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Math 140 Quiz 1 - Summer 2004 Solution Review

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(Small white numbers next to problem number represent its difficulty as per cent getting it wrong.)

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

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

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Problem 3 (38) Simplify the radicals and combine any like terms. Assume all variables represent positive real numbers. 192 1 3 - 3(2 27) 1 3 = 192 1 3 - 32 1 3 (3 3 ) 1 3 = 192 1 3 - 92 1 3 =

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

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_______________ __________ (10 1/2 ) 2 - 3 2 Problem 5 (55) Rationalize the denominator: 10.

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Problem 6 (55) Use rational exponents to simplify the radical. Assume that all variables represent positive numbers. (99) 1 12 = (3 4 ) 1 2 = 3 4 2 = 3 1 =

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Problem 7 (52) Solve the equation:.

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

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Problem 8 (66) Solve the equation by a u-substitution and factoring. x 4 + x 2 – 2 = 0 Let u = x 2. Then the equation is u 2 + 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. u 2 + u – 2 = (u + 2) (u - 1) = 0 => u = 1 or -2. Since u= x 2 > 0, drop –2 case & deduce x 2 = u = 1. Hence, x 2 – 1= (x + 1) (x - 1) = 0 => x = -1 or 1.

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Problem 9 (62) Use radical notation to write the expression. Simplify if possible:. Note: (-1) = (-1) 3 & 512 = 2 9. Thus, -512 x 12 = (-1) 3 2 9 x 12 => (-512 x 12 ) 1/3 = [(-1) 3 2 9 x 12 ] 1/3 = (-1) 3/3 2 9/3 x 12/3 = - 2 3 x 4 = - 8x 4

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Problem 10 (41) 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? Let W = width & L = length = W + 94 Perimeter = 2L + 2W = 236 2(W + 94) + 2W = 236 4W=236-188=48 => W=12" & L=106"

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Problem 11 (38) Solve by completing the square: x 2 + 8x = 3. x 2 + 8x + (8/2) 2 = 3 + (8/2) 2 x 2 + 8x + 16 = (x + 4) 2 = 19 x + 4 = 19 1/2 or x + 4 = -19 1/2 x = -4 + 19 1/2 or x = -4 - 19 1/2 {-4± }

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Problem 12 (31) Solve the equation: 18n 2 + 78n = 0. 18n 2 + 78n = 0 6n(3n + 13) = 0 n = 0 or (3n + 13) = 0 n = 0 or n = -13/3 {-13/3, 0}

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Problem 13 (34) Solve the equation by factoring: x 3 + 6x 2 - 7x = 0. x(x 2 + 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}

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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

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

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

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Problem 17 (48) Solve the equation. p 2 - 5p + 81 = (p + 4) 2 = p 2 + 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: YES!

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

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Problem 19 (45) Simplify the complex fraction.

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

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1.5 Solving Inequalities. Write each inequality using interval notation, and illustrate each inequality using the real number line.

1.5 Solving Inequalities. Write each inequality using interval notation, and illustrate each inequality using the real number line.

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