Practice Test Unit 2 (Part 2)

x2 – 6x + 8 = 0 (x – 2)(x – 4) = 0 , x – 4 = 0 x – 2 = 0 x = 2 x = 4 If x2 – 6x + 8 = 0 and y = x + 3, then what are the possible values of y ? 1 x2 – 6x + 8 = 0 (x – 2)(x – 4) = 0 , x – 4 = 0 x – 2 = 0 x = 2 x = 4 Let x = 2: y = x + 3 Let x = 4: y = x + 3 = 2 + 3 = 4 + 3 = 5 = 7

2 (2x + 3y)2 – (2x – 3y)2 (2x + 3y)(2x + 3y) – (2x – 3y) (2x – 3y) (4x2 + 6xy + 6xy + 9y2) – (4x2 – 6xy – 6xy + 9y2) (4x2 + 12xy + 9y2) – (4x2 – 12xy + 9y2) 4x2 + 12xy + 9y2 – 4x2 + 12xy – 9y2 24xy

(x – y)2 = (x – y)(x – y) = x2 – xy – xy + y2 = x2 – 2xy + y2 3 If x2 + y2 = 37 and xy = 24, what is the value of (x – y)2? (x – y)2 = (x – y)(x – y) = x2 – xy – xy + y2 x2 + y2 = 37 = x2 – 2xy + y2 = x2 + y2 – 2xy = 37 – 2xy xy = 24 = 37 – 2(24) = 37 – 48 = –11

If (–8x + 3)(–4x2 + 4x + 6) = ax3 + bx2 + cx + d for all real values of x, what is the value of c ? 18 32x3 – 44x2 – 36x + 18 (Add matching colors)

(y – 5)2 = 0 Find y2 – 2y when y = 5 (y – 5)(y – 5) = 0 (5)2 – 2(5) If (y – 5)2 = 0, then find the value of y2 – 2y ? (y – 5)2 = 0 Find y2 – 2y when y = 5 (y – 5)(y – 5) = 0 (5)2 – 2(5) y – 5 = 0 , y – 5 = 0 25 – 10 y = 5 y = 5 15

6 If x and y are positive integers, then which of the following must be equal to ?

7 Step 1 Step 2 Step 3

8 If and ab  0, then 1  a3 = b  x Reciprocal a3 = bx

9 LCD = 4m2 m2 – 8m + 12 = 0 (m – 2)(m – 6) = 0 m2 + 12 = 8m Solve for m. 9 LCD = 4m2 m2 – 8m + 12 = 0 (m – 2)(m – 6) = 0 m2 + 12 = 8m m – 2 = 0 , m – 6 = 0 m2 – 8m + 12 = 0 m = 2 m = 6

7 + 4 11 Find x + 4 when x = 7 4x – 3 = 25 +3 +3 4x = 28 x = 7 If , then find the value of x + 4 ? 10 Find x + 4 when x = 7 7 + 4 4x – 3 = 25 +3 +3 11 4x = 28 x = 7

49 < x – 2 < 64 51 < x < 66 +2 +2 +2 If x is an integer and , how many different values of x are possible? 11 First, solve inequality. 49 < x – 2 < 64 +2 +2 +2 51 < x < 66

51 < x < 66 If x is an integer and , how 11 many different values of x are possible? 11 51 < x < 66 Integers between 51 and 66 OR 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 – 51 = 15 15 – 1 = 14 Answer: 14 Values

11 Try this strategy How many values of m are possible? Use an inequality with two numbers that are close together. 2 < m < 7 Integers between 2 and 7 3 4 5 6 How many values of m are possible? 10 < m < 45 Answer: 4 Values OR 45 – 10 = 35 35 – 1 = 34 7 – 2 = 5 5 – 1 = 4

12 –16 –16 w = 81

If and , then what is the value of b ? 13 a = 64 b = 16

14 –3x + 13 < –14 –13 –13 –3x < –27 x > 9

10 < x x 10 < x x –10 10 < –10 –5 10 < –5 –1 < –10 If , then which of the following values could be x ? 15 Strategy: Test each answer by substituting for x. A. –10 B. –5 10 < x x 10 < x x –10 10 < –10 –5 10 < –5 –1 < –10 –2 < –5 NO NO

10 < x x 10 < x x 10 < x x 1 10 < 1 2 10 < 2 5 10 If , then which of the following values could be x ? 15 C. 1 D. 2 E. 5 10 < x x 10 < x x 10 < x x 1 10 < 1 2 10 < 2 5 10 < 5 10 < 1 5 < 2 2 < 5 NO NO YES

16 2x = 1 NO NO If , then which of the following statements must be true? 16 Strategy: Substitute ¼ for x in each answer. 2x = 1 NO NO

16 2x > 1 NO NO If , then which of the following statements must be true? 16 Strategy: Substitute ¼ for x in each answer. 2x > 1 NO NO

z < 0 17  y = ( + ) positive y > 0 x > y  If x > y and y > 0 and xz < 0, then which of the following must be true about all the values of z? 17  y = ( + ) positive y > 0 x > y  x = ( + ) positive xz < 0  xz = ( – ) negative (+)z = ( – ) z < 0 (+)(–) = ( – ) z is negative

18 If the sum of two integers x and k is less than x, which of the following must be true? x + k < x –x –x k < 0

Which of the following equations represents the statement above? 19 Twice the difference between a certain number and its square root is 15 more than twice the number. Which of the following equations represents the statement above? 19

If a number is doubled and then increased by 10, the result is 5 less than the square of the number. Which of the following equations represents the statement above? 20 2N + 10 = N2 – 5 2N + 10 = N2 + 5

x – 2 3(x – 2) 3(x – 2) = x + 6 2x – 6 = 6 3x – 6 = x + 6 +6 +6 –x –x If 2 is subtracted from a number and this difference is tripled, the result is 6 more than the number. Find the number. 21 Let number = x ‘2 is subtracted from a number’ = x – 2 3(x – 2) ‘The difference is tripled’ = 3(x – 2) = x + 6 2x – 6 = 6 3x – 6 = x + 6 +6 +6 –x –x 2x = 12 2x – 6 = 6 x = 6

Let 1st integer = x 13 Let 2nd integer = x + 2 13+2 = 15 If the sum of two consecutive odd integers is 28, what is the product? 22 Let 1st integer = x 13 Let 2nd integer = x + 2 13+2 = 15 x + x + 2 = 28 Sum = 13 + 15 = 28 2x + 2 = 28 –2 –2 2x = 26 Product = 13  15 = 195 x = 13

–15 3 –5 –2 Let integer = x 2x + 10 = x2 – 5 –2x – 10 –2x – 10 If a positive integer is doubled and then increased by 10, the result is 5 less than the square of the integer. What is the integer? 23 Let integer = x 2x + 10 = x2 – 5 –15 –2x – 10 –2x – 10 3 –5 0 = x2 – 2x – 15 –2 0 = (x + 3)(x – 5) Two numbers with Product of –15 and Sum of –2 x + 3 = 0 , x – 5 = 0 x = –3 x = 5

Cost of pencil = A Cost of pen = B Jon  1A + 2B = 3.50 Lauren  Jon buys one pencil and two pens for $3.50. Lauren buys four pencils and three pens for $5.50. How much would one pencil and one pen cost? 24 Cost of pencil = A Cost of pen = B Jon  1A + 2B = 3.50 (Multiply Jon by –4) Lauren  4A + 3B = 5.50 Jon  –4A – 8B = –14.00 Add Equations Lauren  4A + 3B = 5.50 – 5B = –8.50 B = 1.70 (Pen cost)

the original equations. 1A + 2B = 3.50 1A + 2B = 3.50 4A + 3B = 5.50 Jon buys one pencil and two pens for $3.50. Lauren buys four pencils and three pens for $5.50. How much would one pencil and one pen cost? 24 Cost of pencil = A Cost of pen = B B = 1.70 (Pen cost) Find A. Use one of the original equations. 1A + 2B = 3.50 1A + 2B = 3.50 4A + 3B = 5.50 A + 2(1.70) = 3.50 Cost of pencil and pen = A + B = 0.10 + 1.70 = 1.80 A + 3.40 = 3.50 A = 0.10

If y varies directly as x2, and y = 3 when x = 3, what is the value of y when x is 6? 25 y varies directly as x y varies directly as x2

5  p = 2  100 5p = 200 p = 40 26 Direct Variation Students receive 5 bonus points for every 2 community service projects they perform. If Mark received 100 bonus points, how many projects did he perform? 26 Note: As the bonus points increase, the community service projects should increase. Direct Variation 5  p = 2  100 5p = 200 p = 40

27 If it takes 4 men 3 hours each to pave a playground, how many hours will it take 12 men to complete the same task? Note: Increasing the number of men will decrease the amount of time to complete the task. Inverse Variation M1H1 = M2H2 4 · 3 = 12 · H2 x1y1 = x2y2 12 = 12H2 1 Hour 1 = H2

f(3) = 33 + 3(3) + 30 f(3) = 27 + 9 + 1 f(3) = 37 What is the value of 28 What is the value of f(x) = 3x + 3x + 30 if x = 3 ? f(3) = 33 + 3(3) + 30 f(3) = 27 + 9 + 1 f(3) = 37

29 If f(x) = x + 2x, what is the value of f(–2)? f(–2) = –2 + 2–2

x – 5 > 0 +5 +5 x > 5 Find the domain for 30 Find the domain for Note: We can only evaluate the square root of numbers greater than or equal to zero. Let expression inside radical be > 0. x – 5 > 0 +5 +5 x > 5

31 The amount a restaurant owner pays for coffee beans is directly proportional to the number of pounds of coffee she buys. If she buys n pounds of coffee at d dollars per pound, what is the total amount she pays, in dollars, in terms of n and d. n lb. d dollars/lb. total 1 lb. $3 dollars/lb. 1  3 = $3 2 lb. $3 dollars/lb. 2  3 = $6 3 lb. $3 dollars/lb. 3  3 = $9 n lb. $d dollars/lb. n  d

32 The cost of preparing for a book sale is $30. If each book is sold for $3.00, express the profit as a function of n, where n represents the number of books sold. Book Cost # of books sold Preparation Cost Profit 3 1 30 3(1) – 30 = 3–30 = –27 3 2 30 3(2) – 30 = 6–30 = –24 3 3 30 3(3) – 30 = 9–30 = –21 3 4 30 3(4) – 30 = 12–30 = –18

32 The cost of preparing for a book sale is $30. If each book is sold for $3.00, express the profit as a function of n, where n represents the number of books sold. Book Cost # of books sold Preparation Cost Profit 3 6 30 3(6) – 30 = 18–30 = –12 3 10 30 3(10) – 30 = 30–30 = 0 3 12 30 3(12) – 30 = 36–30 = 6 3 n 30 3n – 30 f(n) = 3n – 30

33 Morgan’s plant grew from 42 centimeters to 57 centimeters in a year. Linda’s plant, which was 59 centimeters at the beginning of the year, grew twice as many centimeters as Morgan’s plant did during the same year. How tall, in centimeters, was Linda’s plant at the end of the year? Step 1 Centimeters Morgan’s plant grew = 57 – 42 = 15 cm. Step 2 Twice centimeters Morgan’s plant grew = 2(15) = 30 cm. Step 3 Height of Linda’s plant = 59 cm + Step 2 = 59 cm + 30 cm = 89 cm.

34 If , then h(x) is Always Positive Never Negative III. Always an Integer False False Note 1 We can only evaluate the square root of numbers greater than or equal to zero. Note 2 Note 3

(31) = 31 + 12 = 3 + 1 = 4 (31)1 = 41 = 41 + 12 = 4 + 1 = 5 35 For all numbers x and y, let xy be defined as xy = xy + y2. What is the value of (31)1 ? (31) = 31 + 12 = 3 + 1 = 4 (31)1 = 41 = 41 + 12 = 4 + 1 = 5

{–5,5} x2 – 25 = 0 (x – 5)(x + 5) = 0 x – 5 = 0 , x + 5 = 0 +5 +5 –5 36 Which values are not in the domain of Let denominator = 0 Solve equation for x. x2 – 25 = 0 (x – 5)(x + 5) = 0 x – 5 = 0 , x + 5 = 0 +5 +5 –5 –5 {–5,5} x = 5 x = –5

37 The sign-up fee at a gym is $50. Members then must pay $25 each month. Express the cost of using the gym as a function of m, where m represents the number of months the member participates. Sign-up Fee # of Months Monthly Cost Gym Cost 50 1 25(1) 50 + 25(1) = 75 50 2 25(2) 50 + 25(2) = 100 50 3 25(3) 50 + 25(3) = 125 50 m 25(m) 50 + 25(m) f(m) = 50 + 25m

38 # of Rings Total rings left 1 7(1) = 7 23 – 7 = 16 NO # of Rings Each time Shannon pushes the button on a machine, a bell rings 7 times. Each time she turns the switch on the machine, the bell rings 3 times. During one hour, Shannon caused the bell on the machine to ring 23 times. How many times did she push the button? # of Rings Total rings left Pushed Button 1 7(1) = 7 23 – 7 = 16 NO There are 3 rings each time a switch is turned. Thus, the number of rings must be a multiple of 3. # of Rings Total rings left Pushed Button 2 7(2) = 14 23 – 14 = 9 YES Turned Switch 3 3(3) = 9

39 Which system of inequalities best represents the graph? A. Wrong B. D. The dotted line has a negative slope. The 2nd inequality for each answer has a negative slope. The inequality sign should be < or > .

39 Which system of inequalities best represents the graph? B. C. D. Wrong The solid line has a positive slope. The 1st inequality for each answer has a positive slope. Inequality sign should be < . The shading is below the line.

  39 Which system of inequalities best represents the graph? B. C. 3 2 - Answers B and C have different slopes for the solid line.  - Find two points on the line. - Use to get to each point on the line. This will determine the slope and equation of the line.

y < –x – 1 –2x + y > –2 y > 2x – 2 40 Negative Slope Determine the solution of the system of inequalities. y < –x – 1 Negative Slope (Shaded Below) –2x + y > –2 y > 2x – 2 Positive Slope (Shaded Above) A. B. C. D.