Chapter 4

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

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

The point of intersection is $\left(a,\text{<mi>a</mi>}\right).$ 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 is on both lines and therefore will have these two characteristics.

Yes

Yes

No

Yes

Increasing

Decreasing

Decreasing

Increasing

Decreasing

2

–2

$y=\frac{3}{5}x-1$

$y=3x-2$

$y=-\frac{1}{3}x+\frac{11}{3}$

$y=-1.5x-3$

perpendicular

parallel

$\begin{array}{l}f(0)=-(0)+2\\ f(0)=2\\ y-\mathrm{int}:(0,2)\\ 0=-x+2\\ x-\mathrm{int}:(2,0)\end{array}$

$\begin{array}{l}h(0)=3(0)-5\\ h(0)=-5\\ y-\mathrm{int}:(0,-5)\\ 0=3x-5\\ x-\mathrm{int}:\left(\frac{5}{3},0\right)\end{array}$

$\begin{array}{l}-2x+5y=20\\ -2(0)+5y=20\\ 5y=20\\ y=4\\ y-\mathrm{int}:(0,4)\\ -2x+5(0)=20\\ x=-10\\ x-\mathrm{int}:(-10,0)\end{array}$

Line 1: m = –10 Line 2: m = –10 Parallel

Line 1: m = –2 Line 2: m = 1 Neither

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

$y=3x-3$

$y=-\frac{1}{3}t+2$

0

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

$y=3x-1$

$y=-2.5$

F

C

A

$y=\text{3}$

$x=-3$

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

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

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

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

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

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

$x=a$

$y=\frac{d}{c–a}x–\frac{ad}{c–a}$

$y=100x–98$

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

$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

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

20.01 square units

2,300

64,170

$P\left(t\right)=75,000+2500t$

(–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

$W(t)=0.5t+7.5$

$\left(-15,\text{<mn>0</mn>}\right)$ : 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

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

$\left(\text{58}.\text{7},\text{<mn>0</mn>}\right):$ In roughly 59 years, the number of people inflicted with the common cold would be 0. $\left(0,\text{12},0\text{25}\right)$ Initially there were 12,025 people afflicted by the common cold.

2063

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

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

$y=\text{3}0t-\text{3}00$

(10, 0) In the year 1990, the company’s profits were zero

Hawaii

During the year 1933

$105,620

P(t) = 190t + 4,360

More than 133 minutes

More than $42,857.14 worth of jewelry

More than $66,666.67 in sales

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

$\text{This value of r indicates a strong negative correlation or slope, so C}$

$\text{This value of r indicates a weak negative correlation, so B}$

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

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

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

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

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

$(\mathrm{-2},\mathrm{-6}),(1,\text{−12}),(5,\mathrm{-20}),(6,\text{−22}),(9,\text{−28});$ Yes, the function is a good fit.

$(\text{189}.8,0)$ 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

$y=-0.2x+21$

More than 250

118,000

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

18,500

$91,625

Extrapolation

2023

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

2027

7,660

Yes

Increasing

y = −1.5x − 6

y = −2x − 1

No

Perpendicular

(−7, 0); (0, −2)

y = −0.25x + 12

150

165,000

y = 875x + 10,625

In early 2018

y = 0.00455x + 1979.5

r = 0.999