1.01 Notes - SharpSchool...

53
Algebra II Indicators p. 33 Answer Key Notes 1.01 A. 21 15 2 3 7 5 5 2 15 + B. 7 2 6 150 3 3 + C. 2 29 20 D. 16 49 E. 15 05 . F. $11,960; $242,000 G. $14,337; $117,800 H. 3.856% I. 12% J. 10 227 14 12 9 log . or e ln . 227 14 12 9 or 227 14.12 9 K. a = 3, b = 2 or a = -3, b = -2 L. 13 11 20 i

Transcript of 1.01 Notes - SharpSchool...

Algebra II Indicators p. 33 Answer Key

Notes1.01

A. 21

15

23 7 5

5

215− − +

B. 7 2 6 1503 3+

C. 2

29

20

D.

16

49

E. 150 5.

F. $11,960; $242,000

G. $14,337; $117,800

H. 3.856%

I. 12%

J. 10

227

14 12

9

log.

or

eln

.

227

14 12

9

or

227

14.129

K. a = 3, b = 2 or a = -3, b = -2

L.

13 11

20

− i

Algebra II Indicators p. 34 Answer Key

Notes1.02

A. 243x5 - 810x4y + 1080x3y2 - 720x2y3

+ 240xy4 - 32y5

B. x6 + 12x5y + 60x4y2 + 160x3y3

+ 240x2y4 + 192xy5 + 64y6

C. 4 11 33

81

3

2x xx

+ + +−

D. x 8x 36

165

x 5

2 − + −+

E. 4(x2 + 4y2)

F. (2x + 3)(4x2 - 6x + 9)

G. (3x - 1)(2x + 5)

H. (x2 + 1)(x + 1)(x - 1)

I. (3y + 4)(x + 2)

J. 2

K.

2

2 1

x

x +

L.

x

x

++

2

6

M.

a

a

++

7

2

N. x - y

O. 6.6 ohms

P. 3.4 seconds

Algebra II Indicators p. 35 Answer Key

Notes2.01

A. (x + 4)2 + (y - 5.5)2 = 6.25

(x + 4)2 + (y - 0.5)2 = 6.25

(x - 1)2 + (y - 5.5)2 = 6.25

(x - 1)2 + (y - 0.5)2 = 6.25

B. vertex: (3.5, -3.25)

y-intercept: (0, 9)

x-intercepts:

7 13

20

+

, ,

7 13

20

,

C. vertex: (-10, 2)

x-intercept: (2, 0)

y-intercepts:

06 30

3,

+

,

06 30

3,

D. x-intercepts: − +( )3 6 0, ,

− −( )3 6 0,

y-intercepts: 0 4 13, +( ) ,

0 4 13, −( )

Algebra II Indicators p. 36 Answer Key

Notes2.02

A. Center: (2, -2)

Foci: 2 13 2+ −( ), ,

2 13 2− −( ),

major axis: 8

Minor axis: 2 3x-intercepts: none

y-intercepts: (0, -0.5), (0, -3.5)

B. Center: (-2, 3)

Vertices: − ±( )2 6,3

Asymptotes: y

7

3x 3 2

7

3= + +

y

7

3x 3 2

7

3= − + −

x-intercepts:

− ±

2

69

7,0

y-intercepts: none

Algebra II Indicators p. 37 Answer Key

NotesC. x-intercepts: (-7.83, 0), (1.83, 0)

y-intercepts: (0, 8.96). (0, -0.96)

D. x-intercepts: (0.76, 0), (5.24, 0)

y-intercepts: (0, 0.38), (0, -8.38)

Algebra II Indicators p. 38 Answer Key

Notes3.01

A. The data is widely scattered between

$0.71 and $1.36. There is not an

algebraic expression that will model year

and gasoline prices very well because of

the data’s scattered nature. Students

should research the petroleum industry

and identify variables that are likely to

affect the prices of petroleum-based

products. Some variables they are likely

to discover are: supplies available from

petroleum producing nations; time of

year; weather; wages and benefits for

industry employees; refinery capacity;

transportation costs; taxes; number of

automobiles. Students should correlate

these variables with prices and

determine if algebraic models are

appropriate.

Gasoline: January Price per Gallon

0.00

0.20

0.40

0.600.80

1.00

1.20

1.40

1.60

1975 1980 1985 1990 1995 2000 2005

Algebra II Indicators p. 39 Answer Key

Notes

PGA Championship

0

200

400

600

800

1000

0 5 10 15 20

Place

B. The best model available

(linear, exponential, quadratic) is

y = 533.81(0.8675)X; this does not fit

very well. A better idea is for a student

to contact the PGA and find out how the

prize money is distributed. Winnings are

dependent upon the place an individual

finishes. According to the model, the

prize awarded depreciates in value about

13% for each place from first. Other

variables that may affect prize money

would include category of play (major

championship event); sponsorships;

ticket prices; television contracts.

Algebra II Indicators p. 40 Answer Key

NotesC. The data is increasing for the period

shown. A linear model,

y = 0.019548x - 1.005658, r = 0.9045,

appears to fit reasonably well for

estimation purposes. Transportation

costs, weather, federal subsidies, and

labor costs are a few variables that affect

the price of food. Have students

investigate variables specific to NC

agriculture.

D. B-40

2L-Tower

Area: 12n - 6 Volume: 3n - 2

Cut Block

Area: 6n2 Volume: 3n2 - 3n + 1

B-41

Blocks

Area: good luck!! Volume: 2n-1

Steps

Area: n2 + 5n Volume: 0.5n2 + 0.5n

Apples: Price per Pound

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1975 1980 1985 1990 1995 2000 2005

Algebra II Indicators p. 41 Answer Key

Notes

Newspaper Circulation

0

10

20

30

40

50

60

70

1900 1920 1940 1960 1980 2000 2020

E. Circulation increases until 1985 and then

begins to decline. Competition (TV,

radio, cable, internet) is the major

variable that affects circulation. Since

the relation is increasing and then

decreasing, a quadratic model is

probably best, y = -0.01x2 + 1.6x - 1.8.

According to the model, newspaper

circulation will drop below 50 million

about 2013.

Algebra II Indicators p. 42 Answer Key

Notes3.02

A. The data is decreasing and should

provide a good opportunity for a

discussion of data and its outliers. Gate

receipts, sponsorships, and TV/radio

contracts are variables that can affect

prize money. Ask a student to contact

NASCAR and find out how prize money

is determined. None of the three (linear,

quadratic, exponential) provide a good

model. If first place is excluded,

exponential becomes the better model,

y = 279901(0.93)X.

Brickyard 400

0

200

400

600

800

1,000

0 5 10 15 20

Place

Algebra II Indicators p. 43 Answer Key

Notes

Food Expenditures Away from Home

0

10

20

30

40

50

1920 1940 1960 1980 2000

B. The data is increasing over the period

shown. A linear model (y = 0.43x - 3.15)

fits well (r = 0.98). According to the

linear model, the portion of food

expenditures that will be spent outside

the home will increase 0.43% annually.

Increase in the number of parents

working (two income families, working

single parent) is probably the biggest

contributor. Students may say more

after school, extracurricular activities

contribute. According to the model,

2023 will be when we spend half of the

money we spend on food outside the

home.

Algebra II Indicators p. 44 Answer Key

Notes3.03

A. Two: -6 ≤ c < 6.125

One: c = 6.125

None: c > 6.125

B. Three: -45 ≤ c < -3

Two: c = -3

One: c > -3

None: there is always at least one real

zero

C. Four: e < 278

Three: 278 ≤ e < 279

Two: 279 < e < 446

One: 446 < e < 447

None: e > 447

3.04

A. 33 - 5x2

B. -25x2 - 30x - 3

C. 13

D. -318

E. 33 - 5x2

F.

x − 3

5

G. 3 5− x

Algebra II Indicators p. 45 Answer Key

Notes3.05

A. As b increases, the curves relocate,

curling right to left.

B. Same shape and y-intercept; reflections

of each other across the y-axis.

C. The curve opens up and the vertex

moves right and down (+x, -y); the

y-intercept remains at (0, 5).

Algebra II Indicators p. 46 Answer Key

NotesD. Open in opposite directions and intersect

at (0, c). The relationship between the

two curves could also be expressed as a

combination of two reflections, one

vertical and one horizontal or rotation

about (0, c).

E. The shape of the curve remains the

same; as c increases, curve moves up the

coordinate plane (+y).

Algebra II Indicators p. 47 Answer Key

NotesF. 1972

surplus: y1 > y

2

deficit: y1 < y

2

G. 1920-80 increasing circulation; 1980-98

decreasing circulation.

Peaked in 1980. According to the

equation, circulation approximated 45

million in 1950 and expects the same in

2011.

H. y-intercept: (0, 26)

x-intercepts: (2, 0), (6.5, 0)

vertex:

17

4

81

8,−

Algebra II Indicators p. 48 Answer Key

Notes3.06

A. 355 seconds; 90 seconds

B. 1997; 1994; Yes, 1999.

Algebra II Indicators p. 49 Answer Key

Notes3.07

A. As d increases, the function moves up

the coordinate plane (+y).

B. The functions have the same y-intercept

and are 180o rotations of the other. The

curves open in opposite directions.

C. As a increases, the turns in the curve

become less pronounced; straighter.

Algebra II Indicators p. 50 Answer Key

NotesD. As b increases, the turns in the curve

become more pronounced.

E. Low in early 1992 (1.3, 0.696); high in

late 1996 (4.9, 1.368); low in late 1998

(7.8, 0.975); increasing 1992-96, after

1998; decreasing 1996-98.

Prices fluctuate over the domain given.

Since the function increases after 1998,

you expect gas prices to increase the

next three years. According to the

function, gas prices will reach $2 in late

2000 (9.7, 2.015). Some variables that

affect the price of gasoline are labor

costs, technology, infrastucture, refining

capacity, availability of crude petroleum,

and political unrest in petroleum

producing countries.

Algebra II Indicators p. 51 Answer Key

NotesF. 3x3 + 2x2 = 9x + 6 given

x2(3x + 2) = 3(3x + 2) distributive property

x2(3x + 2) - 3(3x + 2) = 0 add/sub property of equality

(x2 - 3)(3x + 2) = 0 distributive property

x2 - 3 = 0 or 3x + 2 = 0 multiplicative property of zero

(x + 3 )(x - 3 ) = 0 3x = -2 add/sub property of equality

difference of squares (distributive)

x + 3 = 0 or x - 3 = 0 x = − 2

3mult/div property of equality

mult property of zero

x = − 3 or x = 3add/sub property of equality

Therefore x = 3 , − 3, or − 2

3

Notes

Algebra II Indicators p. 52 Answer Key

NotesG. As c increases, the curve closes and

looks more parabolic.

H. As e increases, the curve moves

vertcally (+y)

I. As a increases positively, the curve

closes and the turns are less pronounced.

Algebra II Indicators p. 53 Answer Key

Notes3.08

A. In the 7th month, $35.0645; in the 20th

month, $17.577; decreasing, then in-

creasing; by month 29, the metal reaches

a value of $69.737

B. 1956; the incidence of measles in 1940;

1940-44 decreasing; 1944-56 increasing;

1956-60 decreasing; 1963; increasing

population, vaccinations.

Algebra II Indicators p. 54 Answer Key

Notes3.09

A. N-shaped curve intersecting the x-axis

once at -2.

B. G(x) will intersect x-axis one unit to the

left of F(x).

C. H(x) will intersect the x-axis at 2, -2,

and -3.

D. Generally increasing functions;

N-shaped; H(x) shares an x-intercept

with F(x) and G(x).

Algebra II Indicators p. 55 Answer Key

Notes3.10

A. The function is decreasing except at

x = 2 where it is undefined.

Asymptotes at y = 0 and x = 2; intercept

at (0, -0.5)

The zero for y = x - 2 is where y

1

x 2=

−is undefined.

B. The verical asymptote moves to the left

(-x direction) as b increases.

.

Algebra II Indicators p. 56 Answer Key

NotesC. The horizontal asymptote moves up

(+y direction) as c increases.

D. For the domain x < 0.5, function is

increasing; for domain x > 0.5, function

is decreasing; undefined at x = -2 and x

= 3. The zeros for y = x2 - x - 6 are are

where y

1

x x 62=

− − is undefined.

Algebra II Indicators p. 57 Answer Key

NotesE. The vertical asymptotte moves to the left

(-x direction) as b increases.

F. The horizontal asymptote moves up

(+y direction) as c increases.

Algebra II Indicators p. 58 Answer Key

NotesG. The function is decreasing except at

x = -2 and x = 3 where it is undefined.

Asymptotes are x = -2, x = 3 and y = 0

for the domain x < -2 and x > 3.

y = x and y

x

x x 62=

− − share an

intercept at the origin; the zeros for

y = x2 - x - 6 are the vertical asymptotes

for y

x

x x 62=

− −;

the zero for y

x

x x 62=

− − is the vertical

asymptote for y

x x 6

x

2

= − −

Algebra II Indicators p. 59 Answer Key

NotesH. The function is increasing for x < 0 and

decreasing for x > 0 except for x = -2

and x = 3 where the function is

undefined. The origin is the only

intercept. The asymptotes are

x = -2, x = 3, and y = 0. y = x2 and

y

x

x x 6

2

2=

− − share an intercept at the

origin; the zeros for y = x2 - x - 6 are the

asymptotes for y

x

x x 6

2

2=

− −; the zero

for y

x

x x 6

2

2=

− − is an asymptote for

y

x x 6

x

2

2= − −

; y

x

x x 6

2

2=

− − and

y

x x 6

x

2

2= − −

share an asymptote,

y = 1, for domain x > 3.

Algebra II Indicators p. 60 Answer Key

NotesI. Decreasing for x < -2.1, increasing for

-2.1 < x < 0.8, and decreasing for x > 0.8

except at x = -3, -1, and 2 where the

function is undefined. Asymptotes at

x = -3, x = -1, x = 2, and y = 0. The

zeros for y = x 2x 5x 63 2+ − − are

where y =

1

x 2x 5x 63 2+ − − is

undefined.

Algebra II Indicators p. 61 Answer Key

NotesJ. Increasing for x < 0.36 and undefined at

x = -3, -1, and 2. Intercepts at (-2, 0)

and (0, − 1

3); asymptotes at x = -3,

x = -1, x = 2, and y = 0 for x > 2 and

x < -3. y = x + 2 shares an x-intercept

with y =

x + 2

x 2x 5x 63 2+ − −;

y =

x + 2

x 2x 5x 63 2+ − − and

y =

x 2x 5x 6

x + 2

3 2+ − − are undefined for

the same x-values; y =

x + 2

x 2x 5x 63 2+ − −is undefined where y = x3 + 2x2 - 5x - 6

has zeros.

Algebra II Indicators p. 62 Answer Key

NotesK. Decreasing for x < -7.4, increasing for

-7.4 < x < -2.2, decreasing for

-2.2 < x < 0.8, increasing for

0.8 < x < 8.8, and decreasing for x > 8.8;

undefined at x =-3, -1, and 2. Intercepts

at (-5, 0) and (5, 0); asymptotes are

x = -3, x = -1, and x = 2. y = x2 - 25 and

y =

x 25

x 2x 5x 6

2

3 2

−+ − −

share x-intercepts;

y =

x 25

x 2x 5x 6

2

3 2

−+ − −

and

y =

x 2x 5x 6

x 5

3 2

2

+ − −− 2

are undefined at

x = -3, -1, and 2. y = x 2x 5x 63 2+ − −has intercepts where

y =

x 25

x 2x 5x 6

2

3 2

−+ − −

is undefined.

Algebra II Indicators p. 63 Answer Key

NotesL.

− =

−2

1

x 42Given

− −( ) =2 x 4 12

Multiplication

property of

equality

x 4

1

2

2 − = − Division property

of equality

x

7

2

2 = Addition property

of equality

x =

7

2± Square root of

equivalent

expressions

M. F(0) = $12 million and

G(0) = $11.8 million; F(x) is increasing

for x > 1, G(x) is increasing for x > 1;

profits are increasing when the functions

are increasing. F is making money

beginning in the third year (x ≥ 3) and G

is making money beginning in the fifth

year (x ≥ 5). F is earning more than G

for the period 2-14 years.

Algebra II Indicators p. 64 Answer Key

Notes3.11

A. The x-intercept moves closer to the

origin and the curve gets steeper.

B. The curve moves to the left

(-x direction).

C. The curve moves up the coordinate plane

(+y direction) as c increases.

Algebra II Indicators p. 65 Answer Key

NotesD. As b increases, the curve moves to the

left (-x direction).

E. As b increases, the y-intercept moves up

(+y direction) and the curve flattens out.

F. Decreasing for x ≤ 2 and increasing for

x ≥ 3. Intercepts and minimums at

(-2, 0) and (3, 0); range only defined for

y ≥ 0; undefined for -2 < x < 3.

y = x x 62 − − and

y = 1

x x 62 − − are undefined for the

same domain.

Algebra II Indicators p. 66 Answer Key

NotesG. x 5x 4 2 x2 + + = − Given

x 5x 4 4 4x x2 2+ + = − +Square equivalent expressions

9x = 0

Add/sub properties of equality

x = 0

Division property of equality

H. 6x 4 x 1+ = + Given

6x 4 x 2x 12+ = + +Square equivalent expressions

0 = x2 - 4x - 3

Add/sub properties of equality

x 2 7= ±Apply quadratic formula; simplify

I. x = 5.3 (approx.)

J. x = -7.1 (approx.)

Algebra II Indicators p. 67 Answer Key

NotesK. No solution

L. Min = $17.80, max = $40.21; 14th

month; in the 27th month (26.2, 50.07)

M. Min = $605.39, max = $1042.05; in the

18th month (17.7, 799.94); in the 28th

month (27.75, 520.92)

Zoom view,

first quadrant

Algebra II Indicators p. 68 Answer Key

Notes3.12

A. 6.1 years; 7 years; 8.4 years

B.

1 1 1 1

216 36 6 1

1331 121 11 1

3375 225 15 1

77

2

3

6

a

b

c

d

1

=

− −− −

− −− −

0.001 0.004 0.005 0.002

0.046 0.12 0.11 0.036

0.459 0.849 0.555 0.165

1.414 0.733 0.45 0.131

C. Answers vary. An example is:

y

11

2x 10= −

y

7

3x

64

3= − +

y

4

5x

3

5= −

D. (3.09, 0.26, 0.27)

E. (1.71, 1.75, 3.08, -0.38)

Algebra II Indicators p. 69 Answer Key

Notes3.13

A. An example: y ≥ -3x - 20

y ≤ -3x - 5

y

3

4x

55

4≤ +

y

3

4x

25

4≥ +

B. 2 ≤ x ≤ 6

10 ≤ y ≤ 40

x + y ≤ 35

I(x,y) = 10x + 6.25y

I(6, 29) = $241.25

Algebra II Indicators p. 70 Answer Key

Notes3.14

A. The graph closes and moves closer to the

y-axis. The vertex approaches the origin

along the x-axis.

B. The graph moves away from the y-axis

(-x direction).

C. rhombus

Algebra II Indicators p. 71 Answer Key

NotesD. Answers vary. An example is:

y ≤ 9 - |4x - 6|

E. |3x - 4| ≤ 17 Given

3x - 4 ≤ 17 and 3x - 4 ≥ -17

Definition of inequality with respect to

absolute value

-17 ≤ 3x - 4 ≤ 17

Re-express inequalities as a compound

inequalitiy

-13 ≤ 3x ≤ 21

Addition property of inequalities

− ≤ ≤13

3x 7

Mult/div property of inequalities

F. Solve by graphing y = |16 - 3x| and

y = |x| + 3; x = {3.25, 9.5}

Algebra II Indicators p. 72 Answer Key

Notes3.15

A. As b increases, the curve becomes

steeper.

B. As b approaches zero, the curve

becomes steeper and approaches

y = 0 more quickly.

C. As a increases, the curve becomes

steeper.

Algebra II Indicators p. 73 Answer Key

NotesD.

log 5

log 3.1,0

( )( )

and (0, -8).

E. (2006, 754,468); 12 years;

y = 429316(1.07302039)x

(x = 0 for 1998)

F. $128.31; 7 months

Algebra II Indicators p. 74 Answer Key

NotesG. Let x = 0 for 1990. The data generates

y = 10498(1.178)x as the exponential

curve of best fit. Assuming that we are

in the 2001 model year, a new vehicle

costs $63,672. The depreciation

function, based on the age of the vehicle

in 2001, is y = 63672(0.849)x.

3.16

A. c 101.662=

B. d

6

e3.25=

C. x e ba= −

D.

x

logc

a

log b=

( )

Algebra II Indicators p. 75 Answer Key

Notes3.17

A. Solve 614.3 = 465.6R8 for R to

determine the annual growth rate.

R = 1.035.

614.3R7 = 782.9 million passengers in

2005.

B. y = 465.6(1.035)x is the model for airline

passengers in millions since 1990

(x = 0 for 1990).

C. $2500; $4.39

D. 350 200e2r= given

350

200e2r=

division property of equality

ln

350

2002r

=

law of logarithms (equivalent

expressions); definition of natural log

ln350

200

2r

=

division property of equality

0.2798 ≈ r

simplify expression

Algebra II Indicators p. 76 Answer Key

NotesE. 960 1 0758= ×A . given

960

1.075A

8=

division property of equality

538.274 ≈ A

simplify expression

F. 663 49 2.165x= × given

663

492.165x=

division property of equality

log

663

49log 2.165x

= ( )law of logarithms

(equivalent expressions)

log

663

49xlog 2.165

= ( )law of logarithms

(exponential expressions)

log663

49

log 2.165

( ) = x

division property of equality

3.372 ≈ x

simplify expression

Algebra II Indicators p. 77 Answer Key

Notes4.01

A. 170; y = 0.5x2 - 1.5x

B. Both sets of data are decreasing over

time, flattening out for the last several

Olympics.

The women’s performance have

improved more over the domain shown.

Training, health and nutrition, and

increased number of swimmers

competing are a few independent

variables affecting the swimmers’

performances.

Men: y = 337.1(0.9956)x

(x = 24 for 1924)

Women: y = 399.5(0.9946)x

(x = 24 for 1924)

Women showed the greater

improvement, 118.35 seconds compared

to the men’s 83.61 seconds, within the

domain shown.

Accoring to the models: men (2004,

216.58), women (2004, 232.20)

The actual results at the 2004 Athens

Olympics: men, 223.10; women, 245.34.

According to the models, the winning

woman at the 2068 or 2072 Olympics

(depends how much you round off

constants in the algebraic model) will

outpreform her male counterpart.

Reality may be a different matter. It

certainly would be an appropriate topic

to address in conjuction with Health/PE,

Anatomy/Physiology, Allied Health

Sciences, Biomedical Technology, or

Sports Medicine classes.

Algebra II Indicators p. 78 Answer Key

NotesC. y = 0.008x - 0.481 (x = 58 for 1958);

(2005, 0.38)

According to the model, first class

postage increases $0.008 annually.

Technology, fuel, and labor costs are

independent variables that affect postage

rates.

Algebra II Indicators p. 79 Answer Key

Notes4.02

A. For ther domain given, the quadratic

best-fit for Imports appears to fit the best

and the correlation coefficient compares

favorably to the exponential (0.9983

compared to 0.9786). Imports:

y = 0.883x2 - 118.696x + 4037.4

(x = 70 for 1970)

Imports: Exponential Best Fit

Imports: Quadratic Best Fit

Algebra II Indicators p. 80 Answer Key

NotesFor ther domain given, the quadratic

best-fit for Exports appears to fit the best

and the correlation coefficient compares

favorably to the exponential (0.9935

compared to 0.9809). Exports:

y = 0.691x2 - 94.406x + 3286.6

(x = 70 for 1970)

Trade was balanced, according to the

best models, in 1972.

Exports: Exponential Best Fit

Exports: Quadratic Best Fit

Algebra II Indicators p. 81 Answer Key

NotesB. The exponential and quadratic curves of

best fit both appear to fit the data well.

The coorelation coefficients are virtually

the same, 0.9980 and 0.9986

respectively. The graph of the residuals

shows both curves fitting well, although

the quadratic has one distinctly bigger

“bump”. The exponential model,

y = 0.394(1.0144)x (x = 0 for 1790),

predicts the 2000 population most

accurately (7.923 million estimated

versus 8.049 million actual).

Exponential Best Fit

Quadratic Best Fit

Exponential Residuals

Quadratic Residuals

Algebra II Indicators p. 82 Answer Key

NotesC. Since the function is always decreasing,

an exponential best-fit is probably the

best model.

y = 23200(0.619)x

(x = 6 for September 6). According to

the model, 38.1% of the customers have

their power restored daily. Power

restored to all customers (<1000) on

September 21.

D. Within the domain provided, the data

increases and then decreases, hence a

quadratic curve as the choice for the best

fit curve.

y = -0.4786x2 + 79.67x - 619.36

(x = 0 for 1900)

According to the model, US petroleum

reserves will be exhausted by 2058-59.

Technology, price, and political/eco-

nomic situations in other petroleum

producing countries are some variables

that affect production.

For 1859, x = -61, the model provides

irrelevant data. The data can be adjusted

to include 1859 (x= 0 for 1859) with the

petroleum production for that year. It

appears there will be no simple algebraic

model for the data

Algebra II Indicators p. 83 Answer Key

Notes4.03

A. Solve 144 = 106.3R12 for R to find the

growth rate for the domain identified.

(R = 1.0256)

Solve 144Rx = 281.4 for x to determine

how long until all Americans are

connected.

x = 26.5 months (October 2002).

B. Solve 44568 = 20102R50 for R to find

the annual family income growth.

(R = 1.016)

Solve 44568Rx = 50000 for x to deter-

mine how for family median income to

reach $50,000. x = 7.2 years (2004).

C. 3454(1.0448)3 = $3939.33

Solve 4500 = 3454(1.0448)x for x to

determine how long it will take tuition to

reach $4500 at the growth rate indicated.

x = 6.0 years (2010-11)

Algebra II Indicators p. 84 Answer Key

Notes4.04

A. Total Value of Goods =

242.8 271.6 290.1 320.0 331.1

46.1 57.3 63.5 75.5 85.5

50.9 59.2 62.4 67.6 76.5

172.7 187.8 182.8 187.2 179.9

100.3 108.4 131.1 157.3 173.7

Balance of Trade =

-14.0 -17.2 -21.7 -16.4 -18.5

-27.5 -33.7 -39.5 -49.7 -56.9

-12.5 -14.4 -15.4 -18.6 -23.2

-65.7 -59.2 -47.6 -56.2 -64.1

1.3 -15.8 -17.5 -14.5 -15.7

B.

1+ a 3 + a 2 + a

2 + b 4 + b 5 + b

;

1+ a 3 + a 2 + a

2 + b 4 + b 5 + b

;

4 6 5

8 10 11

;

1 5 3

2 6 8

Algebra II Indicators p. 85 Answer Key

NotesC.

2 0

0 3

=

1 3 2

2 4 5

2 6 4

6 12 15

D.

21 3 2

2 4 5

7 7 7

1 1 1

+

− − −

or

21 3 2

2 4 5

7 0

0 1

1 1 1

1 1 1

+

E.

1 0

0 1;

1 0

0 1−

F.

1 3 2

2 4 5

9 0

0 6

1 1 1

1 1 1

+

−−