This is a sample test for the elementary algebra diagnostic test at the Harbor College. This is an interactive PowerPoint file. For the best viewing, download it and run it in the viewing–mode. The solution and a link to the review of the topics are provided for each question. The review–links of all the topics may be accessed at here http://www.lahc.edu/math/algebra/ele-alg-place-review.html

How many pieces of candies did Chuck get? 8 Question 1. A bag contained 48 pieces of candies. Joe took 2/3 of the bag, Mary took 3/4 of what was left, and Chuck got the rest. How many pieces of candies did Chuck get? 8 A 6 B 4 C 3 D Answer for Question 1:

Solution to Q1 The correct answer is 4. The statement “(fraction or %) of an amount” is translated to the multiplication operation.

The statement “(fraction or %) of an amount” Solution to Q1 The correct answer is 4. The statement “(fraction or %) of an amount” is translated to the multiplication operation. Joe took ¾ of the bag so ¼ of the bag was left, 1 12 i.e. (48) = 12 pieces were left. 4

The statement “(fraction or %) of an amount” Solution to Q1 The correct answer is 4. The statement “(fraction or %) of an amount” is translated to the multiplication operation. Joe took ¾ of the bag so ¼ of the bag was left, 1 12 i.e. (48) = 12 pieces were left. 4 Mary took 2/3 so Chuck got 1/3 of 12, 1 4 i.e. (12) = 4 pieces. 3 Link to review Next Question

How many pieces of candies did Chuck get? 8 Question 1. A bag contained 48 pieces of candies. Joe took 2/3 of the bag, Mary took 3/4 of what was left, and Chuck got the rest. How many pieces of candies did Chuck get? 8 A 6 B 4 Correct! C 3 D Next Question

2 1 Question 2. 1 3 = 3 3 1 –1 A 3 2 –1 B 3 1 C –2 3 2 –2 D 3 Answer for Question 2:

Solution to Q2 2 The correct answer is . –1 3 1 2 3 1 3 3 1 2 1 = 3 3 3

Solution to Q2 2 The correct answer is . –1 3 1 2 3 1 3 3 1 2 1 = 3 3 3 1 2 = 2 3 3 2 = 1 3 2 1 2 hence Link to review 1 3 = –1 3 3 3 Next Question

2 1 Question 2. 1 3 = 3 3 1 –1 3 2 –1 Correct! 3 1 –2 3 2 –2 3 A 3 2 –1 Correct! B 3 1 C –2 3 2 –2 D 3 Next Question

Question 3. After Joe got a 8% raise he earns $1,350 per week Question 3. After Joe got a 8% raise he earns $1,350 per week. How much was his weekly wage before the raise? A $1,250 B $1,225 C $1,200 D $1,175 Answer for Question 3:

Solution to Q3 The answer is $1,250. Suppose Joe’s wage was x before the raise, then a raise of 8% is 0.08x.

Solution to Q3 The answer is $1,250. Suppose Joe’s wage was x before the raise, then a raise of 8% is 0.08x. Hence his new wage is x + 0.08x or 1.08x, which is $1,350.

Solution to Q3 The answer is $1,250. Suppose Joe’s wage was x before the raise, then a raise of 8% is 0.08x. Hence his new wage is x + 0.08x or 1.08x, which is $1,350. Therefore 1.08x = 1350 so that 1350 x = = 1250. 1.08 Link to review Next Question

Question 3. After Joe got a 8% raise he earns $1,350 per week Question 3. After Joe got a 8% raise he earns $1,350 per week. How much was his weekly wage before the raise? A $1,250 Correct! B $1,225 C $1,200 D $1,175 Next Question

Question 4. 4,000 x (7.5 x 10–9) x (8 x 103) = 240. A 24 B C 2.4 D 0.24 Answer for Question 4:

Solution to Q4 The answer is 0.24. 4,000 x (7.5 x 10–9) x (8 x 103) = 4 x 103 x (7.5 x 10–9) x (8 x 103)

Solution to Q4 The answer is 0.24. 4,000 x (7.5 x 10–9) x (8 x 103) = 4 x 103 x (7.5 x 10–9) x (8 x 103) = 4 x 7.5 x 8 x 103 x 10–9 x 103 Next Question

Solution to Q4 The answer is 0.24. 4,000 x (7.5 x 10–9) x (8 x 103) = 4 x 103 x (7.5 x 10–9) x (8 x 103) = 4 x 7.5 x 8 x 103 x 10–9 x 103 = 240 x 10-3 = 0.24 Link to review Next Question

Question 4. 4,000 x (7.5 x 10–9) x (8 x 103) = 240. A 24 B C 2.4 D 0.24 Correct! Next Question

Question 5. Given the right triangle as shown with c = 12, a = 5, which is the closest approximated value of x? A 12 11 B C 10 D 9 Answer for Question 5:

Solution to Q5 The answer is 11. By the Pythagorean Theorem it must be that x2 + 52 = 122

Solution to Q5 Solution to Q5 The answer is 11. By the Pythagorean Theorem it must be that x2 + 52 = 122 x2 + 25 = 144 x2 = 119 or that x = √119 ≈ 11 Link to review Next Question

Question 5. Given the right triangle as shown with c = 12, a = 5, which is the closest approximated value of x? A 12 11 B Correct! C 10 D 9 Next Question

Question 6. Which of the following percentages is closest to 2/7? 30% A 25% B C 20% 15% D Answer for Question 6:

Solution to Q6 The answer is 30%. 2/7 = 0.285.. ≈ 30% Link to review Next Question

Question 6. Which of the following percentages is closest to 2/7? 30% A Correct! 25% B C 20% 15% D Next Question

Question 7. 4.5 x 0.2 – (3.5 – 2 x 0.85) = 7.2 A B –0.9 C 10 D –4.8 Answer for Question 7:

Solution to Q7 The answer is 7.2. 4.5 x 0.2 – (3.5 – 2 x 0.85) = 0.9 – (3.5 – 1.7)

Solution to Q7 The answer is 7.2. 4.5 x 0.2 – (3.5 – 2 x 0.85) = 0.9 – (3.5 – 1.7) = 0.9 – 1.8 = –0.9 Link to review Next Question

Question 7. 4.5 x 0.2 – (3.5 – 2 x 0.85) = 7.2 –0.9 Correct! 10 –4.8 A B –0.9 Correct! C 10 D –4.8 Next Question

3 15 – 3 26 is approximately Question 8. 13 11 9 7 B 11 C 9 D 7 Answer for Question 8:

Solution to Q8 The answer is 9. 3 15 is approximately 3(4) = 12, and 3 26 is approximately 3 Next Question

Solution to Q8 The answer is 9. 3 15 is approximately 3(4) = 12, and 3 26 is approximately 3 so that 3 15 – 3 26 ≈ 12 – 3 = 9. Link to review

3 15 – 3 26 is approximately Question 8. 13 11 Correct! 9 7 B 11 Correct! C 9 D 7 Next Question

Question 9. If x = 3, and that xy – x – y = –4, what is y? 3/2 A –3/2 B 1/2 C –1/2 D Answer for Question 9:

Solution to Q9 The answer is -1/2. If x = 3, then xy – x – y = –4 is 3y – 3 – y = –4.

Solution to Q9 The answer is -1/2. If x = 3, then xy – x – y = –4 is 3y – 3 – y = –4 so that 2y = –1 or y = –1/2 Link to review Next Question

Question 9. If x = 3, and that xy – x – y = –4, what is y? 3/2 A –3/2 B 1/2 C –1/2 Correct! D Next Question

Question 10. If x = –3, then –x2 – 2x + 3 is A 6 B 12 C 18 D Answer for Question 10:

Solution to Q10 The answer is 8. If x = –3, then –x2 – 2x + 3 is –(–3)2 – 2(–3) + 3 = –9 + 6 + 3 = 0 Link to review Next Question

Question 10. If x = –3, then –x2 – 2x + 3 is Correct! A 6 B 12 C 18 D Next Question

Question 11. If a = –2, b = –1, c = 3, then b2 – 4ac is 23 A 25 B C –25 –23 D Answer for Question 11:

Solution to Q11 The answer is 25. If a = –2, b = –1, c = 3, then (–1)2 – 4(–2)(3) = 1 + 24 = 25 Link to review Next Question

Question 11. If a = –2, b = –1, c = 3, then b2 – 4ac is 23 A 25 Correct! B C –25 –23 D Next Question

Question 12. If a – bx = c, then x = 𝑐−𝑎+𝑏 𝑐 𝑎–𝑏 C 𝑎−𝑐 𝑏 D Answer for Question 12:

Solution to Q12 The answer is . If a – bx = c then a – c = bx so = x 𝑎−𝑐 𝑏 If a – bx = c then a – c = bx so 𝑎−𝑐 𝑏 = x Link to review Next Question

Question 12. If a – bx = c, then x = 𝑐−𝑎+𝑏 𝑐 𝑎–𝑏 C 𝑎−𝑐 𝑏 Correct! D Next Question

Question 13. –2(3A – 4B) – 5(–7A + 6B) = C 29A – 38B –41A – 38B D Answer for Question 13:

Solution to Q13 The answer is 29A –22B. –2(3A – 4B) – 5(–7A + 6B) = –6A + 8B + 35A – 30B = 29A – 22B Link to review Next Question

Question 13. –2(3A – 4B) – 5(–7A + 6B) = Correct! A –41A – 22B B C 29A – 38B –41A – 38B D Next Question

14. Given the system 2x – y = 1 3x – y = –1 the solution for x is: 2 A 2 B –2 C –4 D Answer for Question 14:

Solution to Q14 The answer is x = –2. Subtract the two equations 3x – y = –1 2x – y = 1 – ) x = –2 Link to review Next Question

14. Given the system 2x – y = 1 3x – y = –1 the solution for x is: 2 A 2 B –2 C Correct! –4 Next Question D

Question 15. 𝑥 2 𝑦 –7 𝑥 –3 𝑦 –4 = x y3 y3/x x5 y3 x5/y3 A y3/x B x5 y3 C x5/y3 D Answer for Question 15:

Solution to Q15 The answer is x5/y3 By the Divide–Subtract rule for the exponents: 𝑥 2 𝑦 –7 𝑥 –3 𝑦 –4 = x 2–(–3) y –7–(–4)

Solution to Q15 The answer is x5/y3 By the Divide–Subtract rule for the exponents: 𝑥 2 𝑦 –7 𝑥 –3 𝑦 –4 = x 2–(–3) y –7–(–4) = x5y–3 = x5/y3 Link to review Next Question

Question 15. 𝑥 2 𝑦 –7 𝑥 –3 𝑦 –4 = x y3 y3/x x5 y3 x5/y3 Correct! A y3/x B x5 y3 C x5/y3 Correct! D Next Question

Question 16. From experience, it’s sunny 5 out of 7 days at Sunny Hills. In a period of 182 days, how many cloudy days are expected? A 130 days 125 days B C 120 days D 115 days Answer for Question 16:

Solution to Q16 The answer is 130 days. Let x be the number of sunny days, using proportion, The number of sunny days x 5 = Total number of days 182 7

Solution to Q16 The answer is 130 days. Let x be the number of sunny days, using proportion, The number of sunny days x 5 = Total number of days 182 7 cross–multiplying so that 7x = 182(5)

Let x be the number of sunny days, using proportion, Solution to Q16 The answer is 130 days. Let x be the number of sunny days, using proportion, The number of sunny days x 5 = Total number of days 182 7 cross–multiplying so that 7x = 182(5) 26 182(5) x = 7 x = 130 Link to review Next Question

Question 16. From experience, it’s sunny 5 out of 7 days at Sunny Hills. In a period of 182 days, how many cloudy days are expected? A 130 days Correct! 125 days B C 120 days D 115 days Next Question

Question 17. y is inversely proportional to x2, if x = 2 then y = 9, what is x if y = 4? 6 B 4 C 3 2 D Answer for Question 17:

Solution to Q17 The answer is 3. k y is inversely proportional to x2 means y = x2

Solution to Q17 The answer is 3. k y is inversely proportional to x2 means y = x2 k if x = 2, y = 9 then 9 = so k = 36, 22 36 and that y = x2

Solution to Q17 The answer is 3. k y is inversely proportional to x2 means y = x2 k if x = 2, y = 9 then 9 = so k = 36, 22 36 and that y = x2 36 so if y = 4, then 4 = x2

Solution to Q17 The answer is 3. k y is inversely proportional to x2 means y = x2 k if x = 2, y = 9 then 9 = so k = 36, 22 36 and that y = x2 36 so if y = 4, then 4 = x2 4x2 = 36 x2 = 9 and that x = 3 or –3. Link to review Next Question

Question 17. y is inversely proportional to x2, if x = 2 then y = 9, what is x if y = 4? 6 B 4 Correct! C 3 2 D Next Question

Question 18. (x – 2)(2x – 3) – (3x + 2)(x – 4) = A B –2x2 + 3x + 14 –2x2 – 17x – 2 C D –2x2 – 3x + 14 Answer for Question 18:

Solution to Q18 The answer is –2x2 + 3x + 14. (x – 2)(2x – 3) – (3x + 2)(x – 4) = (x – 2)(2x – 3) + (–3x – 2)(x – 4)

Solution to Q18 The answer is –2x2 + 3x + 14. (x – 2)(2x – 3) – (3x + 2)(x – 4) = (x – 2)(2x – 3) + (–3x – 2)(x – 4) = x2 – 7x + 6 – 3x2 + 10x + 8 = –2x2 + 3x + 14 Link to review Next Question

Question 18. (x – 2)(2x – 3) – (3x + 2)(x – 4) = A B –2x2 + 3x + 14 Correct! –2x2 – 17x – 2 C D –2x2 – 3x + 14 Next Question

Question 19. Starting with a full tank of gas, we drove at a constant speed for 2½ hr and we used up 2/5 of the tank. How much more time we can travel at the same speed before we run out of gas? A 4 hr 3¾ hr B C 3½ hr 3¼ hr D Answer for Question 19:

Use proportion to solve the problem. Solution to Q19 The answer is 3 3 hrs. 4 Use proportion to solve the problem. Let x be the number of hours that we may travel on the remaining 3/5 tank of gas.

Use proportion to solve the problem. Solution to Q19 The answer is 3 3 hrs. 4 Use proportion to solve the problem. Let x be the number of hours that we may travel on the remaining 3/5 tank of gas. Hours of traveling time: x 5/2 = 3/5 2/5 Fraction of the tank:

Use proportion to solve the problem. Solution to Q19 The answer is 3 3 hrs. 4 Use proportion to solve the problem. Let x be the number of hours that we may travel on the remaining 3/5 tank of gas. Hours of traveling time: x 5/2 = cross–multiply 3/5 2/5 Fraction of the tank: 2x 3 5 = 5 5 2

Use proportion to solve the problem. Solution to Q19 The answer is 3 3 hrs. 4 Use proportion to solve the problem. Let x be the number of hours that we may travel on the remaining 3/5 tank of gas. Hours of traveling time: x 5/2 = cross–multiply 3/5 2/5 Fraction of the tank: 2x 3 5 3 = = 5 5 2 2

Use proportion to solve the problem. Solution to Q19 The answer is 3 3 hrs. 4 Use proportion to solve the problem. Let x be the number of hours that we may travel on the remaining 3/5 tank of gas. Hours of traveling time: x 5/2 = cross–multiply 3/5 2/5 Fraction of the tank: 2x 3 5 3 = = 5 5 2 2

Use proportion to solve the problem. Solution to Q19 The answer is 3 3 hrs. 4 Use proportion to solve the problem. Let x be the number of hours that we may travel on the remaining 3/5 tank of gas, Hours of traveling time: x 5/2 = cross–multiply 3/5 2/5 Fraction of the tank: 2x 3 5 3 = cross again = 5 5 2 2 4x = 15 Link to review x = 15/4 = 3 3 Next Question 4

Question 19. Starting with a full tank of gas, we drove at a constant speed for 2½ hr and we used up 2/5 of the tank. How much more time we can travel at the same speed before we run out of gas? A 4 hr 3¾ hr Correct! B C 3½ hr Next Question 3¼ hr D

Question 20. Which of the following is a factor of 3x – 3y + ax – ay B a – 3 y + x C x – y D Answer for Question 20:

Solution to Q20 The answer is (x – y) Factoring by grouping: 3x – 3y + ax – ay = 3(x – y) + a(x – y)

Solution to Q20 The answer is (x – y) Factoring by grouping: 3x – 3y + ax – ay = 3(x – y) + a(x – y) = (x – y) (3 + a) Link to review Next Question

Question 20. Which of the following is a factor of 3x – 3y + ax – ay B a – 3 y + x C x – y D Correct! Next Question

Question 21. 2x – 1 x – 1 Combine and simplify – x – 3 x + 2 Answer for Question 21:

Solution to Q21 The answer is x2 + 7x – 5 (x – 3)(x + 2) Multiply the expression by 1 in the form of LCD/LCD distribute and simplify: 2x – 1 x – 1 (x – 3)(x + 2) – (x – 3)(x + 2) x – 3 x + 2

Solution to Q21 The answer is x2 + 7x – 5 (x – 3)(x + 2) Multiply the expression by 1 in the form of LCD/LCD distribute and simplify: (x + 2) (x – 3) 2x – 1 x – 1 (x – 3)(x + 2) – (x – 3)(x + 2) x – 3 x + 2

Solution to Q21 The answer is x2 + 7x – 5 (x – 3)(x + 2) Multiply the expression by 1 in the form of LCD/LCD distribute and simplify: (x + 2) (x – 3) 2x – 1 x – 1 (x – 3)(x + 2) – (x – 3)(x + 2) x – 3 x + 2 = (2x – 1) (x + 2) – (x – 1)(x – 3) (x – 3)(x + 2) x2 + 7x – 5 = (x – 3)(x + 2) Link to review Next Question

Question 21. 2x – 1 x – 1 Combine and simplify – x – 3 x + 2 Correct! (x – 3)(x + 2) Next Question

Question 22. The solutions of 3x(x – 1) = x2 + 9 are B x = –3/2, 2 x = 2/3, –3 C x = –2/3, 3 D Answer for Question 22:

Solution to Q22 The answer is x = –3/2, 3 3x(x – 1) = x2 + 9 3x2 – 3x = x2 + 9 2x2 – 3x – 9 = 0

Solution to Q22 The answer is x = –3/2, 3 3x(x – 1) = x2 + 9 3x2 – 3x = x2 + 9 2x2 – 3x – 9 = 0 (2x + 3)(x – 3 ) = 0 so x = –3/2, 3 Link to review Next Question

Question 22. The solutions of 3x(x – 1) = x2 + 9 are B x = –3/2, 2 x = 2/3, –3 Correct! C x = –2/3, 3 D Next Question

1 Question 23. Solve for x if a = 1 – x 1 1 – a 1 1 + a –1 1 + a 1 B 1 + a –1 C 1 + a 1 D a – 1 Answer for Question 23:

Solution to Q23 The answer is x = 1 1 – a 1 a = 1 – x 1 = 1 – a x 1 x = 1 – a Link to review Next Question

1 Question 23. Solve for x if a = 1 – x 1 Correct! 1 – a 1 1 + a –1 B 1 + a –1 C 1 + a 1 D a – 1 Next Question

Question 24. (1 – x2) (x2 – 3x) Simplify * x2 (x2 – 4x + 3) x + 1 x A x –x – 1 B x x – 1 C x 1 – x D x Answer for Question 24:

Solution to Q24 The answer is –x – 1 x (1 – x2) (x2 – 3x) * x2 (x2 – 4x + 3) (1 – x)(1 + x) x(x – 3) = x2(x – 3)(x – 1)

Solution to Q24 The answer is –x – 1 x (1 – x2) (x2 – 3x) * x2 –1 (1 – x)(1 + x) x(x – 3) = x2(x – 3)(x – 1) –x – 1 = x Link to review Next Question

Question 24. (1 – x2) (x2 – 3x) Simplify * x2 (x2 – 4x + 3) x + 1 x A x –x – 1 Correct! B x x – 1 C x 1 – x D x Next Question

Question 25. The vertices of a rectangle are (1, 1), (5, 1), (1, 7), and (5, 7). What is the area of the rectangle? A 16 20 B 24 C 28 D Answer for Question 25:

So the area of the rectangle is Δx Δy = 6×4 = 24. Link to review Solution to Q25 The answer is 24. (1, 5) We have (7, 5) Δy = 5 – 1 = 4 Δx = 7 – 1 = 6 (1, 1) (7, 1) So the area of the rectangle is Δx Δy = 6×4 = 24. Link to review Next Question

Question 25. The vertices of a rectangle are (1, 1), (5, 1), (1, 7), and (5, 7). What is the area of the rectangle? A 16 20 B 24 Correct! C 28 D Next Question

Question 26. The solution for the inequality 3 > 1 – 2x ≥ –3 is –1 2 B –1 2 C –1 2 D –1 2 Answer for Question 26:

and reverse the inequality Question 26. The answer is –1 2 3 > 1 – 2x ≥ –3 subtract 1 2 > – 2x ≥ –4 divide by –2 and reverse the inequality –1 < x ≤ 2 or –1 2 Link to review Next Question

Question 26. The solution for the inequality 3 > 1 – 2x ≥ –3 is Correct! A –1 2 B –1 2 C –1 2 D –1 2 Next Question

With the given measurements, what is x? Question 27. Following are two prints of different sizes of the same picture of Tomo. With the given measurements, what is x? 2 A 2 ft 3 2 ft x B 3 ft 3 ft 5 ft 2 3 ft C 3 1 3 ft D 3 Answer for Question 27:

Solution to Q27 The answer is 3 1/3 ft. The ratios of the measurements must satisfies the proportion x : 5 = 2 : 3 or that x 2 = 2 ft x 5 3 3 ft 5 ft Next Question

Solution to Q27 The answer is 3 1/3 ft. The ratios of the measurements must satisfies the proportion x : 5 = 2 : 3 or that x 2 = so 2 ft x 5 3 1 3 ft 2 x = (5) = 3 3 5 ft 3 Link to review Next Question

With the given measurements, what is x? Question 27. Following are two prints of different sizes of the same picture of Tomo. With the given measurements, what is x? 2 A 2 ft 3 2 ft x B 3 ft 3 ft 5 ft 2 3 ft C 3 1 3 ft D Correct! 3 Next Question

Which of the following pictures represents Question 28. Which of the following pictures represents A, B and C on the real line most accurately if A = 3, B = 9 and C = 12? A B C D Answer for Question 28:

C = 12 must be the right most point. Solution to Q28 The answer B. C = 12 must be the right most point. 12

C = 12 must be the right most point. Solution to Q28 The answer B. C = 12 must be the right most point. By dividing from 0 to 12 into 4 pieces 12

C = 12 must be the right most point. Solution to Q28 The answer B. C = 12 must be the right most point. By dividing from 0 to 12 into 4 pieces we see that A = 3, B = 9 and C = 12 must be 3 6 9 12 Link to review Next Question

Which of the following pictures represents Question 28. Which of the following pictures represents A, B and C on the real line most accurately if A = 3, B = 9 and C = 12? A Correct! B C D Next Question

Question 29. Let point A = (–10, 20), the coordinate of the point that is 40 to the right and 40 below A is (40, –30) A (30, –30) B (40, –20) C D (30, –20) Answer for Question 29:

Solution to Q29 The answer (30, –20). the point that’s 40 to the right and 40 below of (–10, 20) is (–10 + 40, 20 – 40) = (30, –20) Link to review Next Question

Question 29. Let point A = (–10, 20), the coordinate of the point that is 40 to the right and 40 below A is (40, –30) A (30, –30) B (40, –20) C D (30, –20) Correct! Next Question

Question 30. The slope of the line that has x–intercept at 4 and y intercept at 3 is A 3/4 –3/4 B C 4/3 D –4/3 Answer for Question 30:

Solution to Q30 The answer is –3/4. The slope is “rise/run”.

Solution to Q30 The answer is –3/4. The slope is “rise/run”. The rise and run for (4, 0) and (0, 3) is (4, 0) (0, 3) 4,–3 run rise Hence the slope is –3/4. Link to review Next Question

Question 30. The slope of the line that has x–intercept at 4 and y intercept at 3 is A 3/4 –3/4 Correct! B C 4/3 D –4/3 Next Question

Question 31. The slope of a vertical line is Undefined B C 1 D –1 Answer for Question 31:

Solution to Q31 The answer is “undefined”. The slope is “rise/run”. The run for two points on a vertical line is 0 hence the slope = rise/0 is not defined. Link to review Next Question

Question 31. The slope of a vertical line is Undefined Correct! B C 1 D –1 Next Question

Question 32. Find the y–intercept of the line that contains the points (4, 15) and (9, 45). –8 –9 B C –10 D –11 Answer for Question 32:

Solution to Q32 The answer is –9. The slope of the line that contains the points (4, 15) and (9, 45) is (45 –15)/(9 – 40) = 6

Solution to Q32 The answer is –9. The slope of the line that contains the points (4, 15) and (9, 45) is (45 –15)/(9 – 40) = 6 so the equation of the line containing these points is y = 6(x – 4) + 15

Solution to Q32 The answer is –9. The slope of the line that contains the points (4, 15) and (9, 45) is (45 –15)/(9 – 40) = 6 so the equation of the line containing these points is y = 6(x – 4) + 15 or y = 6x – 9 Hence the y–intercept is at y = –9. Link to review

Question 32. Find the y–intercept of the line that contains the points (4, 15) and (9, 45). –8 –9 B Correct! C –10 D –11