R6 Math 209 Learning Team Problems Chapter 6 (1-4) (Rockswold & Krieger)

Section 6.1: P. 367-368: 82, 88

Section 6.2: P. 375: 90, 92

Section 6.3: P. 383: 78, 80

Section 6.4: P. 390: 84, 88

Rev 1: 4/30/2012

Section 7.1: P. 442: 102

Section 7.2: P. 450: 78

Section 7.3: P. 458: 74

Section 7.4: P. 467: 106

Section 10.1: P. 660: 120

Section 10.2: P. 669: 111

Section 10.3: P. 680: 118

Section 10.6: P. 715: 80

6.1 #82: Flight of a Golf Ball (Refer to the preceding exercise.)

Repeat the previous exercise if the height of a golf ball in feet after t seconds is given

by 128t – 16t2.

(a) Identify the greatest common factor.

(b) Factor this expression.

(a) GCF = 16t

(b) 128t – 16t2 = 16t(8 – t)

6.1 #88: Volume of a Box(Refer to the preceding exercise.)

[ A box is constructed by cutting out square corners of a rectangular
piece of cardboard and folding up the sides. If the cutout corners

have sides with length x, then the volume of the box is given by the
polynomial: 4x3 -60x2 + 200x ]

An (open) box is constructed from a square piece of metal that

is 20 inches on a side.

(a) If the square corners of length x are cut out, write

a polynomial that gives the volume of the box.

(b) Evaluate the polynomial when x = 4 inches.

polynomial expression.

(a) V = l*w*h so l = 20 – 2x, w = 20 – 2x, h = x

V = (20 – 2x)(20 – 2x)x

= 4x(10 – x)(10 – x)

= 4x(100 -20x + x2)

V = 4x3 – 80x2 + 400x

(b) V = 4*43 – 80*42 + 400*4

= 256 - 1280 + 1600

V = 576 cubic inches

6.2 #90: A cube has a surface area of 6x2+ 36x + 54. Find

the length of a side.

The Surface Area Of a cube = 6x2 + 36x + 54

= 6(x2 + 6x + 9) = 6(x+3)(x+3).

So Length of a side = (s + 3)

6.2 #92: Write a polynomial in factored form that represents

the total area of the figure.

Use Box Method of factoring:

x+4

x x24x

+44x16

x2 + 8x + 16 = (x + 4)2

6.3 #78: A rectangle has an area of 2x2 + 5x +3 . Find possible

dimensions for the rectangle. Make a sketch of the

rectangle.

Using AC Method 2*3 = 6 contains factor pairs 2, 3 which
sum to 5, so

2x2 + 5x + 3 = 2x2 + 2x + 3x + 3

Using Box Method: x +1

2x2x22x

+33x 3

Factors are: (2x + 3)(x + 1) which represent length and width

6.3 #80: Write a polynomial in factored form that represents

the total area of the figure.

Using Box Method: 3x +2

x3x22x

+26x 4

So we have (x+2)(3x+2)

6.4 #84: Factor this expression completely.2y4 + 24y3+ 72y2 .

Extract the common factor: 2y2to get: 2y2(y2 + 12y + 36)

Using AC Method, AC = 36 = (+6)(+6). So we have:

2*y*y(y + 6)(y + 6)

6.4 #88: A square has an area of 9x2 + 30x + 25 . Find the

length of a side. Make a sketch of the square.

Using AC Method, AC = 225. Factor pairs of 225 which add

To 30 are: 15, 15, so we get: 9x2 + 15x + 15x + 25

Using Box Method, 3x +5

3x9x215x

+515x 25

So the length of a side is 3x + 5; sketch needed

7.1 #102: Standing in Line A worker at a poolside store can
serve 20 customers per hour. If children arrive randomly

at an average rate of x per hour, then the average

number of children N waiting in line is given by

N = x2/(400 – 20x) for x < 20 (Source: N. Garber.)

(a) Complete the table.

x510181919.99920

N0.08330.58.118.0519998undefined

(1/12)

(b) Compare the number of children waiting in line if

the average rate increases from 18 to 19 children per hour.

It goes from 8.1 to 18.05, an increase of nearly 10 children

7.2 #78: Stopping on Hills If a car is traveling at 50 miles

per hour on a hill with wet pavement, then its stopping

distance D is given by

D = (2500/30)•(1/(x + 0.3) )

where x equals the slope of the hill. (Source: L. Haefner.)

(a) Multiply and simplify the formula for D.

(b) Compare the stopping distance for an uphill slope

of x = +0.1 to a downhill slope of x = –0.1.

(a) So we have D = (250/3)•(1/(x + 0.3) ) = 250/(3x + 0.9)

(b) when x = +0.1, then D = 250/1.2 = 208.33

when x = –0.1, then D = 250/0.6 = 416.67

So stopping means going twice as far when going downhill.

7.3 #74: Intensity of a Light Bulb The farther a person is

from a light bulb, the less intense its light is. The

equation: I = 19/4d2 approximates the intensity of light

from a 60-watt light bulb at a distance of d meters,

where I is measured in watts per square meter.

(Source: R. Weidner.)

(a) Find I for d = 2 meters and interpret the result.

(b) The intensity of light from a 100-watt light bulb

is about I = 32/4d2 Find an expression for the sum

of the intensities of light from a 100-watt bulb

and a 60-watt bulb.

a) I = 19/(4*22) = 19/(4*4) = 19/16 watts per square meter

b) I = 19/(4d2) + 32/(4d2) = 51/(4d2)

7.4 #106: Geometry Find the sum of the areas of the two

rectangles shown in the figure. Write your answer in

factored form.

Sum of L*W = (x – 2)*(1/(x – 1) ) + (x)*(2/(x + 1) )

So we have (x – 2) /(x – 1) + 2x / (x + 1)

The LCD is (x – 1)(x + 1) so we have [(x – 2)(x + 1) + 2x(x – 1) ]/[(x – 1)(x + 1)] =

(x2 + x – 2x -2 + 2x2 -2x)/[(x – 1)(x + 1)] =

(3x2 – 3x – 2)/[(x – 1)(x + 1) ]

10.1 #120: Animal Pulse Rate According to one model, an

animal’s heart rate varies according to its weight. The

formula N(w) = 885*(w)-1/2 gives an estimate for the

average number N of beats per minute for an animal

that weighs w pounds. Use the formula to estimate

the heart rate for a horse that weighs 800 pounds.

(Source: C. Pennycuick.)

N(800) = 885*(800)-1/2

= 885/sqrt(800) = 885/ 28.28427 = 31.28948 bpm

So the answer is: 31 bpm

10.2 #111: Bird Wings Heavier birds tend to have larger

wings than lighter birds do. For some birds the relationship

between the surface area A of the bird’s

wings in square inches and its weight W in pounds

can be modeled by

A = 100 cbrt(W2)

(Source: C. Pennycuick, Newton Rules Biology.)

(a) Find the area of the wings when the weight is

8 pounds.

(b) Write this formula with rational exponents.

(a)

A = 100 [cbrt(8)]2

= 100*2*2 = 400

(b) A = 100*W(2/3)

10.3 #118: Perimeter Find the exact perimeter of the rectangle.

Then approximate your answer.

√8 cm by √18 cm

P = 2L + 2W = 2sqrt(8) + 2sqrt(18)

= 2sqrt(23) + 2sqrt(2*32)

= 2(2)sqrt(2) + 2(3)sqrt(2)

= (4 + 6)sqrt(2)

=10sqrt(2)

= 10*1.414 = 14.14

10.6 #80: Exercises 79 and 80: Corrosion in Airplanes Corrosion

in the metal surface of an airplane can be difficult to

detect visually. One test used to locate it involves passing

an alternating current through a small area on the plane’s

surface. If the current varies from one region to another, it

may indicate that corrosion is occurring. The impedance

Z (or opposition to the flow of electricity) of the metal is

related to the voltage V and current I by the equation

Z = V/I , where Z, V, and I are complex numbers. Calculate

Z for the given values of V and I. (Source: Society for Industrial

and Applied Mathematics.)

V = 10 + 20i ; I = 3 + 7i

Z = (10 + 20i)/(3 + 7i) = 10(1+2i)/(3+7i)

Z = 10(1 + 2i)*(3-7i)/(3+7i)(3-7i)

= 10(3 – 7i + 6i +14)/(9 + 49)

= 10(17 – i )/58

Z = 5(17 - i)/29