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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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Section 8.1 Completing the Square. Factoring Before today the only way we had for solving quadratics was to factor. x 2 - 2x - 15 = 0 (x + 3)(x - 5) =

Section 8.1 Completing the Square. Factoring Before today the only way we had for solving quadratics was to factor. x 2 - 2x - 15 = 0 (x + 3)(x - 5) =

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