,x=\mathrm{-10},\mathrm{10}=\mathrm{-i}$

viii.

$\mathrm{10}=-\frac{\mathrm{ii}}{3},$ $x=\frac{1}{3}$

2.half dozen Other Types of Equations

iv.

$0,$ $\frac{1}{2},$ $-\frac{1}{2}$

v.

$1;$ inapplicable solution $-\frac{2}{9}$

6.

$\mathrm{-2};$ inapplicable solution $\mathrm{-one}$

ten.

$\mathrm{-i},$ $0$ is not a solution.

2.7 Linear Inequalities and Absolute Value Inequalities

two.

$\left(-\infty ,\mathrm{-ii}\right)\cup \left[3,\infty \right)$

6.

$\left[-\frac{3}{\mathrm{fourteen}},\infty \right)$

seven.

$\mathrm{vi}<\mathrm{10}\le \mathrm{ix}\text{}\text{or}\left(6,9\right]$

viii.

$\left(-\frac{\mathrm{i}}{8},\frac{1}{\mathrm{two}}\right)$

x.

$k\le 1$ or $k\ge 7;$ in interval annotation, this would exist $(-\infty ,\mathrm{one}]\cup [7,\infty ).$

2.1 Section Exercises

1.

Answers may vary. Yeah. It is possible for a point to be on the x-axis or on the y-axis and therefore is considered to Not be in ane of the quadrants.

iii.

The y-intercept is the point where the graph crosses the y-axis.

5.

The x-intercept is $\left(2,0\right)$ and the y-intercept is $\left(0,\mathrm{half-dozen}\right).$

7.

The ten-intercept is $\left(2,0\right)$ and the y-intercept is $\left(0,\mathrm{-3}\right).$

9.

The 10-intercept is $\left(3,0\right)$ and the y-intercept is $\left(0,\frac{\mathrm{ix}}{\mathrm{eight}}\right).$

23.

$\left(\mathrm{three},\frac{-3}{2}\right)$

31.

not collinear

33.

$\text{A:}\left(\mathrm{-3},2\right),\text{B:}\left(\mathrm{one},3\right),\text{C:}\left(4,0\right)$

35.

$x$	$y$
$\mathrm{-3}$	1
0	2
3	3
6	4

49.

$x=0y=\mathrm{-2}$

53.

$x=-\mathrm{one.667}y=0$

55.

$15-11.2=\mathrm{3.viii}\text{mi}$ shorter

59.

Midpoint of each diagonal is the same point $(2,\mathrm{–2})$ . Notation this is a characteristic of rectangles, merely not other quadrilaterals.

2.2 Section Exercises

1.

It means they accept the same slope.

3.

The exponent of the $\mathrm{10}$ variable is 1. It is called a first-degree equation.

5.

If we insert either value into the equation, they make an expression in the equation undefined (nada in the denominator).

17.

$x\ne \mathrm{-4};$ $x=\mathrm{-3}$

19.

$\mathrm{ten}\ne \mathrm{i};$ when we solve this we get $x=\mathrm{ane},$ which is excluded, therefore NO solution

21.

$x\ne 0;$ $x=-\frac{5}{\mathrm{ii}}$

23.

$y=-\frac{\mathrm{four}}{5}x+\frac{\mathrm{fourteen}}{5}$

25.

$y=-\frac{\mathrm{three}}{4}x+2$

27.

$y=\frac{1}{2}x+\frac{5}{2}$

37.

Parallel

39.

Perpendicular

45.

$k_1=-\frac{1}{3},{\mathrm{thousand}}_{\mathrm{two}}=\mathrm{iii};\text{Perpendicular}\text{.}$

47.

$y=0.245x-\mathrm{45.662.}$ Answers may vary. $y_{\text{min}}=\mathrm{-l},y_{\text{max}}=\mathrm{-40}$

49.

$y=-2.333x+\mathrm{6.667.}$ Answers may vary. $y_{\min }=\mathrm{-10},y_{\max }=10$

51.

$y=-\frac{A}{B}x+\frac{C}{B}$

53.

$\begin{array}{l}\text{The slope for}(\mathrm{-ane},\mathrm{one})\text{to}(0,4)\text{is}\mathrm{three.}\\ \text{The gradient for}(\mathrm{-one},\mathrm{ane})\text{to}(2,0)\text{is}\frac{-\mathrm{i}}{3}.\\ \text{The slope for}(\mathrm{ii},0)\text{to}(\mathrm{iii},3)\text{is}\mathrm{three.}\\ \text{The slope for}(0,4)\text{to}(3,3)\text{is}\frac{-1}{\mathrm{three}}.\end{array}$

Yes they are perpendicular.

2.3 Section Exercises

ane.

Answers may vary. Possible answers: We should define in words what our variable is representing. We should declare the variable. A heading.

seven.

Ann: $23;$ Beth: $46$

21.

She traveled for 2 h at 20 mi/h, or 40 miles.

23.

$five,000 at 8% and $15,000 at 12%

25.

$B=100+.05x$

33.

$W=\frac{P-2L}{\mathrm{two}}=\frac{58-2(15)}{2}=\mathrm{xiv}$

35.

$f=\frac{pq}{p+q}=\frac{\mathrm{eight}(\mathrm{xiii})}{8+\mathrm{xiii}}=\frac{104}{21}$

39.

$h=\frac{2A}{b_1+b_2}$

41.

length = 360 ft; width = 160 ft

45.

$A=88\text{in}{.}^{\mathrm{two}}$

49.

$h=\frac{\mathrm{Five}}{\pi r^2}$

2.4 Section Exercises

1.

Add the real parts together and the imaginary parts together.

three.

Possible answer: $i$ times $i$ equals -1, which is not imaginary.

9.

$-\frac{23}{29}+\frac{15}{29}i$

33.

$\frac{\mathrm{ii}}{5}+\frac{11}{\mathrm{v}}i$

45.

${\left(\frac{\sqrt{3}}{2}+\frac{\mathrm{ane}}{2}i\right)}^{\mathrm{half\; dozen}}=\mathrm{-1}$

55.

$\frac{\mathrm{nine}}{\mathrm{ii}}-\frac{9}{\mathrm{two}}i$

2.5 Department Exercises

1.

It is a second-degree equation (the highest variable exponent is 2).

3.

We want to have reward of the zero property of multiplication in the fact that if $a\cdot b=0$ then information technology must follow that each factor separately offers a solution to the product being zero: $a=0or\text{b}=0.$

five.

One, when no linear term is present (no x term), such as $x^{\mathrm{ii}}=16.$ 2, when the equation is already in the form ${(ax+b)}^{\mathrm{ii}}=d.$

ix.

$x=\frac{-5}{2},$ $x=\frac{-1}{3}$

13.

$x=\frac{-3}{2},$ $x=\frac{3}{2}$

17.

$x=0,$ $x=\frac{-\mathrm{iii}}{\mathrm{vii}}$

25.

$x=\mathrm{-2},$ $\mathrm{ten}=\mathrm{eleven}$

29.

$z=\frac{\mathrm{ii}}{\mathrm{iii}},$ $z=-\frac{\mathrm{one}}{2}$

31.

$x=\frac{3\pm \sqrt{17}}{\mathrm{iv}}$

39.

$\mathrm{10}=\frac{-\mathrm{one}\pm \sqrt{17}}{\mathrm{ii}}$

41.

$x=\frac{5\pm \sqrt{\mathrm{thirteen}}}{6}$

43.

$x=\frac{-1\pm \sqrt{17}}{8}$

45.

$x\approx 0.131$ and $x\approx 2.535$

47.

$\mathrm{ten}\approx -\mathrm{6.seven}$ and $x\approx 1.7$

49.

$\begin{array}{ccc}\hfill a{\mathrm{10}}^{\mathrm{ii}}+bx+c& =& 0\hfill \\ \hfill {\mathrm{ten}}^2+\frac{b}{a}x& =& \frac{-c}{a}\hfill \\ \hfill {\mathrm{ten}}^2+\frac{b}{a}\mathrm{10}+\frac{b^2}{\mathrm{four}{a}^2}& =& \frac{-c}{a}+\frac{b}{4a^2}\hfill \\ \hfill {\left(x+\frac{b}{2a}\right)}^{\mathrm{two}}& =& \frac{b^2-4ac}{4a^{\mathrm{ii}}}\hfill \\ \hfill x+\frac{b}{2a}& =& \pm \sqrt{\frac{b^2-\mathrm{iv}ac}{4a^2}}\hfill \\ \hfill x& =& \frac{-b\pm \sqrt{b^2-\mathrm{iv}ac}}{2a}\hfill \end{array}$

51.

$\mathrm{10}(x+10)=119;$ 7 ft. and 17 ft.

55.

The quadratic equation would exist $(100\mathrm{10}\mathrm{-0.v}{x}^2)-(\mathrm{sixty}x+300)=300.$ The two values of $x$ are twenty and 60.

2.6 Section Exercises

i.

This is not a solution to the radical equation, information technology is a value obtained from squaring both sides and thus irresolute the signs of an equation which has caused it not to exist a solution in the original equation.

3.

He or she is probably trying to enter negative 9, but taking the square root of $\mathrm{-ix}$ is not a real number. The negative sign is in front end of this, so your friend should be taking the foursquare root of 9, cubing it, and then putting the negative sign in front, resulting in $\mathrm{-27.}$

5.

A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.

11.

$\mathrm{ten}=8,x=27$

15.

$y=0,\frac{\mathrm{three}}{2},\frac{-\mathrm{iii}}{\mathrm{two}}$

19.

$x=\frac{2}{\mathrm{five}},\mathrm{\pm 3}i$

31.

$x=\frac{-5}{4},\frac{7}{4}$

37.

$\mathrm{ten}=\mathrm{one},\mathrm{-ane},\mathrm{three},-3$

45.

$x=\mathrm{iv},6,\mathrm{-vi},\mathrm{-8}$

two.vii Section Exercises

1.

When we split both sides past a negative it changes the sign of both sides so the sense of the inequality sign changes.

v.

Nosotros start by finding the x-intercept, or where the role = 0. Once we have that bespeak, which is $(3,0),$ we graph to the right the straight line graph $y=x\mathrm{-3},$ and then when we draw it to the left we plot positive y values, taking the accented value of them.

vii.

$\left(-\infty ,\frac{3}{4}\right]$

9.

$\left[-\frac{13}{2},\infty \right)$

thirteen.

$\left(-\infty ,-\frac{37}{\mathrm{iii}}\right]$

15.

All real numbers $\left(-\infty ,\infty \right)$

17.

$\left(-\infty ,-\frac{10}{\mathrm{iii}}\right)\cup \left(4,\infty \right)$

19.

$\left(-\infty ,\mathrm{-4}\right]\cup \left[\mathrm{viii},+\infty \right)$

27.

$\left[\mathrm{-10},12\right]$

29.

$\begin{array}{ll}x>-\mathrm{six}\text{and}\mathrm{ten}>-2\hfill & \text{Accept the intersection of ii sets}.\hfill \\ x>-\mathrm{ii},(-2,+\infty )\hfill & \hfill \end{array}$

31.

$\begin{array}{ll}\mathrm{10}<-3\mathrm{or}x\ge \mathrm{i}\hfill & \text{Take the matrimony of the two sets}.\hfill \\ (-\infty ,-\mathrm{iii}){{{\displaystyle \cup }}^{\text{}}}^{\text{}}[1,\infty )\hfill & \hfill \end{array}$

33.

$\left(-\infty ,\mathrm{-1}\right)\cup \left(3,\infty \right)$

35.

$\left[\mathrm{-11},\mathrm{-3}\right]$

37.

Information technology is never less than zero. No solution.

39.

Where the blueish line is above the orange line; point of intersection is $x=-3.$

$\left(-\infty ,\mathrm{-iii}\right)$

41.

Where the bluish line is above the orangish line; always. All real numbers.

$(-\infty ,-\infty )$

47.

$\{x|x<6\}$

49.

$\{\mathrm{10}|\mathrm{-iii}\le \mathrm{10}<\mathrm{five}\}$

55.

Where the blue is below the orangish; ever. All real numbers. $(-\infty ,+\infty ).$

57.

Where the blue is beneath the orange; $\left(1,7\right).$

63.

$\begin{array}{l}\mathrm{lxxx}\le T\le 120\\ 1,600\le 20T\le \mathrm{ii},400\end{array}$

$\left[\mathrm{one},600,2,400\right]$

Review Exercises

1.

x-intercept: $\left(3,0\right);$ y-intercept: $\left(0,\mathrm{-4}\right)$

9.

midpoint is $\left(2,\frac{23}{2}\right)$

19.

$y=\frac{1}{6}x+\frac{4}{3}$

21.

$y=\frac{2}{\mathrm{iii}}x+6$

27.

$x=-\frac{3}{\mathrm{four}}\pm \frac{i\sqrt{47}}{\mathrm{iv}}$

29.

horizontal component $\mathrm{-2};$ vertical component $\mathrm{-i}$

47.

$x=\frac{-1\pm \sqrt{5}}{4}$

49.

$x=\frac{2}{\mathrm{v}},\frac{-\mathrm{one}}{3}$

59.

$x=\frac{11}{\mathrm{ii}},\frac{\mathrm{-17}}{2}$

63.

$\left[\frac{-10}{3},\mathrm{two}\right]$

67.

$\left(-\frac{4}{3},\frac{\mathrm{one}}{5}\right)$

69.

Where the blue is below the orangish line; betoken of intersection is $x=\mathrm{3.5.}$

$\left(3.5,\infty \right)$

Practice Test

one.

$y=\frac{3}{2}\mathrm{ten}+\mathrm{two}$

3.

$\left(0,\mathrm{-3}\right)$ $\left(\mathrm{four},0\right)$

nine.

$x\ne \mathrm{-iv},2;$ $x=\frac{-\mathrm{five}}{\mathrm{two}},1$

15.

$y=\frac{\mathrm{-5}}{9}x-\frac{\mathrm{two}}{\mathrm{ix}}$

17.

$y=\frac{\mathrm{v}}{2}\mathrm{ten}-\mathrm{iv}$

21.

$\frac{5}{\mathrm{xiii}}-\frac{\mathrm{fourteen}}{\mathrm{thirteen}}i$

25.

$x=\frac{1}{2}\pm \frac{\sqrt{2}}{2}$

29.

$x=\frac{1}{2},\mathrm{ii},\mathrm{-2}$

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Source: https://openstax.org/books/college-algebra/pages/chapter-2

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