Martens_WA1

# Martens_WA1 - 3x 4y 2 = 0 B Perpandicular line y – 4...

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Section 1.2 24. 6x – 5y = 15 Y = 6/5x -3 The slope is m = 6/5 and the y intercept is 0, -3 26. y = -1 Slope = 0 y intercept is (0, -1) 28. Point (-1, 2), Slope m = Undefined. x = -1 y = 2 y -2 = 0 x+1 = 0 Slope is undefined therefore the slope is vertical. 32. Point (-2, 4), Slope m = -3/5 Y – 4 = -3/5 ( x + 2 ) 5y – 20 = -3(x + 2)

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5y – 20 = -3x – 6 0 = 3x + 5y – 14 38. (-3, 6), (1, 2) m = (6-2)/(-3-1) m = 4/-4 m = -1 y – 2 = -1 (x - 1) y – 2 = -x + 1 x + y – 3 = 0 --2nd way of doing it (just for fun)-- -3-1 = m(6-2) -4 = m(4) -4 = -1*4, m=-1 60. Point(-3, 2), Line x + y = 7 y = -x + 7 m = -1 A. Parallel line y – 2 = -1 ( x + 3 ) y – 2 = -x – 3
x+y+1 = 0 B. Perpandicular line y – 2 = 1 ( x + 3 ) y – 2 = x + 3 -x+y-5 = 0 62. Point(-6, 4), Line 3x + 4y = 7 y = -3/4x + 7/4 m = -3/4 A. Parallel line y – 4 = -3/4(x + 6) 4y – 16 = -3x -18

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Unformatted text preview: 3x +4y + 2 = 0 B. Perpandicular line y – 4 = -3/4(x+6) 3y -12 = 4x +24 4x – 3y + 36 = 0 Section 1.3 10. f(x) = 3x -1--= = 3, x does not equal 1 32. g(x) = 4/x Domain = (-, 0) U (0, ) ∞ ∞ Range = (-, 0) U (0, ) ∞ ∞ Section 1.5 34. f(x) = x 2 , x ≥ x = y √ f-1 (x) = x √ 56. f(x) = cot x, interval = (0, ) π Function f passes the horizontal line test on (0, ) therefore it is one to one. π 64. Part a.) f(x) = 16 – x 4 The function is one to one where x is greater than or equal to 0. Part b.) f(x) = 16 – x 4 16 – x 4 = y 16 – y – x 4 = 0 16 – y = x 4 x = 4 16 - √ 4 y √ y = 4 16 - √ 4 x √ f-1 (x) = 4 16 - √ 4 x, and x 16 √ ≤ 76. (g-1 ° g-1 )(-4) g-1 (g-1 (-4)) g-1 ( 3-4) √ – 9-4 √ 94. arccot(- 3) √ 5 π /6...
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## This note was uploaded on 12/30/2009 for the course MAT 231 taught by Professor Thurber during the Spring '09 term at Thomas Edison State.

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Martens_WA1 - 3x 4y 2 = 0 B Perpandicular line y – 4...

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