•

Standard 19.0

42. Toni is solving this equation by completing the square

.

ax 2 +bx +c = 0 (where a

≥ 0) Step 1:

ax 2 +bx = - c

• •

Step 2: x 2 + b / a x = c / a Step 3: ?

•

Which should be Step 3 in the solution?

a)

 

c b b a x

b)

x



b a



b b c b

c)

x

2 

x

   

a

2

a a

2

a



c ax

d)

x

2 

b a x



b

2

a

2  

c a



b

2

a

2

• • • • • The key to answering this question is knowing the formula for “completing the square” The next step in completing the square is

b

2 to add to both sides of the equation. 2

a

By doing this, you will form a perfect square on the left side of the equation.

At this point, solve the problem for the question is asking for step 3. ∴ “d” is the answer since it is the only one adding ( b / 2 ) 2 to each side. The next two steps are to find the square root of each side and solve the equation.

Standard 19.0

43.Four steps derive the quadratic formula are shown below.

What is the correct order for these steps?

a) I, IV, II, III b) I, III, IV, II c) II, IV, I, III d) II, III, I, IV

• You must first look over what they give you and what they ask you. The question is telling you that the quadratic formula is where you end up at. • The question is also helping you on the test. Notice steps I and IV were in the previous question for “completing the square”. Use this to your advantage. Knowing those two steps and knowing the actual formula of • 

b



b

2  4

ac

is very close to step III (or 2

a

• where you want to end up at) you can assume step II is 3 rd . ∴ “a” is the answer

Standard 20.0

44. Which is one of the solutions to the equation 2x 2 – x – 4 = 0?

4  1 4  33 1  4  1  4 33

• Using the quadratic formula, solve.

2

x

2

x

0 ( ) 1

2



• ∴ “c” is the answer

1  1 4 1  4 1  4 8 33 )( 4 )  4 )

Standard 20.0

45. Which state best explains why there is no real solution to the quadratic equation 2x 2 + x + 7 = 0 a) The value of 1 2 - 4 • 2 • 7 is positive.

b) The value of 1 2 - 4 • 2 • 7 is equal to 0.

c) The value of 1 2 - 4 • 2 • 7 is negative.

d) The value of 1 2 - 4 • 2 • 7 is not a perfect square.

• • • • •

1 st , this is a tricky question so you want to take it slow. You are dealing with a quadratic equation which means they are asking for what is inside the radical sign “√ “ in the quadratic formula. Taking what is inside “b²- 4ac”, you plug in the numbers given to solve. This step has already been done for you in the choices given. Begin eliminating what you know. If you find the value of the numbers given, “a” and “b” are false statements ruling them out. “c” and “d” are true statements (answer is -55). Now, if the discriminant is a negative number your answer is always no real solution.

∴

“c” is the answer

Standard 20.0

46. What is the solution set of the quadratic equation 8x 2 + 2x + 1 = 0?

a) b)

,  1 2 4    1  2 ,  1  2 

c)

    1  8 7 ,  1  8 7   

d) No real solution

•

Using the quadratic formula, you would plug in the values of “a”, “b”, and “c”.

•

When you begin to solve you will notice the discriminant (b² - 4ac) is negative, this equation has no solution in the real number system.

∴ “d” is the answer  2  2 2  ( 4 )( 8 )( 1 ) ( 2 )( 8 )  2  16 4  32  2  16  28

Standard 21.0

47.The graph of the equation y = x 2 – 3x – 4 is shown below.

For what value of x is y = 0?

a) x = -1 only b) x = -4 only c) x = -1 and x = 4 d) x = 1 and x = - 4

• • •

The question is simply asking for the “x” value if y = 0. In this case, the parabola intersects “x” in two places. Both places will be the value of “x”.

Since the line intersects “x” in two places look at “c” and “d” first. At points (-1, 0) and (4, 0), notice “y” equals 0 and the line intersects “x”. Therefore (

∴) x = -1 and x = 4 and your answer is “c”.

Standard 21.0

48. Which best represents the graph of y = - x 2 + 3?

• If the coefficient of x

² is positive, the parabola (the “u” shaped line) opens upward, if it is negative it opens downward. In this case, its negative so “c” and “d” are eliminated.

•

Finding the vertex (curve of the parabola).

 2

a b y



ax

2 

bx



c

 0

y

 1 ( 0 ) 2  3 2 ( 1 )

x

 0

y

 3 • In this type of problem, if you are missing the “bx”, then “c” (3) will be your y intercept.

• ∴ “b” is the answer

Standard 22.0

49. How many times does the graph of y = 2x 2 – 2x + 3 intersect the x-axis?

a) none b) one c) two d) three

  2

y

  2  1 2 2   2  1 2  3 2 ( 2 ) 2

y

  2  1 4  1  3 4

x

 1

y

 1 2  1  3 2

y

 2 1 2 •

With no negative sign in front of the coefficient 2 ( 2 x 2 – 2x + 3) the parabola opens upward. With “y” being a positive number,

•

the parabola will not intersect the x-axis.

∴ “a” is the answer

Standard 23.0

50. An object that is projected straight downward with initial velocity v feet per second travels a distances s = vt + 16t 2 , where t = time in seconds. If Ramon is standing on a balcony 84 feet above the ground and throws a penny straight down with an initial velocity of 10 feet per second, in how many seconds will it reach the ground?

a) 2 seconds b) 3 seconds c) 6 seconds d) 8 seconds

• • • They give you the formula s= vt + 16t

² .

You need to read and dissect the problem and place it into the formula. “s” = 84 feet. Remember it tells you “t” represents seconds.

•

“vt” = 10 ft per second 10/1 (“t” is unknown)

•

“16t²” = unknown seconds that the “vt”

84

increased by time it hits the ground.

 10

t

 16

t

2 8

t

 21  0

t

0  2   84   21 2 0  16

t

2 10

t

   5

t

 0  2 ( 8

t

2 ( 8

t

 5

t

21 )(

t

42 )  2 ) 8

t t

 2 8   2 2

t

   2 0 0  2 • ∴

, since you don’t go back in time (negative number), your answer is 2 seconds or “a”.

Standard 23.0

51. The height of a triangle is 4 inches greater than twice its base. The area of the triangle is 168 square inches. What is the base of the triangle? a) 7 in.

b) 8 in.

c) 12 in.

d) 14 in.

• Using the formula A=½bh, plug in the given to solve • Set it up as a quadratic equation equal to 0.

168 2  168   1

b

( 4  2

b

( 2

b

 2 4 )

b

)

b

 14  0

b

 14  14   14  0

b

  14 336  2

b

2  4

b

0  2

b

2  4

b

 336 0  2 (

b

2  2

b

 168 ) 0  2 (

b

 14 )(

b

 12 )

b

 12  0

b

 12  12  12  0

b

 12 • -14 is not an answer nor does it make sense to be an answer. 12, however is given.

• ∴ “c” is the answer

Standard 12.0

lowest terms.

a) b) c)

x

 2

y

3

x

 2

y

3

y x

 2

y

3 d)

x

 2

y

3

y

3

xy

 6

y

2



x

2

x

 4

xy

 4

y

2 3  3 

y xy

2 

y x x

    2 6

y

2

x

2

y

 

y

 2

y

 3

y

• 1 st , factor the numerator. • Next, reduce the denominator by factoring out what is common, 3y.

• Finally, reduce what you have in common in the numerator and denominator.

• ∴ your answer is “b”

a) b) c) d) terms.

Standard 12.0

4

x

2  1 3 

x

2

x

 1   1 3 

x

 3  2

x

 1 3  2 4 

x x

  1 3   3 

x

2

x

 3   1

6

x

2  4

x

2 21

x

 1  9 3 (  2 2

x x

 2 1  7  2

x x

  3 ) 1  3 (  2 2

x x

  1 1 )(  2

x x

  3 ) 1  3 (

x

( 2

x

 3 )  1 ) • After factoring you are left with answer “b”

c) a) terms.

x

 2

x x

  1 2

x

 1

Standard 12.0

x

 4

x

 4

x

2  3

x

 2

b) d)

x x

 2  1

x x

 2  1

• UNFOIL or factor the numerator and denominator • Reduce • ∴ “a” is the answer

x

2  4

x

 4

x

 

x x

2     2 2

x

3   

x x x

2     2 2 1   (

x

 1 )

55. a)

Standard 13.0

7

z

2 4

z

 7

z

 8 

z

3

z

2  2 

z

2 4 

z

7 

z

4 (

z

 2   1 )

b)

7 (

z

 2 ) 4 (

z

 1 )

z

4 (

z z

  2 ) 7

z

(

z

 1 ) 4 (

z

 2 )

• Factor, reduce, then multiply the (numerator •numerator) and the (denominator • denominator)  7 2 7   ( 2  (

z

4

z

  8 )

z

(

z

3

z

 2

z

2 4 ) 

z

) 7

z

(

z

4 (

z

  1 ) 2 )  (

z

 2 )(

z z

7  (

z z

4 (

z

  1 )( 2   1 )

z

 2 )  1 ) • ∴ “a” is the answer

Standard 13.0

56. Which fraction equals the product?

x

 5  2

x

 3 3

x

 2  

x

 5

a)

2

x

 3

b)

3

x

 2 3

x

 2 4

x

 3 6

x

2

x

 25  5

x

 6 3 2

x

2

x

2  7

x

 15  13

x

 10

• To solve this problem you use FOIL for the numerators and denominators • Numerators = (x + 5)(2x – 3) • Denominator = (3x + 2)(x – 5) • Numerator Denominator • (x + 5)(2x – 3) (3x + 2)(x – 5) • 2x ² - 3x + 10x – 15 3x² - 15x + 2x - 10 • 2x ² + 7x – 15 3x² - 13x - 10 • ∴ “d” is the answer

57.

x

2

Standard 13.0



x

8 

x

3  16  2

x

2

x

  8 9 (

x

 3

x

)( 

x

 2 3 ) 2 2 (

x

 3 )(

x

 3 )

x

 4 (

x

 4 )(

x

 3 )

x

 2 4 )(

x

 3 ) 2 2 (

x

 3 )



x

2  8

x

(

x

  16   3 ) 2

x x

2  8  9 

x



x

 4  (

x

(

x

  3 ) 4  (

x

 (

x

 3 )  4 ) 4 )   (

x

(

x

2 (

x

 3 )(

x

 3 )(

x

2 (

x

   3 ) 4 ) 4 )  3 ) (

x

 4 )(

x

 3 ) 2 ∴ “c” is the answer.

Standard 15.0

58. A pharmacist mixed some 10% saline solution with some 15% saline solution to obtain 100 mL of a 12% saline solution. How much of the 10% saline solution did the pharmacist use in the mixture?

a) 60 mL b) 45 mL c) 40 mL d) 25 mL

• Create a chart to keep track of all the information then set up equation to solve.

Amount of Solution (mL) Percent Saline Amount of Saline

10% Solution 15% Solution 12% Solution x 100 – x 100 10% 15% 12% .1x

15 - .15x 12 • The answer is 60mL • ∴ “a” is the answer 15  .

15

x

 .

1

x

15  .

05

x

 12  12 15  15  .

05

x

 .

05

x

 .

05

x

 .

05

x x

 12  15   3   3  .

05  60

Standard 15.0

59. Andy’s average driving speed for a 4 hour trip was 45 miles per hour. During the first 3 hours he drove 40 miles per hour. What was his average speed for the last hour of his trip?

a) 50 miles per hour b) 60 miles per hour c) 65 miles per hour d) 70 miles per hour

• Setup a chart to keep track of the given.

Part of trip

First 3 hours

Rate

40

Time

3

Distance

120 miles Total Trip 45 4 180 miles • Next, use what you know. The total trip is 180 miles by using Rate •Time = Distance (rt=d). Also using the same formula, Andy drove for 120 miles. To find the average speed for miles per hour you subtract the two distances. ( 45  4 )  ( 40  3 ) 

x

180  120  60 

x x

• ∴ “b” is the answer

Standard 15.0

60. One pipe can fill a tank in 20 minutes, while another takes 30 minutes to fill the same tank. How long would it take the two pipes together to fill the tank?

a) 50 min b) 25 min c) 15 min d) 12 min

• 1 st , reduce your choices with what you know. You can eliminate “a” and “b” since one pipe alone can fill up the tank in 20 minutes, thus two tanks will not take as long. Then set up as fractions. 1 tank 20 minutes  1 tank 30 minutes 

x

1 20  3 3  1 30  2 2 

x

3 60  2 60 

x

5 60  Reduce  1 12 

x

• 1 tank per 12 minutes. ∴ “d” is the answer

Standard 15.0

61.Two airplanes left the same airport traveling in opposite directions. If one airplane averages 400 miles per hour and the other airplane averages 250 miles per hour, in how many hours will the distance between the two planes be 1625 miles?

a) 2.5

b) 4 c) 5 d) 10.8

• In finding distance the formula is rt=d (rate • time = distance) Airplane 1 Airplane 2

Rate

400 250 t t

Time

400

t

 250

t

 1625 650

t

 1625 650

t

650

t

 1625 650  2 .

5

Distance

400t 250t • 2.5 hours would be your answer. ∴ “a” is the answer

Standard 16.0

62. Which relation is a function?

a) {(-1, 3), (-2, 6), (0, 0), (-2,-2)} b) {(-2, -2), (0, 0), (1, 1), ( 2, 2)} c) {(4, 0), (4, 1), (4, 2), (4, 3)} d) {(7, 4), (8, 8), (10, 8), (10, 10)}

• Knowing what a “relation” and a “function” are is key.

• A relation is grouping of inputs (x) with outputs (y) normally shown as ordered pairs (x, y). • A function is a relation however, each input has only one output. No inputs will be used twice.

• Notice in “a”, -2 is used twice, “c” 4 is used more than once, and “d” 10 is used twice. Only “b” does not have a repeating input. Perform the vertical line test to verify.

• ∴ “b” is the answer

Standard 17.0

63. For which equation graphed below are all the y-values negative?

• This is the type of problem you want to do quickly and not over think it. • Notice the question asked for the y-values that are all negative . To solve, you look at the lines in the graphs. In short, if the line goes above or on the x-axis it is not the answer. Thus, answers “b”, “c”, and “d” all go past the x-axis. When they do this y≥ 0 (y is greater than or equal to zero) making those three choices incorrect. • ∴ “a” is the answer

Standard 17.0

64. What is the domain of the function shown on the graph below?

a) {-1, -2, -3, -4} b) {-1, -2, -4, -5} c) {1, 2, 3, 4} d) {1, 2, 4, 5}

• This problem is clearly a trick trying to determine if you know the difference between domain (it’s “x”) and range (it’s “y”). (x, y) or (domain, range) • Now, knowing that domain is “x”, and all the points are to the right of zero (positive numbers), answers “a” and “b” are eliminated (negative numbers) • Then use what you see. There is a point at (5, -2). Answer “c” has no 5 while “d” does.

• ∴ “d” is the answer