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Math 140 Quiz 1 - Summer 2004 Solution Review

Math 140 Quiz 1 - Summer 2004 Solution Review. Math 140 Quiz 1 - Summer 2004 Solution Review. Math 140 Quiz 1 - Summer 2004 Solution Review. (Small white numbers next to problem number represent its difficulty as per cent getting it wrong.). Problem 1 (3).

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Math 140 Quiz 1 - Summer 2004 Solution Review

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  1. Math 140Quiz 1 - Summer 2004 Solution Review

  2. Math 140Quiz 1 - Summer 2004 Solution Review

  3. Math 140Quiz 1 - Summer 2004 Solution Review (Small white numbers next to problem number represent its difficulty as per cent getting it wrong.)

  4. Problem 1 (3) -7.3q + 1.7q = -59.7 – 1.9 -5.6q = -61.6 q = 11 Thus, the answer is E). Solve the equation: -7.3q + 1.9 = -59.7 – 1.7q.

  5. Problem 2 (69) Divide and simplify. Assume that all variables represent positive real numbers.

  6. Problem 3 (38) Simplify the radicals and combine any like terms. Assume all variables represent positive real numbers. 19•21/3 - 3•(2 • 27)1/3 = 19•21/3 - 3•21/3(33)1/3 = 19•21/3 - 9•21/3 =

  7. Problem 4 (79) Perform the indicated operation and simplify: 4 + 2w . w - 2 2 - w

  8. _______________ Problem 5 (55) Rationalize the denominator: 10 . __________ (101/2)2 - 32

  9. Problem 6 (55) Use rational exponents to simplify the radical. Assume that all variables represent positive numbers. (9•9)1/12 = (34)1/12 = 34/12 = 31/3=

  10. Problem 7 (52) Solve the equation: .

  11. Problem 7 Continued Alternate approach: Multiply by LCD = 12x. Then, 12x – 7(12x)/(3x) = (10/4)(12x) 12x – 28 = 30x -18x = 28 x = -28/18 = -14/9 Solve the equation: .

  12. Problem 8 (66) Let u = x2 . Then the equation is u2 + u – 2 = 0 There are only two factoring possibilities: (u - 1) (u + 2) & (u + 1) (u - 2). But only the combination (u + 2) (u - 1) works. u2 + u – 2 = (u + 2) (u - 1) = 0 => u = 1 or -2. Since u= x2> 0, drop –2 case & deduce x2= u = 1. Hence, x2– 1= (x + 1) (x - 1) = 0 => x = -1 or 1. Solve the equation by a u-substitution and factoring. x4 + x2 – 2 = 0

  13. Problem 9 (62) Note: (-1) = (-1)3 & 512 = 29. Thus, -512 x12 = (-1)3 29 x12 => (-512 x12 )1/3 = [(-1)3 29 x12 ]1/3 = (-1)3/3 29/3 x12/3 = - 23 x4 = - 8x4 Use radical notation to write the expression. Simplify if possible: .

  14. Problem 10 (41) Let W = width & L = length = W + 94 Perimeter = 2L + 2W = 236 2(W + 94) + 2W = 236 4W=236-188=48 => W=12" & L=106" A rectangular carpet has a perimeter of 236 inches. The length of the carpet is 94 inches more than the width. What are the dimensions of the carpet?

  15. Problem 11 (38) x2 + 8x + (8/2)2 = 3 + (8/2)2 x2 + 8x + 16 = (x + 4)2 = 19 x + 4= 191/2 or x + 4= -191/2 x = -4 + 191/2 or x = -4 - 191/2 {-4± } Solve by completing the square: x2 + 8x = 3.

  16. Problem 12 (31) 18n2 + 78n = 0 6n(3n +13) = 0 n = 0 or (3n +13) = 0 n = 0 or n = -13/3 {-13/3, 0} Solve the equation: 18n2 + 78n = 0.

  17. Problem 13 (34) x(x2 + 6x - 7) = 0 x(x + 7)(x - 1) = 0 x = 0 or x + 7= 0 or x - 1= 0 x = 0 or x = - 7 or x = 1 {-7, 0, 1} Solve the equation by factoring: x3 + 6x2 - 7x = 0.

  18. Problem 14 (69) The manager of a coffee shop has one type of coffee that sells for $10 per pound (lb) and another type that sells for $15/lb. The manager wishes to mix 40 lbs of the $15 coffee to get a mixture that will sell for $14/lb. How many lbs of the $10 coffee should be used? Let t = amt of $10/lb & f = amt of $15/lb = 40 To have value equal: 10t +15f = 14(40+t). 10t +15(40) = 560+ 14t or 10t +600 – 14t = 560 -4t = -40 => t = 10 pounds

  19. ( 5 6 7 8 9 10 5 6 7 8 9 10 Problem 15 (38) • Recall rules: + => Open(+) right/left(-) end Note: x > 6 => 6 < x so only left end is determined. • Open end Closed end • Left side: a < x => (a, a <x => [a, • (6, ............................) Write each expression in interval notation. Graph each interval. x > 6

  20. ( ] -3 -2 -1 0 1 2 -3 -2 -1 0 1 2 Problem 16 (21) • Recall notation rules: • Open end Closed end • Left side: a < x => (a, a <x => [a, • Right side: x < a => , a) x <a => , a] • (-2, 1] Write each expression in interval notation. Graph each interval. -2 < x< 1

  21. Problem 17 (48) p2 - 5p + 81 = (p + 4)2 = p2 + 8p + 16 -13p + 65 = 0 -13p = -65 p = 5 ________________________ {5} is solution set. Always check when squaring radical equations since spurious roots can be introduced. Check: Solve the equation. YES!

  22. Problem 18 (17) 5m + 4 = 2 or 5m + 4 = -2 5m = -2 or 5m = -6 m = -2/5 or m = -6/5 {-2/5, -6/5} Solve the equation: |5m + 4| + 8 = 10 .

  23. Problem 19 (45) Simplify the complex fraction.

  24. Problem 20 (41) • r + 4 > 2 or r + 4 < -2 • r > 2 - 4 or r < -2 - 4 • > - 2 or r < -6 (-, -6] or [-2, ) Solve the inequality. Write answer in interval notation. |r + 4| > 2

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