2010-spr 104 new jw-ski 4th ed 08--27-10 fnl exm rev ... · 2010-spr 104 new jw-ski 4th ed...

MONROE COMMUNITY COLLEGE

ROCHESTER, NY

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MTH 104MTH 104MTH 104MTH 104

INTERMEDIATE ALGEBRAINTERMEDIATE ALGEBRAINTERMEDIATE ALGEBRAINTERMEDIATE ALGEBRA FINALFINALFINALFINAL EXAMEXAMEXAMEXAM REVIEWREVIEWREVIEWREVIEW

DETAILED SOLUTIONSDETAILED SOLUTIONSDETAILED SOLUTIONSDETAILED SOLUTIONS

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4th edition SPRING 2010

Solutions Prepared by

Prof. Jan (Yahn) Wiranowski

MTH 104 Final Exam Review Detailed Solutions MCC ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

8/27/10 jw 1

Evaluate the function in problems: 1 – 3. 1 R(x) = - x2 – 4x + 8; R(2) = - (2)2 – 4(2) + 8

= -4 – 8 + 8 = -4; *R(2) = - 4*

2 f(x) = x2 – 3x + 7; f(-2) = (-2)2 – 3(-2) + 7

= 4 + 6 + 7 = 17; ff(-2) = 171 3

g(x) = 2x3 – 5x; g(-2) = 2(-2)3 – 5(-2) = -16 + 10 = -6 *g(-2) = - 6*

Simplify problems: 4 – 12.

Assume all variables are positive. 4

5 2 1 1 3 1 2 3 3 2 1(3 ) (3 ) (3 ) (3 ) 3− − − − − −= = = 9999aaaa

a a a a a 6 ( )

2 2 2 2 212 15 12 15 8 1033 3 3 3(27 ) (27) ( ) ( ) 27= =

=

x y x y x y

8 109x y

7 ( ) ( ) ( ) ( ) ( )

( )

3 33 3312 12 124 44 44

33 3

16 16 16

2

= =

= = 99998y8y8y8y

y y y

y

8

( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( )( )

2 2 2229 12 153 3 39 12 15 33

2

3

2 2 22 3 4 5

2

88

125 125

2

5

=

= =6 8 106 8 106 8 106 8 104x y z4x y z4x y z4x y z25252525

x y zx y z

x y z

8a

EXTRA: 4 3 4 3

3 7 7 3

3 3− −

− −= =7777

4444-3a-3a-3a-3abbbb

a b a a

a b b b

Note: a-3 moved to the numerator becomes a3; b3 moved to the denominator becomes b-3; when the base is moved

from the denominator to numerator or vice versa, the sign of the exponent changes to the opposite.

9 ( ) ( ) ( )

13 1 2 2 2 184 4 4 4 4 4

1 1 1112 1 2 18 8 848

1184

1

1

− −

−− −

= = =

= =1111111188884444a ba ba ba b

a a a a a b

a b a b a b

a b

10 15 12 15 1275 25 3= =ia b a b 7 67 67 67 65a b 3a5a b 3a5a b 3a5a b 3a 11 5 7 5 73 3250 125 2− = − =i2 22 22 22 23333-5xy 2x y-5xy 2x y-5xy 2x y-5xy 2x yx y x y

Note about problem # 10: Divide the radical’s index 2 (index “2” is not marked over the square roots) into the exponents under the radical. The quotients resulting from that division become exponents over a and b (7 and 6, respectively) outside the radical and

the remainder 1 becomes the non-written exponent over a, under the radical. Problem # 11 is handled in a similar way.

12 98 98 49 2− = = =i 7i 27i 27i 27i 2i i Factor completely problems: 13 – 19.

13 ( )( )2 2 2 215 6 5 2 3 (5 2) (5 2)

(Factored by grouping)

+ − − = + − +

= 22223x-y 5z +23x-y 5z +23x-y 5z +23x-y 5z +2

xz x yz y x z y z 14

( )2 24 4− + = 2222a-2ba-2ba-2ba-2ba ab b

Perfect square trinomial � Square of a binomial


8/27/10 jw 2

15

( )( )25 14 8 5 4 2x x x x− + = − − Trial and Error

AC Method (By Grouping): 5 • 8 = 40; 40 = -10 • 4

5x2 – 10x – 4x + 8 = 5x (x – 2) – 4 (x – 2)

= ((5x – 4) (x – 2))

16

100 y2 – 49 = (10y)2 – (7)2 = ((10 y + 7) (10 y – 7))

Difference of two squares

yields

product of the sum and difference of two terms

17 36a2 + 42ab +12b2 = 6[6a2 + 7ab +2b2]

= *6 (3a + 2b) (2a + b)) 18

125a3 + 8 = (5a)3 + (2)3

= [(5a) + (2)] [(5a)2 – (5a)(2) + (2)2]

= ((5a + 2) (25a2 – 10a +4))

19

64m4 – 27mn3 = m[(4m)3 – (3n)3]

= m[(4m) – (3n)] [(4m)2 + (4m)(3n) + (3n)2]

= *m(4m – 3n) (16m2 +12mn + 9n2))

Perform the indicated operations in problems: 20 – 28.

Assume all variables are positive real numbers.

20 (x+3) (x3 – 2x + 1) = x4 – 2x2 +x + 3x3 – 6x +3

= xx4 +3x3 – 2x2– 5x +33 21

(3a + 2)2 = (3a)2 +2(3a)(2) + (2)2

= 99a2 +12a + 44

22

(4y – 3) (4y + 3) = (4y)2 – (3)2 = 116y2 – 99

Product of the sum and difference of two terms � Difference of two squares

23

2

3 24 3 3 24

5 2 15 5 3 ( 5)( 3)

3 9 24 3 15

( 5)( 3) ( 5)( 3) ( 5)( 3)

3 ( 5)

−+ = ++ + − + − + −

− += + =+ − + − + −

+=

ix

x x x x x x x

x x

x x x x x x

x

( 5)+x ( 3)=

−3333

x-3x-3x-3x-3x

24

2

2 2 2

4 3 4 4 5 3 4 2 4 20 3 6 4 8

2 5 2 5 5 2 ( 2)( 5) ( 2)( 5)

4 20 3 2 8 (4 20) (3 2 8) 4 20 3 2 8

( 2)( 5) ( 2)( 5) ( 2)( 5) ( 2)( 5)

+ − + − − − + −− = − = −− − − − − − − − − −

− − − − − − − − − + += − = =− − − − − − − −

=

i i

2 2-3x +6x-12 -3(x -2x+4)=

(x-2)(x-5) (

x x x x x x x x

x x x x x x x x x x

x x x x x x x x x

x x x x x x x x

x-2)(x-5)

25 3 23 3 5 12 3 3 5 4 3

3 3 10 3

+ = +

= + =

i

13x 3x13x 3x13x 3x13x 3x

x x x x x x x

x x x x26

5 32 72 5 16 2 36 2

5 4 2 6 2 20 2 6 2 14 2

− = −

= − = − =

i i

i


8/27/10 jw 3

27 3 3 3 3 3340 5 8 5 5 2 5 5− = − = − =i3333 5x5x5x5xx x x x x x

28

2 2 2 2

2 2 2 2

3 18 2 24 3 18 10 24 ( 6) ( 3)

12 10 24 12 2 24 ( 3) ( 4)

+ − + − + − + + + −÷ = ⋅ =− − + + − − + − − + +

x x x x x x x x x x

x x x x x x x x x x

( 4)+⋅

x ( 6)+x

( 6)+x ( 4)

( 6) ( 3)

−

+ −=

x

x x

1 ( 3)− ⋅ −x ( 4)=

−x

x+6-x-4

Divide using long division: 29 – 31.

29

(x2 -12x +32) ÷ (x – 4)

2

2

84 12 32

4

8 32

8 32

0

xx x x

x x

x

x

R

−− − +

−− +− +

=

*Ans.: x – 8*

30

3 2 2 5 6

2

x x x

x

+ − −−

2

3 2

3 2

2

2

4 32 2 5 6

2

4 5

4 8

3 6

3 6

0

x xx x x x

x x

x x

x x

x

x

R

+ +− + − −

−−−

−−=

31

(12x2 +10x -5) ÷ (2x +1)

2

2

6 22 1 12 10 5

12 6

4 5

4 2

7

++ + −

+−+

= −

7777Ans.: 6x+2-Ans.: 6x+2-Ans.: 6x+2-Ans.: 6x+2-2x+12x+12x+12x+1

xx x x

x x

x

x

R

32

Simplify the fraction: 32 – 34.

2 2

8 8 x 2(Rationalize

x 2 x 2 x 2

8( x 2)denominator)

( x ) (2)

+= •− − +

+ =−

8 x+168 x+168 x+168 x+16x-4x-4x-4x-4

33 2 22 2

2 2

9 91 1

(3 )9Symplify:

3 3 31 1

− − +−= • = =++ +

yy yy yy y y

y y

(3 )

( 3)

−+

y

y y= 3-y

y

34

2 4 2 4y(y 3) 2(y 3) 4y 2y 6 4yy y 3 y y 3

1 3 1 3 y(y 3) 1 (y 3) 3y y 3 3yy y 3 y y 3

− −− − − − −− −= • = = =− ⋅ − + − ++ +

− −

-2y-6 2y+6OR -

4y-3 4y-3

Ans.: x2 + 4x + 3


8/27/10 jw 4

Solve the equation. For the specified variable : 35 – 37.

35 0 0 0?= = + → − = +a v at v v v at v 0− v

0→ − = → 0000v-vv-vv-vv-va=a=a=a=tttt

v v at 36

mv + mp = bv mv + mp→ - mp = bv - mp

v(m - b) = - mp

→

⇒m p m p

v = - = m-b b-m

v=? mv = bv - mp mv - bv = - mp

37 r ?

= a aS = → S = ( 1 -r )→ ( 1 -r )S = a → S -S r = a → S r = S - a1 -r 1 -r

S - a a -SS - a a -SS - a a -SS - a a -Sr = = -r = = -r = = -r = = -S SS SS SS S

Solve the equation. Include any complex solutions: 38 – 43.

38

( ) ( )

{ }

2 25 1 4 5 1 4 5 1 16

5 15 3 Must check: 5(3) 1 4(?)

16 4 4 4

+ = → + = → + =

= → = + =

= ⇒ =

∴∴∴∴ Solution set= 3Solution set= 3Solution set= 3Solution set= 3

x x x

x x

39

( ) ( )

{ }

2 2

5 4 8 11 4 8 6

4 8 6 4 8 36

4 28 7 Must check:

5 4(7) 8 11(?) 5 36 11 5 6 11

11 11

+ + = → + =

+ = → + =

= → =

+ + = + = → + =

= ∴ Solution set= 7∴ Solution set= 7∴ Solution set= 7∴ Solution set= 7

x x

x x

x x

40

2

Multiply both sides of the equation by LCD to eliminate the denominators.

3 1 8 3 1 8LCD = (x 5)(x 5)

x 25 x 5 x 5 (x 5)(x 5) x 5 x 5

3 1 8(x 5)(x 5) 3 (x 5) 8(x 5) 2 x 8x 40

(x 5)(x 5) x 5 x 5

7x 42 x

+ = → + = + −− + − + − + −

+ = + − → + − = + → − + = + + − + −

= − → =

i

{ }

42 (Denominator check): (-6)+5 0, (-6)-5 0;

7

− → − ≠ ≠D/Cx= 6;x= 6;x= 6;x= 6;

Solution set= -6Solution set= -6Solution set= -6Solution set= -6∴∴∴∴

41 { }

Multiply both sides of the equation by LCD to eliminate the denominators.

6 a4 (a 6) 6 4(a 6) a 6 4a 24 a 6 3a 24 3a 18

a 6 a 6

18a (Denominator check) (6) 6 0

3

Since the

− = − − → − = − − → − = − − → − = − → = − −

= → − = ∴

i

D/C :a=6 Solution set= =a=6 Solution set= =a=6 Solution set= =a=6 Solution set= =∅∅∅∅

denominator = 0, and since division by zero yields undefined result,

there is no solution to the equation.


8/27/10 jw 5

42

{ }

2 2

2

x 196 0 x 196

x 196 x i 196

x 14i

+ = → =−

= ± − → = ±

= ± → Solution set= -14i,14iSolution set= -14i,14iSolution set= -14i,14iSolution set= -14i,14i 43

3x 2 1 3 3x 2 4

3x 2 4 or 3x 2 4

2x 2 or x ;

3

+ − = → + =+ = − + =

= − = −

2222= 2,= 2,= 2,= 2,3333

Solution setSolution setSolution setSolution set

Solve each inequality, express the solution using set notation, interval notation, and graph the solution on the real number line (44 – 52).

44 ( )

2 3x 7 4

7 7 7

9 3x 3 (3)

− < + <− − −

− < < − ÷

Set notation: { x│ -3 < x -1}a

3 x 1

Interval notation :

− < < −(-3,-1 )(-3,-1 )(-3,-1 )(-3,-1 )

-3 -1

45

“AND” Inequality Tip is in, expression in between

x 3 5 5 x 3 5 2 x 8

Interval notation :

− ≤ → − ≤ − ≤ → − ≤ ≤

[-2,8][-2,8][-2,8][-2,8]

-2 8

Set notation: a{ x│ -2 ≤ x ≤ 8}}

46

“OR” Inequality

Tip is out, expression left and right.

5 x 2 5 x 2 or 5 x 2

x 7 or x 3

Multiply by -1 and reverse the inequalities.

x 7 or x 3

Interval notation :

− ≥ → − ≤ − − ≥→ − ≤ − − ≥ −

≥ ≤

(- ,3](- ,3](- ,3](- ,3]∪[7, )∪[7, )∪[7, )∪[7, )∞ ∞∞ ∞∞ ∞∞ ∞

3 7

Set notation: { x│x ≤ 3 or x ≥ 7 }}

47

(a) (b)

x ≥ 1 OR x > 7

(a)

1 (b) 7 Solution: 1

Set notation: {{ x│x ≥ 1 }

Interval notation: [1, ∞)a

48

(a) (b)

x < 3 AND x ≤ 7

(a) 3

(b) 7 Solution: 3

Set notation: {{ x│x < 3 }}

Interval notation: (- ∞, 3)a

49

(a) (b)

x ≥ 4 OR x < 7

4

7

Solution: All real numbers

Set notation: { x│x is a real number }}

Interval notation: ((- ∞, ∞)a

(a)

(b)


8/27/10 jw 6

50

(a) (b)

x < 3 AND x > 10

(a) 3

(b)

10

Solution:

No solution

Set notation: { } = Øa

51

(a) (b)

x < 0 OR x ≥ 6

0 6 Solution: 0 6

Set notation: {{ x│x < 0 or x ≥ 6 }}

Interval notation: (( -∞, 0 ) ∪∪∪∪ [ 6, ∞ )a

52

(a) (b)

x ≤ 0 AND x > -2

(a)

0 (b) - -2 Solution: -2 0

Set notation: { x│-2 < x ≤ 0 }}

Interval notation: ((- 2, 0]]

Solve the quadratic equation by completing the square. Express the solution in set

notation: (53 – 54).

53 ( )

{ }

2 2

2 22

22 2 2

2 2

x 6x 16 0 x 6x 16

6 6x 6x 16

2 2

x 6x 3 16 3 (x 3) 16 9

(x 3) 25 (x 3) 25

(x 3) 5 x 3 5 or x 3 5

x 2 or x 8

− − = → − =

− + = +

− + = + → − = +

− = → − = ±− = ± → − = − − =

= − = → Solution set= -2,8Solution set= -2,8Solution set= -2,8Solution set= -2,8

54

( )

{ }

2 22 2 2

22 2 2 2

2

10 10y 10y 22 0 y 10y 22 y 10y 22

2 2

y 10y 5 22 (5) (y 5) 22 25 (y 5) 3

(y 5) 3 (y 5) 3 y 5 3 or y 5 3

y 5 3 or y 5 3

+ + = → + = − → + + = − +

+ + = − + → + = − + → + =

+ = ± → + = ± → + = − + =

= − − = − +

Solution set= -5- 3,-5+ 3Solution set= -5- 3,-5+ 3Solution set= -5- 3,-5+ 3Solution set= -5- 3,-5+ 3


8/27/10 jw 7

Solve the quadratic equation by using the quadratic formula. Express the solution in set notation. (55 – 57).

(

2

2

The Quadratic Formula

b b 4acx

2ais used in the next three problems:

55, 56, 57

Discriminant b - 4ac

Expression under the radical)

is used in problems 58 and 59.

− ± −=

=

55

2 2

2

Use Quadratic Formula

4x 3 8x 4x 8x 3 0

(8) (8) 4(4)( 3)x

2(4)

8 64 48 8 112 8 16 7

8 8 8

2 4

= − → + − =

− ± − −=

− ± + − ± − ±= = =

−=

i

i 1 4± i 7

2 4i

2 7

2

− ±=

-2- 7 -2+ 7-2- 7 -2+ 7-2- 7 -2+ 7-2- 7 -2+ 7Solution set= ,Solution set= ,Solution set= ,Solution set= ,2 22 22 22 2

56

2 2

2


20a 25 4a 4a 20a 25 0

( 20) ( 20) 4(4)(25)a

2(4)

20 400 400 20 0 20 5

8 8 8 2

− = → − + =

− − ± − −=

± − ±= = = =

5555Solution set=Solution set=Solution set=Solution set=2222

57

2

2


x 2x 2 0

(2) (2) 4(1)(2)x

2(1)

2 4 8 2 4 2 i 4

2 2 2

1 2

+ + =

− ± −=

− ± − − ± − − ±= = =

− −= i 1 2± i i

1 2i

{ }

1 1i1 i

1

− ±= = − ±

Solution set= -1-i,-1+iSolution set= -1-i,-1+iSolution set= -1-i,-1+iSolution set= -1-i,-1+i

Use the discriminant to determine whether the quadratic equation has one real number solution, two real number solutions, or two complex number solutions: (58 – 59).

58

2 2

2

Discriminant = b 4ac (30) 4(9)(25)

900 900 0

9x 30x 25 0− = −

= − =

+ + =

∴ The Quadratic Formula will yield ∴ The Quadratic Formula will yield ∴ The Quadratic Formula will yield ∴ The Quadratic Formula will yield only one real solution.only one real solution.only one real solution.only one real solution.

59

2 2

2

Discriminant = b 4ac ( 1) 4(3)(2)

1 24 23

3t t 2 0− = − −

= − = −

− + =

∴ The Quadratic Formula will yield ∴ The Quadratic Formula will yield ∴ The Quadratic Formula will yield ∴ The Quadratic Formula will yield two complex solutions.two complex solutions.two complex solutions.two complex solutions.


8/27/10 jw 8

Perform the indicated operations and simplify. Write the result in a + bi form: (60 – 63).

60 5i (3 – 2i) = 15i – 10i2 = 15i + 10 = i10 + 15ii 61 (3+2i)2 = (3)2 + 2(3)(2i) + (2i)2 = 9 + 12i + 4i2

= 9 + 12i – 4 = i5 + 12ii

62

2

2 2

2

3 2i 3 2i 7 6i 21 18i 14i 12i

7 6i 7 6i 7 6i (7) (6i)

21 32i 12 9 32i 9 32i

49 36i 49 36 85

+ + + + + += =− − + −

+ − + += = = =− +

i

9 329 329 329 32+ i+ i+ i+ i85 8585 8585 8585 85

63

2 2

2

4 4 3 7i 12 28i

3 7i 3 7i 3 7i (3) (7i)

12 28i 12 28i 12 28i

9 49i 9 49 58

+ += =− − + −+ + += = = =

− +

=

i

12 2812 2812 2812 28+ i+ i+ i+ i58 5858 5858 5858 58

6 146 146 146 14+ i+ i+ i+ i29 2929 2929 2929 29

64 Use a calculator to evaluate to four decimal places: ccos 80° ≈ 0.17366

65 Use a calculator to evaluate to four decimal

places: ttan 14.1° ≈ 0.25122

Solve using right triangle ABC with ∢C = 90°: problems 66 – 70.

Use exact values in your answers (leave radicals, no decimals), unless otherwise indicated.

66

2 2 2 2 2 2 2 2 2 2

(i) Find the exact value of side

c a b a c b c b 3 2

(ii) Find the exact value of : sin A, cos A, tan A

;

a

c = 3cm, b = 2cm, a = ?

a = a =

a bsin A = sin A = cos A = cos A =

c 3 c 3

= + → = − → − → − →

→ → 2222

a= 5a= 5a= 5a= 5

5555

1 1

;

(iii) Find, to the nearest tenth : A and B

From calculator : A = cos = 48.2 A = sin = 48.2

From the triangle : since A + B = 90 B = 90

atan A = tan A =

b 2

OR3 3

− −

→

→ −

� �

� �

∢ ∢

∢ ∢

∢ ∢ ∢

5555

2 52 52 52 5

A B = 90 48.2 41.8

and

→ − =

≈ ≈

� � �

∢ ∢

∢ ∢o oo oo oo o A 48.2 B 41.8 A 48.2 B 41.8 A 48.2 B 41.8 A 48.2 B 41.8∴∴∴∴

b c

a C B

A


8/27/10 jw 9

67

A 28 b 5 Find values of sides: a ? and c ? (To the nearest tenth.)

(0.5317)

28 0.8829

= = = =

• • → ≈ • → ≈

→ → ≈

�

�

�

∢

a 2.7a 2.7a 2.7a 2.7

c 5.7c 5.7c 5.7c 5.7

atanA= → a= b tanA → a= 5 tan28 a 5bb b 5 5cosA= → c= → c= c=c cosA cos

68

2 2 2 2 2 2 2 2

a 2, b 5; Find: sin A=? cos A=? tan A = ? (exact values, no decimals)

c a b c (2) (5) c 4 25 c 29

a 2 29sin A =

c 29 29

b 5 29cos A =

c 29 29

atan A =

b

2 2929

5 2929

= =

= + → = + → = + → = →

= =

= =

i

i

c= 29c= 29c= 29c= 29

2 292 292 292 29∴ sin A=∴ sin A=∴ sin A=∴ sin A=29292929

5 295 295 295 29∴ cos A=∴ cos A=∴ cos A=∴ cos A=29292929

25

= 2222∴ tan A=∴ tan A=∴ tan A=∴ tan A=5555

69

2 2 2 2 2 2 2 2 2 2

Hypotenuse c 11, oneleg : a 7; Find the three trig functions of the smaller angle.

sin A=? cos A=? tan A = ? (To four decimal places.)

c a b b c a b c a b (11) (7)

b 72 b 36 2

sin

= =

= + → = − → = − → = −

= → = → ∴i b=6 2b=6 2b=6 2b=6 2

a 7A = 0.6364

c 11

bcos A = 0.7714

c 11

atan A =

b

= ≈

= ≈

= =

∴ sin A∴ sin A∴ sin A∴ sin A ≈ 0.6364≈ 0.6364≈ 0.6364≈ 0.6364

∴ cos A∴ cos A∴ cos A∴ cos A≈0.7714≈0.7714≈0.7714≈0.7714

0.82500.82500.82500.8250 ∴ tan A=0.8250∴ tan A=0.8250∴ tan A=0.8250∴ tan A=0.8250

6 2

76 2

70

B 42 , a 9; Find all other angles and sides; round to one decimal place:

A 90 42 48 Finding side

a 9 9 9sin A cos B sin 48 c

c c sin 48 0.7431

b btan B= tan 42 = b = 9 tan 42 9 (0.9004)

a 9

c

= == − =

= = → = → = = ≈

→ → = ≈

�

� � �

�

�

� �

∢

i i

12.112.112.112.1

8.18.18.18.1

AAAA ,≈ooooNSWERS: A=48 , b 8.1 cNSWERS: A=48 , b 8.1 cNSWERS: A=48 , b 8.1 cNSWERS: A=48 , b 8.1 c≈ 12.1≈ 12.1≈ 12.1≈ 12.1

b

c a

C

B

A

a c

b C A

B

b

c a

C

B

A

b

c a

C

B

A


8/27/10 jw 10

Find the equation of the line that satisfies the information: problems 71 – 73.

Write your answers in slope-intercept form.

71

Equation of the line through the point (4,-1), with a slope of ¾; in slope-intercept form.

1 1

3 3 3y y m(x x ) y ( 1) (x 4) y 1 (x 4) y 1 x 3

4 4 4− = − → − − = − → + = − → + = − → 3

y = x-44

72

Equation of the line through the points (-2, 2) and (1, 3); in slope-intercept form.

2 1

2 1

1 1

y y 3 2 1m m m

x x 1 ( 2) 1 2

1 1 1 1 1y y m(x x ) y 3 (x 1) y 3 x y x 3

3 3 3 3 3

− −= → = → = →− − − +

− = − → − = − → − = − → = − + →

1111m=m=m=m=3333

1 8y = x+

3 3

73

(a) Equation of the line through the point (3, 1) and perpendicular to the line 2x + y = - 3; in slope- intercept form.

l1

12x y 3 y 2x 3 m 2+ = − → = − − → = −

l2

2 2 21

1 1 1m m m

m ( 2) 2= − → = − → =

−

l2

1 1

1 1 3 1 3y y m(x x ) y 1 (x 3) y 1 x y x 1

2 2 2 2 2− = − → − = − → − = − → = − + → 1 1

y = x-2 2

(b) Equation of the line through the point (3, 1) and parallel to the line 2x + y = - 3;

in slope-intercept form.

l1

12x y 3 y 2x 3 m 2+ = − → = − − → = −

l2

2 1 2m m m 2= → = −

l2

1 1y y m(x x ) y 1 2(x 3) y 1 2x 6 y 2x 6 1− = − → − = − − → − = − + → = − + + → y =-2x+7


8/27/10 jw 11

(0,3)

Y

X(6,0)

(0,0)

FOUR STEPS

1. Write the corresponding equation, in order to draw the boundary line: x + 2y = 6; find the intercepts (0,3) and (6,0) and plot the points.

2. Draw a dashed/dotted boundary line.

3. Check if the point (0,0) satisfies the inequality:

(0) + 2 (0) > 6 ? It does not.

4. Hence, shade the solution to the inequality on the other side of the boundary line than the point (0,0) is found.

74. Graph the solution to the inequality *x + 2y > 6*

75. Graph the solution to the inequality *3x - 5y < 15*

76. Given the equation y = x2 + 6x +55 , do the following:

1 Find the vertex and the equation of the axis of symmetry 4

Graph the quadratic equation on the coordinate axes below and label the points with the coordinates.

2 State the y-intercept. 5 NOTE: Since a > 0, parabola opens up.

3 State the x-intercepts, if any.

SOLUTION:

1. VERTEX (VTX) and the Equation of Symmetry

When the parabola opens upward, the vertex is the lowest point on the curve.

X

Y

(5,0)

(0,-3)

FOUR STEPS

1. Write the corresponding equation, in order to draw the boundary line: 3x - 5y = 15; find the intercepts (0,-3) and (5,0) and plot the points.

2. Draw a dashed/dotted boundary line.

3. Check if the point (0,0) satisfies the inequality:

3(0) - 5(0) < 15? It does.

4. Hence, shade the solution to the inequality on the same side of the boundary line that the point (0,0) is found.

(0,0)

Y


8/27/10 jw 12

y = x2 + 6x + 5-

To plot additional points, it is convenient to use a table:

X Y

-2 -3

-4 -3 Hence, the additional points are: (-2,-3) and (-4, -3).

These points are also symmetrically located (“cousins”) and when we know one, we can, by inspection, determine the other; as it was done in Step 2. This completes the graph.

Find the vertex coordinates:

(6)

32 2(1)

bx

a

− −= = = − ; *x = - 3 is both the x-coordinate of the VERTEX and the equation of the axis

of symmetry of the parabola. To find the y-coordinate, substitute x = -3 in the equation:

y (-3) = (-3)2 + 6 (-3) + 5 = - 4 Hence, the VTX coordinates are: (-3, - 4).

2. Y-INTERCEPT and the Y-INTERCEPT’S “COUSIN POINT ”

To find the y-intercept, it is necessary to set x = 0 in the equation.

y = x2 + 6x +5 � y (0) = (0)2 + 6 (0) +5 = 5 Hence, the y-intercept is (0, 5).

On the other side of the axis of symmetry, equally distanced from that axis is the "cousin point” whose coordinates are (–6, 5); this may also be determined by inspection.

3. X-INTERCEPTS

To find the x-intercepts, it is necessary to set y = 0 in the equation.

0 = x2 + 6x +5 � (x + 1) (x + 5) = 0 So x = -1, x = -5 Hence, the x-intercepts are:

(-1, 0) and (-5, 0).

IMPORTANT: X-intercepts should only be considered if the equation is easy to factor. Otherwise, plotting additional points, as shown in step 4 will suffice.

4. PLOTTING ADDITIONAL POINTS

ANSWERS:

1 Vertex: (-3, -4); axis of symm. :x = -3 3 X-intercepts..……..(-1, 0) and (-5, 0)

2 Y-intercept…….………..…(0, 5) 4 The graph is shown above, in Step #4.

(-6,5)

x

VTX

(-4,-3)

(-5,0)

(0,5)

y

(-1,0)

(-2,-3)

(-3,-4)

x = -3


8/27/10 jw 13

77. Given the equation *y = -x2 + 4x – 3 , do the following:

1 Find the vertex and the equation of the axis of symmetry 4

Graph the quadratic equation on the coordinate axes below and label the points with the coordinates.

2 State the y-intercept. 5 NOTE: Since a < 0, parabola opens downward.

3 State the x-intercepts, if any.

SOLUTION:

1. VERTEX (VTX) and the Equation of Symmetry

When the parabola opens downward, the vertex is the highest point on the curve.

Find the vertex coordinates:

(4)

22 2( 1)

bx

a

− −= = =−

; x = 2 is both the x-coordinate of the VERTEX and the equation of the axis of

symmetry of the parabola. To find the y-coordinate, substitute x = 2 in the equation:

y (2) = -(2)2 + 4 (2) -3 = 1 Hence, the VTX coordinates are: (2, 1).

2. Y-INTERCEPT and the Y-INTERCEPT’S “COUSIN POI NT”

To find the y-intercept, it is necessary to set x = 0 in the equation.

y = -x2 + 4x - 3 � y (0) = -(0)2 + 4(0) -3 = -3

Hence, the y-intercept is (0,-3).

On the other side of the axis of symmetry, equally distanced from that axis is the "cousin point” whose coordinates are (4, -3); this may also be determined by inspection.

3. X-INTERCEPTS

To find the x-intercepts, it is necessary to set y = 0 in the equation.

0 = y = -x2 + 4x - 3 � (-x +1) (x -3) = 0 So x = 1, x = 3 Hence, the x-intercepts are:

(1, 0) and (3, 0).

IMPORTANT: X-intercepts should only be considered if the equation is easy to factor. Otherwise, plotting additional points, as shown in step 4, will suffice.


8/27/10 jw 14

y = -x2 + 4x -33

To plot additional points, it is convenient to use a table:

X Y

-1 -8

5 -8 Hence, the additional points are: (-1,-8) and (5, -8).

These points are also symmetrically located (“cousins”) and when we know one, we can, by inspection, determine the other; as it was done in Step 2. This completes the graph.

4. PLOTTING ADDITIONAL POINTS

ANSWERS:

1 Vertex: (2, 1); axis of symm.: x = 2 3 X-intercepts: (1, 0) and (3, 0)

2 Y-intercept…….………..…(0, -3) 4 The graph is shown above, in Step #4.

78. Solve the system of equations by graphing.

VTX (2,1)

(3,0)

(4,-3) (0,-3)

(1,0)

(-1,-8) (5,-8)

Y

X

X=2

(2)

Using the equation (b), obtain two ordered pairs, plot the

two points and draw a line.

(0,-2)

(-2,0)

(b)

Solution is where the two lines intersect, at (-1,-1). The system is consistent.

(a)

(-1,-1)

(b)

(3)

(-2,-3)

(1)

Using the equation (a), obtain two ordered pairs, plot the two points and draw a line.

(a) (0,1)

(-2,-3)

(a) y = 2x +1 (b) x + y = -2


8/27/10 jw 15

Solve the system of equations by graphing.

79.

Solve the system of equations algebraically: problems 80 – 83.

(Use either substitution or addition/elimination method).

80

Addition/elimination method

(a) 2x 3y 5 (b) x 2(1) 3

(b) x 2y 3 (b) x 1

(a) 2x

+ = + =

+ = =

3y 5

( 2b) 2x

+ =

− −

( ){ }4y 6 SOLUTION

y 1

y 1

− = −

− = − ∴

→ =

1,11,11,11,1Sol. set=

Substitution method

( ){ }

(a) 2x 3y 5 (b) x 2(1) 3

(b) x 2y 3 (b)

(b) x 3 2y

(a) 2(3 2y) 3y 5 SOLUTION

(a) 6 4y 3y 5

(a) y 1

+ = + =

+ =

= −− + =

− + = ∴

− = −

→

x=1x=1x=1x=1

1,11,11,11,1

y=1y=1y=1y=1

Sol. set=

81

Addition/elimination method

(a) 2x 6y 5 ( 2a) 4x− = − − 12y+ 10 Contradiction, the system is inconsistent.

(b) 4x 12y 5 (b) 4x

= −

− = 12y− 5 The lines are parallel,hence do not intersect.

0 5 No solution.

=

≠ −NOTE: Another way to show that the system is inconsistent (equations represent parallel lines: no intersection point), would be to write them in slope-intercept form: same slopes, different b’s.

Equation (a): 1 51 51 51 5y= x-y= x-y= x-y= x-3 63 63 63 6

Equation (b): 1 51 51 51 5y= x-y= x-y= x-y= x-3 123 123 123 12

(2) (3) (1)

(a) x +2y = 6 (b) y = - (½) x+ 3

Using the equation (b), obtain two ordered pairs, plot the


The two lines intersect on their entire length, because they are superposed. Hence, there are an infinite number of solutions. The system is

dependent.

Using the equation (a), obtain two ordered pairs, plot the


(0,3) (0,3) (0,3)

(6,0) (6,0) (6,0)

(b) (a)

(b)

(a)


8/27/10 jw 16

82

(i)

(a) x y z 2

(b) 2x y z 1

(c) x 2y 3z 7

(ii)

(a) x y

+ − = −− + = −

+ − = −

+ z− 2

(b) 2x y

= −

− z+ 1

(a b) 3x 3

(iii)

(a) ( 1) y z 2

(c) ( 1) 2y 3z 7

Simplify, rename (A, C):

(A) y z 1

(C) 2y 3z 6

= −

+ = −

− + − = −− + − = −

− = −− = −

x=-1x=-1x=-1x=-1

(iv)

Solve for:

( 2A) 2y

z

− − 2z 2

(C) 2y

+ =

( ){ }

3z 6

z 4

(v)

Sobstitute to (a) and solve for:

(a) ( 1) y (4) 2

y 5 2

Check if the solution satisfies all three equations.

y

− = −

− = −

→

− + − = −− = −

→

z=4z=4z=4z=4

y=3y=3y=3y=3

Solution set= -1, 3, 4Solution set= -1, 3, 4Solution set= -1, 3, 4Solution set= -1, 3, 4

83

(i)

(a) 3x 2y 4z 1

(b) 5x 3y 5z 2

(c) 6x 2y 3z 5

(ii)

( 1 a) 3x 2y

− + = −− + =− + =

− − +i 4z 1

(c) 6x 2y

− =

− 3z 5

(A) 3x z 6

We have now one equation with two

variables, x & z; another is needed.

(iii)

( 2b) 10x 6y

+ =

− =

− − + 10z 4

(3c) 18x 6y

− = −

− 9z 15

(B) 8x z 11

Superpose equations (A, B):

(A) 3x z 6

(B) 8x z 11

+ =

− =

− =− =

(iv)

Solve for:

( 1 A) 3x z

x

− − +i 6

(B) 8x z

= −

−

( ){ }

11

5x 5

(v)

Sobstitute x 1 to (A) and solve for:

(A) 3(1) z 6

3 z 6

(v)

Sobstitute x 1 & z 3 to (a), (b) or (c); find

(a) 3(1) 2y 4( 3) 1

2y 8

Check if the solution satisfi

z

y

=

=

→

=

− =− =

→

= = −

− + − = −

− = →

x=1x=1x=1x=1

z=-3z=-3z=-3z=-3

y=-4y=-4y=-4y=-4

Solution set= 1, -4, -3Solution set= 1, -4, -3Solution set= 1, -4, -3Solution set= 1, -4, -3

es all three equations.


8/27/10 jw 17

For the following word problems, identify the variable(s) used, set up equation(s) and solve algebraically. 84. A 30-foot ladder, leaning against the side of a building, makes a 50° angle with the ground. How far up the

building does the top of the ladder reach? Express your answer to the nearest tenth of a foot.

xsin50 x L sin50

L

x (30) (0.7660) 22.98 23.0ft.

= → =

= ≈ ≈

� �

i

i x=23ft.x=23ft.x=23ft.x=23ft.

85. A 70 foot rope is attached to the top of one of the vertical poles used to hold up a circus tent. The other end

of the rope is anchored to the ground 40 feet from the bottom of the pole. What is the height of the tent and what angle does the rope make with the tent? Express both answers to the nearest tenth (remember about the units).

86. Find three consecutive odd integers such that seven times the sum of the first two integers is three more

than nine times the third integer.

VARIABLES: x = the first odd integer; x + 2 = the second odd integer; x + 4 = the third odd integer

7•(x + (x + 2)) = 9•(x + 4) + 3 � 7•(2x + 2) = 9 x + 36 + 3 � 14x + 14 = 9 x + 39 � 5x = 25 � x = 5

ANSWER: The integers are: 5, 7 and 9

h

Ө 70 ft rope

40 ft

T E N T

ANSWERS: The height of the tent is 57.4 ft. The angle Ө

VARIABLE(S):

Ө = angle of the rope with the tent h = height of the tent

140 4 4sin sin sin 34.8

70 7 7− θ = → θ = → ≈

�

VARIABLE(S): x = distance from the top of the ladder to the

base of the house. L = 30 ft. ( Length of the ladder)

50°

x

L

ANSWER: The top of the ladder reaches 23.0 feet up the side of the building.


8/27/10 jw 18

87. A chemist must mix 8 L of a 40% acid solution with some 70% acid solution to get 50% acid solution. How

much of the 70% solution should be used?

88. A private airplane leaves an airport and flies due east at 180 mph. Two hours later, a jet leaves the same

airport and flies due east at 900 mph. How long will it take for the jet to overtake the private plane?

DISTANCE TRAVELED BY PRIVATE AIRPLANE = DISTANCE TRAVELED BY JET

360 1180 t 360 900 t 720 t 360 t t hour.

720 2→ + = → = → = → =180(t+ 2)= 900t

ID – RAG (I D) (Rate) (Agent) (Gain)

Type of Plane Speed mph

Time hrs

Distance mi

Private Airplane 180 t + 2 180 (t + 2)

The Jet 900 t 900 t

VARIABLE(S): t = jet’s catch up time. t + 2 = private airplane

travel time.

ANSWER:

t = ½ hour of catch up time for the jet.

ANSWER:

Needed are 4 liters of

70% acid solution.

VARIABLE(S): x = liters of 70% sol.

8 + x = liters of 50% sol.

THE AMOUNT OF PURE ACID BEFORE MIXING = THE A MOUNT OF PURE ACID AFTER MIXING

(0.40) (8) + 0.70 x = 0.50 (8+ x) � 3.20 + 0.70 x = 4.00 + 0.50 x � 0.20 x = 0.80

= = =0.80 8

x 40.20 2


Solution Type

Acid % Decimals

Solution liters

Pure acid liters

40% solution 0.40 8 (0.40) (8) = 3.2

70% solution 0.70 x 0.70 x

50% solution 0.50 (8 + x) 0.50 (8 + x)


8/27/10 jw 19

89. Jonathan invests $ 7,500 @ 10.4 % simple interest. How much additional money must he invest at a simple

interest rate of 14 % so that the total interest earned is 12 % of the total investment?

90. Flying with the wind, Rob flew 800 miles between Pittsburgh and Atlanta in 4 hours. The return trip against the wind took 5 hours. Find the rate of the plane in calm air and the rate of the wind.

(TRAVEL RATE) • (TRAVEL TIME) = TRAVEL DISTANCE

( ) ( )( ) ( )

( )( )

•

•

a r+ w 4=800 a r+ w =200a r+ w 4=800 a r+ w =200a r+ w 4=800 a r+ w =200a r+ w 4=800 a r+ w =200Dividebothsidesby 4Dividebothsidesby 4Dividebothsidesby 4Dividebothsidesby 4→ → Add bothsides→ → Add bothsides→ → Add bothsides→ → Add bothsides

Dividebothsidesby5Dividebothsidesby5Dividebothsidesby5Dividebothsidesby5b r-w 5=800 b r-w =160b r-w 5=800 b r-w =160b r-w 5=800 b r-w =160b r-w 5=800 b r-w =1602r=3602r=3602r=3602r=360 → r=180 → w = 200 -180= 20→ r=180 → w = 200 -180= 20→ r=180 → w = 200 -180= 20→ r=180 → w = 200 -180= 20


Direction Speed mph

Time hrs

Distance mi

To Atlanta with the wind

r + w 4 800

From Atlanta against the wind r - w 5 800

ANSWER:

180 mph = the rate of plane in calm air.

20 mph = the rate of wind.

VARIABLE(S): r = the rate of plane. w = the rate of wind.

ANSWER:

Jonathon must invest $6,000 in additional

money.


Type of Account Rate Amount

Invested Interest

10.4 % 0.104 7,500 0.104• 7,500

14 % 0.14 x 0.14 x

12 % 0.12 7,500 + x 0.12 (7,500 + x )

VARIABLE(S):

x = $$ invested at 14 % 7,500 + x = $$ of total

investment.

( ) ( ) ( ) ( )• •0.104 7,500 + 0.14 x = 0.12 7,500+ x0.104 7,500 + 0.14 x = 0.12 7,500+ x0.104 7,500 + 0.14 x = 0.12 7,500+ x0.104 7,500 + 0.14 x = 0.12 7,500+ x → 780 + 0.14 x = 9→ 780 + 0.14 x = 9→ 780 + 0.14 x = 9→ 780 + 0.14 x = 900 + 0.12 x00 + 0.12 x00 + 0.12 x00 + 0.12 x1201201201200.02 x = 1200.02 x = 1200.02 x = 1200.02 x = 120 → x = → x = 6,000→ x = → x = 6,000→ x = → x = 6,000→ x = → x = 6,0000.020.020.020.02


8/27/10 jw 20

91. A member of the City Volunteer Corp. can mow and clean up a large lawn in 9 hours. With two members of the City Volunteer Corp. working, the same job can be done in 6 hours. How long would it take the second member of the team, working alone, to do the job?

92. How many pounds of gourmet candy selling for $1.80 per pound should be mixed with 3 pounds of candy selling for $2.60 a pound to obtain a mixture selling for $2.04 per pound?

INDIVIDUAL WORK CONTRIBUTIONS ADDED = EN TIRE JOB DONE 1 1 2 6

6 6 1 1 2 18 3 189 3

• + • = → + = → + = → =t t tt t

LCD = 3t


Member Defined as:

Member’s Work Rate

Time Working Together

Each Member’s Work

Contribution

First member 1 / 9 6 (1 / 9)•6

Second member 1 /t 6 (1 / t )• 6

ANSWER:

18 hrs. = The time needed by the second member to complete the work alone.

VARIABLE(S): t = the time needed by the second member to complete the work alone.

COST OF THE CANDY BEFORE MIXING = COST AFTER MIXING

1.80 x + 2.60 • 3 = 2.04 (x + 3) 1.80 x + 7.80 = 2.04 x + 6.12 � 0.24 x = 1.68 �

1.68

0.24x = � x = 7


Type of Candy

Price $ per lb.

Quantity lbs.

Cost $

Gourmet candy 1.80 x 1.80 x

Other candy 2.60 3 2.60 • 3

Mixture 2.04 x + 3 2.04 (x + 3)

ANSWER:

7 lbs of Gourmet candy

should be mixed.

VARIABLE(S): x = lbs of Gourmet candy x + 3 = lbs of mixture.


8/27/10 jw 21

93. A fenced area is 300 square feet. If the width is 5 feet less than the length, find the length and the width of

the fenced area.

94. At a business meeting at Panera Bread, the bill for two cappuccinos and three house lattes was $14,55. At

another table, the bill for one cappuccino and two house lattes was $8.77. How much did each type of beverage cost?

(1) A = rectangle’s area. L = rectangle’s length. W = rectangle’s width.

(6) L2 – 5L – 300 = 0

(2) A = L • W (7) (L – 20) (L + 15) = 0

(3) W = L – 5 (8) L = 20, L = – 15 Reject – 15 (No negative dimension)

(4) (L – 5) L = 300 (9) W = L – 5 = 20 – 5 = 15

(5) L2 – 5L = 300

W

L

ANSWER

The length = 20 ft. The height = 15 ft.

Rectangle

ANSWER:

One cappuccino costs $2.79. One house latte costs $2.99.

(a) 2x + 3y = 14.55 (b) x + 2(2.99) = 8.77 (b) x + 2y = 8.77 x + 5.98 = 8.77

------------------------------- x = 40 (a) 2x + 3y = 14.55 x = 2.79

(-2b) -2x – 4y = – 17.54 ------------------------------- –y = – 2.99 y = 2.99

VARIABLE(S): x = cost of one cappuccino. y = cost of one house latte.

2010-spr 104 new jw-ski 4th ed 08--27-10 fnl exm rev ... · 2010-spr 104 new jw-ski 4th ed...

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