R6 Math 209 Learning Team Problems Chapter 6(6), 8(4), 10(4),11(1-5), 12(1,3) (Rockswold

R6 Math 209 Learning Team Problems Chapter 6(6), 8(4), 10(4),11(1-5), 12(1,3) (Rockswold & Krieger)

Section 6.6: P. 403: 66

Section 8.4: P. 565: 116

Section 10.4: P. 672: 130

Section 11.1: P. 726: 96

Section 11.2: P. 741: 92

Section 11.3: P. 753: 116

Section 11.4: P. 767: 124

Section 11.5: P. 776: 64

Section 12.1: P. 808: 94

Section 12.3: P. 837: 112

Rev 1: 4/30/2012

6.6 #66: Use the Pythagorean Theorem as described above to find

the value of x in the figure.

| \

x | \ x+8

| \

------

x+7

x2 + (x + 7) 2 = (x + 8) 2

x2 + x2 + 14x + 49 = x2 + 16x + 64

2x2 + 14x + 49 = x2 + 16x + 64

x2 – 2x – 15 = 0 (Use AC method to factor, if necessary)

x2 – 5x + 3x – 15 = 0 (factors are -5 and +3)

(x – 5)(x + 3) = 0

So x = 5, and x = -3 is extraneous for this problem

8.4 #116: Heart Rate of an Athlete The following table lists

an athlete’s heart rate after the athlete finishes exercising

strenuously.

Time (in Minutes) 0 2 4 6

Heart Rate (bpm) 180 137 107 90

(a) Does P(t) = (5/3) t2- 25t + 180

model the data in the table exactly? Explain.

(b) Does P provide a reasonable model for the athlete’s

heart rate?

What should be the domain of P?

(a) No

Time 0 2 4 6

Heart Rate (bpm) 180 137 107 90

P(t) 180 136.67 106.67 90

(b) Yes

10.4 #130: Geometry A square has a diagonal that is 10 feet

long. Find the exact perimeter of the rectangle and

simplify your answer.

So the side of a square would be D/sqrt(2) = 10/sqrt(2)

and the perimeter would be 4*10/sqrt(2) = 40/sqrt(2) = 20*sqrt(2)

11.1 #96: Maximizing Area A rectangular pen being constructed

for a pet requires 60 feet of fence.

(a) Write a formula f(x) that gives the area of the

pen if one side of the pen has length x.

(b) Find the dimensions of the pen that give the

largest area. What is the largest area?

(a) f(x) = x(30 – x) since 2x + 2(30 – x) = 60

(b) The largest area occurs when the rectangle is a square, so x = 15 and F(15) = 225 sq. feet

11.2 #92: Head Start Enrollment The table lists numbers of

students in thousands enrolled in Head Start for

selected years.

Year 1966 1980 1995

Students (Thousands) 733 376 750

(a) Determine f (x) = a(x – h)2 + k, so that f

models these data.

(b) Estimate Head Start enrollment in 1990 and compare

it to the actual value of 541 thousand.

(a) f(x) = a(x-h)2 + k Since lowest data are: h=1980, k = 376 we have:

f(1980) = 376

f(1995) = a(1995-1980) 2 + 376 = 750

So a(15*15) +376 = 750 and 225a + 376 = 750, so 225a = 374 and
a = 1.662222

So f(x) = 1.662222 (x – 1980) 2 + 376

f(1966) = 1.662222 (-14) 2 + 376 = 701.7955

(b) f(1990) = 1.662222 (10) 2 + 376 = 166.2222 + 376 = 542.2222 thousand

compared with 541 thousand

11.3 #116: Falling Object If a metal ball is thrown downward

with an initial velocity of 22 feet per second (15 mph)

from a 100-foot water tower, its height h in feet above

the ground after t seconds is modeled by

h(t) = –16t2 – 22t + 100.

(a) Determine symbolically when the height of the

ball is 62 feet.

(b) Support your result in part (a) either graphically or

numerically.

then its height is given by h(t) = –16t2 + 22t + 100

Determine when the height of the ball is 80 feet.

(a) 62 = -16t2 -22t + 100 =

16t2 + 22t - 38 = 0

8 t2 +11t – 19 = 0

So ac = -152, (- 8 , + 19)

8 t2 +19t – 8t – 19 = 0

t - 1

8t 8 t2 – 8t

19 +19t 19

So (t – 1)(8t + 19) = 0 and t = 1 second

(b) h(1) = -16 -22 + 100 = 62 Ft

(c)

80 = -16t2 +22t + 100 =

16t2 - 22t - 20 = 0

8 t2 - 11t – 10 = 0

So ac = -80, (+ 5 , - 16)

8 t2 +5t – 16t – 10 = 0

t - 2

8t 8 t2 – 16t

+5 +5t 10

(t – 2) (8t + 5) = 0 and t = 2 seconds

11.4 #124: Hospitals The general trend in the number of hospitals

in the United States from 1945 through 2000 is

modeled by

f(x) = -1.38x2 + 84x + 5865,

where x = 5 corresponds to 1945, x = 10 to 1950,

and so on until x = 60 represents 2000. See the scatterplot

and accompanying graph.

(a) Describe any trends in the numbers of hospitals

from 1945 to 2000.

(b) What information does the vertex give?

of hospitals in 1970. Compare your result with

that shown in the graph.

(d) Use the formula for to estimate the year (or

years) when there were 6300 hospitals. Compare

your result with that shown in the graph.

(a) The Number of Hospitals increased until x = 30 and then decreased as x approached 60.

(b) The vertex gives the maximum point (x, y) or about (1970, 7200) of the parabola which approximates the scatter chart.

= -1.38*900 + 84*30 + 5865 = -1242 +2520 + 5865 = 7143 vs 7150

(d) for f(x) = 6300 = -1.38x2 + 84x + 5865

so -1.38x2 + 84x + 5865 – 6300 = -1.38x2 + 84x – 435 = 0

and is approximately = 5.715183 or Summer 1945 as well as = 55.15437 or Winter 1995

The Graph generally corroborates this, but is not accurate enough.

11.5 #64: Early Cellular Phone Use Our society is in transition

from an industrial to an informational society.

Cellular communication has played an increasingly

large role in this transition. The number of cellular

subscribers in the United States in thousands from

1985 to 1991 can be modeled by

f(x) = 163x2 – 146x + 205,

where x is the year and x = 0 corresponds to 1985,

x = 1 to 1986, and so on. (Source: M. Paetsch, Mobile

Communication in the U.S. and Europe.)

(a) Write a quadratic inequality whose solution set

represents the years when there were 2 million

subscribers or more.

(b) Solve this inequality.

(a) Then f(x) = 2000 ≤ 163x2 –146x + 205 and

(b) 0 ≤ 163x2 –146x – 1795, so x > 3.796411 = late 1988 ( x < -2.900706 is before cellphones). 3.796411 is the positive root using the quadratic formula where x = 0 means 1985.

12.1 #94: Skin Cancer and Ozone Ozone in the stratosphere

filters out most of the harmful ultraviolet (UV) rays

from the sun. However, depletion of the ozone layer is

affecting this protection. The formula U(x) = 1.5x

calculates the percent increase in UV radiation for an

x percent decrease in the thickness of the ozone layer.

The formula C(x) = 3.5x

calculates the percent increase in skin cancer cases when
the UV radiation increases by x percent.
(Source: R. Turner, D. Pierce, and I. Bateman, Environmental Economics.)

(a) Evaluate U(2) and C(3) and interpret each result.

(b) Find (C • U)(2) and interpret the result.

(a) U(2) = 1.5(2) = 3; C(3) = 3.5(3) = 10.5

(b) (C•U)(2) = C( U(2) ) = C(3) = 10.5

12.3 #112: Path Loss for Cellular Phones For cellular phones

to work throughout a country, large numbers of cellular

towers are necessary. How well the signal is

propagated throughout a region depends on the

location of these towers. One quick way to estimate

the strength of a signal at x kilometers is to use the

formula

D(x) = –121 – 36 log x

This formula computes the decrease in the signal,

using decibels, so it is always negative. For example,

D(1) = –121

means that at a distance of 1 kilometer

the signal has decreased in strength by 121 decibels.

(Source: C. Smith, Practical Cellular & PCS Design.)

(a) Evaluate D(3) and interpret the result.

(b) Graph D in [1, 10, 1] by [–160, –120, 10].

(a) D(3) = –121 – 36 log 3 = –121 – 36(0.4771213) = -121 – 17.17637 = -138.1764

(b)