# Intermediate Algebra Appeals Practice Test

Document Sample

```					                  Intermediate Algebra Appeals Practice Test 6/07
2   3
1.   Evaluate 3  4 .
3   7
3   1
2.   Evaluate 5  2 .
5   2
1   3
3.   Evaluate 9     4 .
4   8
4.   What is 15.2 percent of 76? Round answer to the nearest tenth.
5.   30 is 20% of what number?
6.   12 is what percent of 108? Round answer to the nearest tenth of a percent.
7.   If 2.5 pounds of steak costs \$12.50, how much will 4 pounds cost?
8.   Evaluate -3 + 2 ∙ -4.
9.   Evaluate (-2)2 – 3 ∙ -1.
10. Simplify 3x  5  2 x  4 .
11. Simplify 23x  1  5 x  6 .

 3x 4 y 3
12. Simplify              .
12 xy 5
13. Determine the common solution: x + y = 4 and 2x – y = 11.
14. Determine the common solution: 3x - 5y = 14 and 2x – 3y = 9.
9
16
16. Determine l if P  2l  w when P  280 feet and w  40 feet.
17. Determine the slope of the line through:
a. (2,-3) and (-1,5)
b. (4,-1) and (4,6)
c. (3,5) and (-3,5)
18. Write in slope-intercept form  y  mx  b  the equation of the line through:
a. (-4,-2) and (0,-3)
b. (-3,2) and (4,5)
c. (5,-1) and (2,-1)
19. Determine the slope for:
a. the line y  5
b. the line 2 x  3 y  4
c. the line x  8
Intermediate Algebra Appeals Practice Test 6/07
20. Write the equation of the line graphed below.

6

-6                                          6

-6

21. Determine (a) the domain and (b) the range of the function graphed below.

6

-6                                          6

-6
22. Graph the line 3x  4 y  12 .
23. Determine the distance between the points (-5,2) and (1,-5). Round your answer to the hundredths
place.

24. State which of the following relations are functions (there may be more than one correct answer):
a.  2,0,  2,4,  2,8
b.    3,2, 5,6, 5,3, 8,9
c.   1,3,  1,3, 5,2
d.   5,2, 4,6
2
Intermediate Algebra Appeals Practice Test 6/07

25. Let f x   2  4 x . Determine (a) f 0 , (b) f 1 , (c) f  1

26. Let g x   9  2 x 2 . Determine (a) g 0 , (b) g 2 , (c) g(-2)

27. Solve for x by factoring:
a. 2 x 2  9 x  5
b. x 2  6 x  9  0
c. 9 x 2  4  0

28. Solve for x using the quadratic formula:
a.   x 2  3x  2  0
b. 2 x 2  4 x  1
c.   x 2  2x  5  0

29. Let Gt   8  2 t , represent the number of rabbits in a barn after t weeks. How many rabbits are there
after (a) 1 week? (b) 3 weeks?

30. Simplify each of the following. Express each answer using positive exponents only.
a.   x 5  y  z 4  x 3  y 2  z 7

b. x 4  y 2  x 4  y 0

c.   x 2  y 5  z  x 3  y 2  z 2

31. Compute 3  811 2 .

2 x 2  12 x  32 x 2  10 x  16
32. Multiply, simplifying as much as possible:                                  .
x 2  16 x  64 x 2  3x  10

x 2  2 x  24 x 2  7 x  6
33. Divide, simplifying as much as possible:                          2            .
x 2  6x  8   x  x6

9x  2        7
34. Add, simplifying as much as possible:                          2         .
3x  2 x  8 3x  x  4
2

3
Intermediate Algebra Appeals Practice Test 6/07
4x  2    2
35. Subtract, simplifying as much as possible:                      .
x  x  20 x  4
2

36. Draw a graph of y  x 2  2 x  15 . In your graph, clearly label and give both coordinates of the x-

intercepts, the y-intercept, and the vertex.

37. Compute log 3 27 .

38. Compute log 64 4 .

x2
39. Determine the domain of the function y                   .
x  2x  8
2

3x
40. Solve for x:         2.
x 1

2x   6    28
41. Solve for x:           2    .
x3 x3 x 9

4
Intermediate Algebra Appeals Practice Test 6/07
Answers/Solutions for Intermediate Algebra Appeals Practice Test
2   3                                                         3     1
1.    Evaluate 3  4 .                                          2.   Evaluate 5  2 .
3   7                                                         5     2
To add or subtract fractions as mixed numbers, you must         To add or subtract fractions as mixed numbers, you must
find a common denominator for the fractions. The least          find a common denominator for the fractions. The least
common denominator for 3 and 7 is 21.                           common denominator for 5 and 2 is 10.
2     2x7         14                                             3     3x2          6
3 3             3                                                5 5          5
3     3x7         21                                             5     5x2        10
3     3x3          9                                             1       1x 5        5
4 4            4                                              2  2         2
7     7x3         21                                             2       2x5        10
Add the fractions: [Add numerators, keep common                 Subtract the fractions: [subtract numerators, keep common
denominators]                                                   denominators]
14 9 23            2                                             6 5      1
          1                                                  
21 21 21           21                                           10 10 10
Add the whole numbers: 3 + 4 = 7                                Subtract the whole numbers: 5 - 2 = 3
2     2
Add the two results: 7  1  8
21     21
2                                                              1
21
1   3                                        4.   What is 15.2 percent of 76? Round answer to the
3. Evaluate 9  4 .                                                nearest tenth.
4   8
To add or subtract fractions as mixed numbers, you must
find a common denominator for the fractions. The least          Translate into a mathematical equation.
common denominator for 4 and 8 is 8.                            What is 15.2 percent of 76?
1     2                                                      ↓ ↓ ↓               ↓ ↓
9 9                                                          N = 0.152            • 76
4     8
3       3                                                  Multiply 0.152 and 76 to get 11.552
 4  4
8       8
Subtract the fractions: You cannot subtract 3 from 2; you       11.552 rounded to the nearest tenth is 11.6.
8            2     10
must ―borrow‖ 1 from 9 [ 1  ] to get 8  1  8
8            8      8
10 3 7
Now, subtract the fractions:      
8 8 8
Subtract the whole numbers: 8 - 4 = 4
7
8
5. 30 is 20% of what number?                                  6.   12 is what percent of 108? Round answer to the
nearest tenth of a percent.
Translate into a mathematical equation.                         Translate into a mathematical equation.
30 is 20% of what number?                                       12 is what percent of 108
↓ ↓ ↓ ↓ ↓                                                      ↓ ↓       ↓         ↓ ↓
30 = 0.20 • N or 0.2N = 30                                      12 =      N         • 108 or 108N = 12
Solve for N: N = 30 ÷ 0.2 = 150                                 Solve for N: N = 12 ÷ 108 = 0.1111
Write N as a percent and round to the nearest tenth.
N = 0.1111 = 11.11% = 11.1%
7. If 2.5 pounds of steak costs \$12.50, how much will            8. Evaluate -3 + 2 ∙ -4.
4 pounds cost?
Let x = the cost of 4 pounds of steak.                          Order of operations: Multiply first, then add
 ; 2.5x  412.50  50
2.5      4
12.50 x                                                         -3 + 2 ∙-4 = -3 + (-8) = -11
x  50  2.5  20
Answer: The cost of 4 pounds of steak is \$20.00                 Answer: -11

5
Intermediate Algebra Appeals Practice Test 6/07

9. Evaluate (-2)2 – 3 ∙ -1.                                        10. Simplify 3x  5  2 x  4  3x  35  2 x  4
 4  31  4  3  7                                             3x  15  2 x  4  3x  2 x  15  4
11. Simplify 23x  1  5x  6 .                                                    3x 4 y 3
 23x   21  15x   16
12. Simplify
12 xy 5
 6 x  2  5x  6  6 x  5x   2  6
 3 1       x4                     y3                           1
   ;         x 41  x 3 ;         5
 y 3 5  y  2 
12   4        x                     y                            y2
 1x 3
4y2
13. Determine the common solution:                                  14. Determine the common solution: 3x – 5y = 14 and
x + y = 4 and 2x – y = 11.                                      2x – 3y = 9.
To find the common solution, eliminate one of the                  To solve for y, multiply the first equation by -2 and the
unknowns by addition or subtraction; then solve for the            second equation by 3; then add the equations
remaining unknown                                                   23x  5 y  14  6 x  10 y  28
x y  4                                                           32 x  3 y  9  6 x  9 y  27
2 x  y  11                                                                                       y  1
3x  0  15                                                       Substitute -1 for y in the first equation:
So x = 15 ÷ 3 = 5. Substitute 5 for x in first equation: 5 + y     3x – 5(-1) =14  3x + 5 = 14  3x = 9  x = 3
= 4  y = 4 – 5 = -1                                               Check in the second equation: 2(3) -3(-1) = 6 + 3 = 9
9                                                  16. Determine l if P  2l  w when P  280 feet
15. Convert       to a percent. Do not round your
16                                                       and w  40 feet.
answer.                                                       P  2l  w  280  2l  40  2l  80
To convert a fraction to a percent, divide the numerator by
the denominator; move the decimal point two places to the          2l  80  280  2l  280  80  200
9                                      2l  200  l  100
right and affix the % sign.         9 16  0.5625  56.25%       Answer: l = 100 feet
16
17. Determine the slope of the line through:                      18. Write in slope-intercept form  y  mx  b the
a. (2,-3) and (-1,5)                                         equation of the line through:
b. (4,-1) and (4,6)                                               a. (-4,-2) and (0,-3)
c. (3,5) and (-3,5)                                               b. (-3,2) and (4,5)
The slope between 2 points, (x1, y1) & (x2,y2) is                                c. (5,-1) and (2,-1)
y  y1                                                                                    3  2  1
m 2          ; if x2 – x1 =0, the slope is undefined              a) Find the slope: m                  
x 2  x1                                                                                  0   4     4
1
5  3 8                        8                        Find b:  2        4   b  2  1  b  b  2 1  3 .
a) m                         Answer:                                             4
1  2  3                       3                                       1
4
6  1 7
b) m              Undefined Answer: Undefined                                                  52       3
44                                                       b) Find the slope: m                 
4   3 7
0
9             9 23
Find b: 2    3  b  2 
55                                                                    3
c) m        
0
0          Answer: 0                                                                      b b  2         .
33 6                                                                7                        7             7     7
3        23
7     7
1  1 0
c) Find the slope: m                 0
25       3
Find b: 1  0  5  b  b  1
Answer: y  0x 1 or y  1

6
Intermediate Algebra Appeals Practice Test 6/07

19. Determine the slope for:                                     20. Write the equation of the line graphed below
a. the line y  5
6
b. the line 2x  3 y  4
c. the line x  8
Write each equation in slope-intercept form: y  mx  b , if
possible.
a) y  0 x  5  m  0 , Answer: Slope is 0.

2    4     2
b) 2 x  3 y  4  3 y  2 x  4  y      x    m                   -6                                                 6
3    3      3
2
3

c) x = 8 cannot be written in slope-intercept form. The
slope is undefined

-6
The x-intercept is (-1,0); the y-intercept is (0,-2).
2  0 2
The slope is                 2
0   1 1
Ans: The equation of the line is y  2x  2
21. Determine (a) the domain and (b) the range of the            22. Graph the line 3x  4 y  12
function graphed below                                   Let x = 0; then 0 – 4y = -12 and y = 3
6                              Let y = 0; then 3x – 0 = -12 and x = -4

-6                                                   6

-6
The domain of the function f is the set of all input values
for which the function is defined. The range of a function
is the set of all output values for the given function.
Looking at the graph we see that the input values, x, for this
function are x  5 and the out put values, y, are y  3
Ans: Domain:    x x  5
Range: y y  3

7
Intermediate Algebra Appeals Practice Test 6/07

23. Determine the distance between the points (-5,2)                   24. State which of the following relations are
and (1,-5). Round your answer to the hundredths                        functions (there may be more than one correct
A function is a relation in which each x-coordinate is
D     x2  x1 2   y 2  y1 2                                    paired with one and only one y-coordinate.
For the relations given, only c and d are FUNCTIONS

So,
D    1   52   5  22      62   72               since they are the only sets in which every x is paired with
a unique y-value.
 36  49  85  9.2195                                           a) is NOT a function because -2 is paired with three
different values of y. b) is NOT a function because 5 is
paired with two different values of y.
25. Let f x   2  4 x .                                                26. Let g x   9  2 x 2
Determine (a) f 0 , (b) f 1 , (c) f  1                            Determine (a) g 0  , (b) g 2  , (c) g  2
a) f 0  2  40  2  0  2 Answer: f 0  2
a) g 0  9  202  9  0  9 Answer: g 0  9

b) f 1  2  41  2  4  2 Answer: f 1  2
b) g 2  9  22  9  8  1 Answer: g 2  1
2

c) f 1  2  41  2  4  6 Answer: f 1  6
c) g  2  9  2 2  9  8  1 Answer: g 2  1
2

27. Solve for x by factoring:                                           28. Solve for x using the quadratic formula:
a. 2 x 2  9 x  5 b. x 2  6 x  9  0 c. 9 x 2  4  0                a. x 2  3x  2  0 b. 2 x 2  4 x  1 c. x 2  2 x  5  0
a) Put equation in ax 2  bx  c  0 form: 2 x 2  9 x  5  0          The quadratic formula gives solution for equations in the
To factor 2 x 2  9 x  5  0 , you must find two numbers               form ax 2  bx  c  0 .
whose sum is -9 [b] and whose product is -10 [ac = 2(-5)].                     b  b 2  4ac
The numbers, -10 and 1, are used as coefficients of x and                x                        Quadratic Formula
2a
substituted for -9x to get 2x 2 10x  1x  5 , a polynomial
with four terms. Now, we factor 2x from the first two                   a) x 2  3x  2  0; a  1, b  3, c  2
terms and 1 from the third and fourth terms to get:
2 xx  5  1x  5  x  52 x  1                                  b 2  4ac  32  41 2  9   8  9  8  17
To solve 2 x 2  9 x  5  x  52 x  1  0 , set each factor             17 cannot be simplified.
equal to 0 and solve for x: x  5  0  x  5 and                              3  17  3  17
x           
2 x  1  0  2 x  1  x 
1
1                     21       2
2                         2
 3  17
2
b) To factor x 2  6 x  9  0 , you must find two numbers
whose sum is -6 and whose product is +9. The numbers are
b) 2 x 2  4 x  1  2 x 2  4 x  1  0; a  2, b  4, c  1
-3 and -3.
x 2  6 x  9  x 2  3x  3x  9  xx  3  3x  3                    b 2  4ac  42  421  16  8  8

 x  3x  3  x  32                                                   8 can be simplified to        4  2   2   2

x 2  6 x  9  x  32  0  x  3  0  x  3                         x
42 2 42 2 2 2
     
2 2
c) 9 x 2  4  3 x   2  is a special product—the
2      2
2
difference of two squares. Its factors are (3x+2) and (3x-2)            c) x  2 x  5  0; a  1, b  2, c  5
2

2
9 x 2  4  3x  23x  2  3x  2  0  3x  2  x                   b 2  4ac  22  415  4  20  16
3
2                                       16 can be simplified to 4i .
and 3x  2  0  3x  2  x 
3                                      2  4i 2  4i
x                 1  2i
2 2                                                         21      2
3 3                                                    Answer: x  1 2i

8
Intermediate Algebra Appeals Practice Test 6/07

29. Let Gt   8  2 t , represent the number of rabbits           30. Simplify each of the following. Express each
in a barn after t weeks. How many rabbits are
there after (a) 1 week? (b) 3 weeks?
5
a) x  y  z 4  x 3  y 2  z 7  x 53  y 1 2  z 4 7
a) t  1  g 1  8  21  16                                                                             y3z3
 x -8  y 3  z 3 Answer:
Answer: There are 16 rabbits after 1 week.                                                                     x8

b) t  3  g 3  8  2 3  64                                      b) x 4  y 2  x 4  y 0  x 44  y 20  x 0  y 2  1 y 2
Answer: There are 64 rabbits after 3 weeks.

c) x 2  y 5  z  x 3  y 2  z 2  x 23  y 52  z 12
 x 1  y 3  z 1
xy 3
z

31. Compute 3 811 2                                                 32. Multiply, simplifying as much as possible:
3  81
12
 3  81  3  9  27                          2 x  12x  32 x 2  10x  16 2x  2x  8 x  2x  8
2
                            
x 2  16x  64 x 2  3x  10   x  8x  8 x  2x  5
Ans: 27                                                                    2x  2
Ans:
 x  5

33. Divide, simplifying as much as possible:                             34. Add, simplifying as much as possible:
2             2              2                2
x  2 x  24 x  7 x  6 x  2 x  24 x  x  6
                                                         9x  2                    7                9x  2             7
2                2              2              2                                                                             
x  6x  8      x x6         x  6x  8 x  7x  6                3x  2 x  8
2
3x  x  4
2               3x  4x  2 3x  4x  1


x  6 x  4   x  3x  2     x3
LCD = 3x  4x  2x  1
x  4 x  2  x  6 x  1       x 1

9 x  2x  1            7 x  2 
x 3
Ans:
x 1
3x  4 x  2 x  1 3x  4 x  2 x  1

2
9x  7x  2                            7 x  14
                                 
3x  4x  2x  1 3x  4x  2x  1

9 x  7 x  2  7 x  14
2                                               2
9 x  16
                                         
3x  4 x  2 x  1              3x  4x  2x  1


3 x  43 x  4   3 x  4 
3x  4x  2x  1 x  2x  1

9
Intermediate Algebra Appeals Practice Test 6/07
35. Subtract, simplifying as much as possible:                     36. Draw a graph of y  x 2  2 x  15 . In your graph,
4x  2      2         4x  2      2
                                                             clearly label and give both coordinates of the x-
x  x  20 x  4 x  5x  4 x  4
2
intercepts, the y-intercept, and the vertex

LCD = x  5x  4                                               y  15  x 2  2 x  y  15  1  x 2  2 x  1
 y  16  x  12
4x  2        2x  5      4 x  2  2 x  10
                                                   Vertex‖ (1,-16); y-int: (0,-15); x-int: (-3,0) and (5,0)
x  5x  4 x  4x  5 x  5x  4
2x  8        2x  4         2
                              
x  5x  4 x  5x  4 x  5

(-3,0)                         (5,0)

(0,-15)          (1,-16)

37. Compute log 3 27                                             38. Compute log 64 4 .
log 3 27  x is equivalent to the exponential equation           log 64 4  x is equivalent to the exponential equation
3 x  27 . Since 3 3  27, x  3                                                                         1
64 x  4 . Since    3
64  4, x 
Ans: log 3 27  3                                                                                      3

Ans: log 64 4 
1
3
39. Determine the domain of the function                                                  3x
40. Solve for x:              2.
x2                                                                         x 1
y 2            .
x  2x  8
The domain of function is the set of values of x for which         LCD = x  1 , So multiply both sides by the LCD.
the function is defined. Since this is a rational function,
you must look at the denominator x 2  2 x  8 and                 x  13x
 2x  1  3x  2 x  2  x  2
determine the values for which the denominator is zero.                  x 1
These are the values that must be excluded from the                        32 6
domain of the function. The function is not defined for            Check:       2
2 1 3
these values since they make the denominator zero.
Ans: x = 2
x 2  2 x  8  0  x  4x  2  0  x  4 or x  2
Ans: Domain: x x  4 and x  2
28                                     28
LCD = x  3x  3
2x     6                            2x     6
41. Solve for x:                                              
x  3 x  3 x2  9                  x  3 x  3 x  3x  3

So multiply both sides (all terms) by the LCD:
x  3x  3 2 x  x  3x  3 6  x  3x  3 28
x 3                 x3                x  3x  3
 2 xx  3  6x  3  28  2 x 2  6 x  6 x  18  28  2 x 2  12x  10  0  2x  5x  1  0  x  5 or x  1
25       6          28           10 6        28       10 24  28          14  28         7      7
x  5 :                                                                                    
 5  3  5  3  52  9           8  2 25  9          8 8       16       8      16      4      4
Check:
2 1     6          28          2 6  28         1 6  28        7 7
x  1 :                                                             
 1  3  1  3  1  9
2           4 2 1 9         2 2 8           2 2
Ans: x  5 or x  1

10

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 7 posted: 11/23/2011 language: English pages: 10
How are you planning on using Docstoc?