Chapter 2

Terry starts at an elevation of 3000 feet and descends 70 feet per second.

3 miles per hour

$d\left(t\right)=100-10t$

Yes.

No.

No.

No.

Increasing.

Decreasing.

Decreasing.

Increasing.

Decreasing.

3

$–\frac{1}{3}$

$\frac{4}{5}$

$f(x)=-\frac{1}{2}x+\frac{7}{2}$

$y=2x+3$

$y=-\frac{1}{3}x+\frac{22}{3}$

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

$-\frac{5}{4}$

$y=\frac{2}{3}x+1$

$y=-2x+3$

$y=3$

Linear, $g(x)=-3x+5$

Linear, $f(x)=5x-5$

Linear, $g(x)=-\frac{25}{2}x+6$

Linear, $f(x)=10x-24$

$f(x)=-58x+17.3$

a. $a=11,900$; $b=1000.1$ b. $q(p)=1000p-100$

$x=-\frac{16}{3}$

$x=a$

$y=\frac{d}{c-a}x-\frac{ad}{c-a}$

$45 per training session.

The rate of change is 0.1. For every additional minute talked, the monthly charge increases by $0.1 or 10 cents. The initial value is 24. When there are no minutes talked, initially the charge is $24.

The slope is $-400.$ This means for every year between 1960 and 1989, the population dropped by 400 per year in the city.

c.

The slopes are equal; y-intercepts are not equal.

The point of intersection is $\left(a,a\right).$ This is because for the horizontal line, all of the $y$ coordinates are $a$ and for the vertical line, all of the $x$ coordinates are $a.$ The point of intersection will have these two characteristics.

First, find the slope of the linear function. Then take the negative reciprocal of the slope; this is the slope of the perpendicular line. Substitute the slope of the perpendicular line and the coordinate of the given point into the equation $y=mx+b$ and solve for $b.$ Then write the equation of the line in the form $y=mx+b$ by substituting in $m$ and $b.$

neither parallel or perpendicular

perpendicular

parallel

$(–2\text{, }0)$; $(0\text{, 4})$

$(\frac{1}{5}\text{, }0)$; $(0\text{, 1})$

$(8\text{, }0)$; $(0\text{, }28)$

$\text{Line 1}:m=8$
$\text{Line 2}:m=–6$
$\text{Neither}$

$\text{Line 1}:m=–\frac{1}{2}$
$\text{Line 2}:m=2$
$\text{Perpendicular}$

$\text{Line 1}:m=–2$
$\text{Line 2}:m=–2$
$\text{Parallel}$

$g(x)=3x-3$

$p(t)=-\frac{1}{3}t+2$

$\left(-2,1\right)$

$\left(-\frac{17}{5},\frac{5}{3}\right)$

F

C

A

$y=\text{3}$

$x=-3$

no point of intersection

$(\text{2},\text{ 7})$

$(–10,\text{ –5})$

$y=100x-98$

$x<\frac{1999}{201}x>\frac{1999}{201}$

Less than 3000 texts

Determine the independent variable. This is the variable upon which the output depends.

To determine the initial value, find the output when the input is equal to zero.

6 square units

20.012 square units

2,300

64,170

$P\left(t\right)=75,000+\mathrm{2,500}t$

(–30, 0) Thirty years before the start of this model, the town had no citizens. (0, 75,000) Initially, the town had a population of 75,000.

Ten years after the model began.

$W\left(t\right)=0.\text{5}t+\text{7}.\text{5}$

$\left(-15,0\right)$: The x-intercept is not a plausible set of data for this model because it means the baby weighed 0 pounds 15 months prior to birth. $\left(0,\text{ 7}.\text{5}\right)$: The baby weighed 7.5 pounds at birth.

At age 5.8 months.

$C\left(t\right)=12,025-205t$

$\text{(58.7, 0)}$: In roughly 59 years, the number of people inflicted with the common cold would be 0. $\text{(0,12,025)}$: Initially there were 12,025 people afflicted by the common cold.

2064

$y=-2t\text{+180}$

In 2070, the company’s profit will be zero.

$y=30t-300$

(10, 0) In 1990, the profit earned zero profit.

Hawaii

During the year 1933

$105,620

More than 133 minutes

More than $42,857.14 worth of jewelry

$66,666.67

When our model no longer applies, after some value in the domain, the model itself doesn’t hold.

We predict a value outside the domain and range of the data.

The closer the number is to 1, the less scattered the data, the closer the number is to 0, the more scattered the data.

61.966 years

No.

No.

Interpolation. About $60°\text{ F}.$

C

B

Yes, trend appears linear because $r=0.\text{985}$ and will exceed 12,000 near midyear, 2016, 24.6 years since 1992.

$y=\text{1}.\text{64}0x+\text{13}.\text{8}00$, $r=0.\text{987}$

$y=-0.962x+26.86,r=-0.965$

$y=-\text{1}.\text{981}x+\text{6}0.\text{197}$; $r=-0.\text{998}$

$y=0.\text{121}x-38.841,r=0.998$

$\left(\text{−2},\mathrm{-6}\right),\left(\text{1},\text{−12}\right),\left(\text{5},\text{−2}0\right),\left(\text{6},\text{−22}\right),\left(\text{9},\text{−28}\right)$; $y=\mathrm{-2}x\mathrm{-10}$

$\left(\text{189}.\text{8},0\right)$ If 18,980 units are sold, the company will have a profit of zero dollars.

$y=0.00587x+\text{1985}.4\text{1}$

$y=\text{2}0.\text{25}x-\text{671}.\text{5}$

$y=-\text{1}0.\text{75}x+\text{742}.\text{5}0$

Yes

Increasing.

$y=-\text{3}x+\text{26}$

3

$y=\text{2}x-\text{2}$

Not linear.

parallel

$(\mathrm{–9},0);(0,\mathrm{–7})$

Line 1: $m=-2;$ Line 2: $m=-2;$ Parallel

$y=-0.2x+21$

250.

118,000.

$y=-\text{3}00x+\text{11},\text{5}00$

a) 800; b) 100 students per year; c) $P\left(t\right)=100t+1700$

18,500

$91,625

Extrapolation.

Midway through 2024.

$y=-1.294x+49.412;r=-0.974$

Early in 2022

7,660

Yes.

Increasing

$y=\mathrm{-1.5}x-6$

$y=-2x-1$

No.

Perpendicular

$\left(-\text{7},0\right)$; $\left(0,-\text{2}\right)$

$y=-0.25x+12$

150

165,000

$y=875x+10,675$

a) 375; b) dropped an average of 46.875, or about 47 people per year; c) $y=-46.875t+1250$

Early in 2018

$y=0.00455x+1979.5$

$r=0.999$