Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Math B70 Midterm I Name: _______________ Answers 1. Which of the following could be the graph of a function? (There may be more than one!) (A) (B) 1. (C) B, C (D) SOLUTION: By the vertical line test (vertical lines cross the graph of a function at most once), (B) and (C) are both the graphs of functions. Problems 2 and 3 refer to the function F graphed in the figure: 2. What is the range of F? 2. {1, 3, 4} SOLUTION: The range of a function consists of all the y-coordinates of the points on the functions graph. This function only has three values for its y-coordinates, 1, 3, and 4. 3. What is the domain of F? 3. [–1, 2] SOLUTION: The domain of a function consists of all the x-coordinates of the points on the functions graph. This function only has all values from -1 to 2 for its x-coordinates, . For this problem, not including the endpoints of the interval would be OK too, since the graph isn't clear on this point. Problems 4 and 5 refer to this function: 4. f (1) = ? SOLUTION: 4. 5 5. –1 Since 1 is less than or equal to x=1, 5. f (0) = ? SOLUTION: Since x=0 is less than 1, 6. What is the range of the function g(x) = | 5 – x | ? 6. SOLUTION: Since the output of function g is an absolute value, it cannot be negative. When x=5, is the input, this function outputs 0. Also this function outputs arbitrarily large numbers (and everything in between). Thus it's range is all nonnegative real numbers, . 7. If f (x) = x + 4 and g(x) = x2+ 1, then ( f – g )(2) = ? 7. 1 8. T=0 SOLUTION: & And so: 8. Solve for T: SOLUTION: First cross multiply (multiply both sides of the equal sign by the least common multiple of all of its denominators): Distribute: We need to get all the terms with T on one side of the equal sign, with all other terms on the opposite side. To do this, we subtract TS from both sides (moving all T terms to the same side), and also subtract RS from both sides (moving all non T terms to the opposite side): Next we factor out the T: Divide both sides by , and remember that anything divided by 0 equals 0: 9. Solve for B: 9. SOLUTION: First multiply both sides of the equal sign by the least common multiple of all of its denominators, AB. This cancels out all the denominators: Thus: 10. If Z is inversely proportional to A, and Z = 5.8 when A = 1.3, find Z when A = 7.0. (Round off to the nearest tenth) SOLUTION: “Z is inversely proportional to A” means that “ Z = 5.8 when A = 1.3”, so So now we know that 6x + 8y = 7 –6x – 5y = 2 12. 7x – 4 y = 5 1.1 . , and multiplying both sides by 1.3 allows us to solve for k: . Solve each system of linear equations. 11. 10. Thus when A=7, For each system, y = ? 11. SOLUTION: Add the equations, and you get 3y = 9, SOLUTION: Using Cramer's rule, we get 3 so y = 3. 12. 5x + 2y = 1 So. 13. x + 2y + z = 1 13. x – 2y + z = 0 x + 2y – z = 1 SOLUTION: Using Cramer's rule, we get → 14. You have a jar of 80 coins, consisting of nickels and dimes worth a total of $5.45. How many dimes are in the jar? 14. 29 SOLUTION: Let n be the number of nickels and d the number of dimes. Then we have the equations: n + d = 80 5n + 10d = 545 Dividing the second equation by 5 yields the equations: n + d = 80 n + 2d = 109 Cramer's rule yields And so, . 15. Calculate how much 23% acid solution must be mixed with how much 57% solution to get 12 liters of 40% acid solution. How much if each solution is needed? NOTE: Set up the equations, but don't solve. T + F = 12 .23T + .57F = .40(12) = 4.8 15. ________________________ 16. Graph the solution set: x + 5y > 5 5x – y < 5 SOLUTION: x + 5y > 5 ← (0, 0) is NOT on shaded side of this line. 5x – y < 5 ← (0, 0) is on shaded side of this line. x + 5y = 5 x | y 0 | 1 5 | 0 5x – y = 5 x | y 0 | –1 1 | 0 17. Solve 9 < 2x + 5 < 11 , and write the solution set in interval notation. SOLUTION: Subtracting 5 from all 3 sides of the inequality, we get 17. (2, 3) For problems 18 and 19, assume A = {a, b, x, y} and B = {a, x, z}. 18. {a, b, x, y, z} 18. 19. {a, x} 4 < 2x < 6 Then divide by 2: 2< x <3 A∪B = ? In interval notation this is (2, 3). 19. A∩B = ? 20. Find the domain of the function: 20. SOLUTION: The radicand must be non-negative: → 21. Find the solution set: SOLUTION: | 4 – 5 x | = 14 4 – 5x = 14 –5x = 10 or 21. 4 – 5x = –14 –5x = –18 x = –2 or x = 18/5 22. Find all solutions, and express in interval notation: | 4 – 5 x | < 14 SOLUTION: 22. –14 < 4 – 5x < 14 Subtract 4 from each side: –18 < –5x < 10 Divide by –5: 18/5 > 23. Graph the solution set: x > –2 | 4 – 5 x | > 14 SOLUTION: Subtract 4 from each side: Divide by –5: 23. 4 – 5x > 14 or 4 – 5x < –14 –5x > 10 or –5x < –18 x < –2 or x > 18/5 24. Write using fractional exponent notation: 24. 25. Write using radical notation: 25. 26. Simplify completely: 26. SOLUTION: 24 divided by 5 equals 4 with remainder 4 18 divided by 5 equals 3 with remainder 3 27. Simplify: 27. 8 SOLUTION: 28. Rationalize the denominator: 28. SOLUTION: 29. Rationalize the denominator: 29. SOLUTION: 30. Solve: 30. x=2 SOLUTION: Start by squaring both sides: Subtract 10 and add 3x to both sides: Factor: → or However, only 2 works when substituted into the original equation. So the only solution is 2. 31. Solve: SOLUTION: 31. Start by multiplying both sides by : Next, square both sides: And so: → 32. Multiply these complex numbers: SOLUTION: (2 – i)(1 + 2i) → 32. Just FOIL it out: ← remember that 33. Divide these complex numbers: 33. SOLUTION: Multiply the numerator and the denominator by the complex conjugate of the denominator. Recall that , so: