Chapter 2

Answers may vary. Yes. 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 one of the quadrants.

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

The x-intercept is $\left(2,0\right)$ and the y-intercept is $\left(0,6\right).$

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

The x-intercept is $\left(3,0\right)$ and the y-intercept is $\left(0,\frac{9}{8}\right).$

$y=4-2x$

$y=\frac{5-2x}{3}$

$y=2x-\frac{4}{5}$

$d=\sqrt{74}$

$d=\sqrt{36}=6$

$d\approx 62.97$

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

$\left(2,\mathrm{-1}\right)$

$\left(0,0\right)$

$y=0$

not collinear

$\text{A:}\left(\mathrm{-3},2\right),\text{B:}\left(1,3\right),\text{C:}\left(4,0\right)$

$d=8.246$

$d=5$

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

$x=0y=<mn>-2</mn>$

$x=0.75y=0$

$x=-1.667y=0$

$15-11.2=3.8\text{mi}$ shorter

$\text{6}.0\text{42}$

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

37mi

54 ft

It means they have the same slope.

The exponent of the $x$ variable is 1. It is called a first-degree equation.

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

$x=2$

$x=\frac{2}{7}$

$x=6$

$x=3$

$x=\mathrm{-14}$

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

$x\ne 1;$ when we solve this we get $x=1,$ which is excluded, therefore NO solution

$x\ne 0;$ $x=-\frac{5}{2}$

$y=-\frac{4}{5}x+\frac{14}{5}$

$y=-\frac{3}{4}x+2$

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

$y=\mathrm{-3}x-5$

$y=7$

$y=\mathrm{-4}$

$8x+5y=7$

Parallel

Perpendicular

$m=-\frac{9}{7}$

$m=\frac{3}{2}$

$m_1=-\frac{1}{3},m_2=3;\text{Perpendicular}\text{.}$

$y=0.245x-\mathrm{45.662.}$ Answers may vary.

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

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

$\begin{array}{l}\text{The slope for }(\mathrm{-1},1)\text{to }(0,4)\text{is }3.\\ \text{The slope for }(\mathrm{-1},1)\text{to }(2,0)\text{is }\frac{-1}{3}.\\ \text{The slope for }(2,0)\text{to }(3,3)\text{is }3.\\ \text{The slope for }(0,4)\text{to }(3,3)\text{is }\frac{-1}{3}.\end{array}$

Yes they are perpendicular.

30 ft

$57.50

220 mi

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

$2,000-x$

$v+10$

Ann: $23;$ Beth: $46$

$20+0.05m$

300 min

$90+40P$

6 devices

$50,000-x$

4 h

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

$5,000 at 8% and $15,000 at 12%

$B=100+.05x$

Plan A

$R=9$

$r=\frac{4}{5}$ or 0.8

$W=\frac{P-2L}{2}=\frac{58-2(15)}{2}=14$

$f=\frac{pq}{p+q}=\frac{8(13)}{8+13}=\frac{104}{21}$

$m=\frac{-5}{4}$

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

length = 360 ft; width = 160 ft

405 mi

$A=88\text{in}{.}^2$

28.7

$h=\frac{V}{\pi r^2}$

$r=\sqrt{\frac{V}{\pi h}}$

$C=12\pi$

Add the real parts together and the imaginary parts together.

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

$\mathrm{-8}+2i$

$14+7i$

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

$8-i$

$\mathrm{-11}+4i$

$2\mathrm{-5}i$

$6+15i$

$\mathrm{-16}+32i$

$\mathrm{-4}\mathrm{-7}i$

25

$2-\frac{2}{3}i$

$4-6i$

$\frac{2}{5}+\frac{11}{5}i$

$15i$

$1+i\sqrt{3}$

$1$

$\mathrm{-1}$

128i

${\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)}^6=\mathrm{-1}$

$3i$

0

$5\mathrm{-5}i$

$\mathrm{-2}i$

$\frac{9}{2}-\frac{9}{2}i$

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

We want to take advantage of the zero property of multiplication in the fact that if $a\cdot b=0$ then it must follow that each factor separately offers a solution to the product being zero:

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

$x=6,$ $x=3$

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

$x=5,$ $x=\mathrm{-5}$

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

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

$x=0,$ $x=\frac{-3}{7}$

$x=\mathrm{-6},$ $x=6$

$x=6,$ $x=\mathrm{-4}$

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

$x=\mathrm{-2},$ $x=11$

$x=3\pm \sqrt{22}$

$z=\frac{2}{3},$ $z=-\frac{1}{2}$

$x=\frac{3\pm \sqrt{17}}{4}$

Not real

One rational

Two real; rational

$x=\frac{-1\pm \sqrt{17}}{2}$

$x=\frac{5\pm \sqrt{13}}{6}$

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

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

$x\approx -6.7$ and $x\approx 1.7$

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

$x(x+10)=119;$ 7 ft. and 17 ft.

maximum at $x=70$

The quadratic equation would be $(100x\mathrm{-0.5}{x}^2)-(60x+300)=300.$ The two values of $x$ are 20 and 60.

3 feet

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

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

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.

$x=81$

$x=17$

$x=8, x=27$

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

$y=0, \frac{3}{2}, \frac{-3}{2}$

$m=1,\mathrm{-1}$

$x=\frac{2}{5},\mathrm{\pm 3}i$

$x=32$

$t=\frac{44}{3}$

$x=3$

$x=\mathrm{-2}$

$x=4,\frac{\mathrm{-4}}{3}$

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

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

$x=\mathrm{-5}$

$x=1,\mathrm{-1},3,-3$

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

$x=1,5$

$x\ge 0$

$x=4,6,\mathrm{-6},\mathrm{-8}$

10 in.

90 kg

When we divide both sides by a negative it changes the sign of both sides so the sense of the inequality sign changes.

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

We start by finding the x-intercept, or where the function = 0. Once we have that point, 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 absolute value of them.

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

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

$\left(-\infty ,3\right)$

$\left(-\infty ,-\frac{37}{3}\right]$

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

$\left(-\infty ,-\frac{10}{3}\right)\cup \left(4,\infty \right)$

$\left(-\infty ,\mathrm{-4}\right]\cup \left[8,+\infty \right)$

No solution

$\left(\mathrm{-5},11\right)$

$\left[6,12\right]$

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

$\begin{array}{ll}x>-6\text{and}x>-2\hfill & \text{Take the intersection of two sets}.\hfill \\ x>-2,(-2,+\infty )\hfill & \hfill \end{array}$

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

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

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

It is never less than zero. No solution.

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

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

Where the blue line is above the orange line; always. All real numbers.

$(-\infty ,-\infty )$

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

$\left(-\infty ,4\right)$

$\{x|x<6\}$

$\{x|\mathrm{-3}\le x<5\}$

$\left(\mathrm{-2},1\right]$

$\left(-\infty ,4\right]$