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12 Feb 2018

1 5 mark . Which one of the following describes the line through the points (2.1) and (3, 4)? A: x + 3y = 5 B: 3.c – y = 2 C: (x, y) = (1, -2) + +(2,6) D: (2,y) = (2,1)+ t(3,1) E: (2, y) = (3, 4)+t(3,-1)

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Tod Thiel
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12 Feb 2018
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Related questions

1 , mark 3. Which one of the following is an equation of the plane through the point (1,4, -5) with normal (1,3, 2)? A: 1 +12y - 10z = 0 B: x + 3y + 2z = 0 C: 1 + 4y - 5z = 0 D: 1 + 4y - 5z = 3 E: x + 3y + 2z=3

Problem 4 Circle your answer. a) (6 points) If f(x) = 2x + 3x2 - 361 + 5, then the tangent line to the graph y= f(x) is horizontal when i) 1 = -2,3 ii) I = 2,-3 iii) r = 0 iv) not listed b) [6 points] Suppose Given h(x) = f(g(T)), where g(2) = 3, g'(2) = 7, f'(2) = -1 and 1'(3) = -2. Then h'(2) is equal to i) - 1 ii) -2 iii) - 14 iv) not listed (x+3)(1 - 1)3 c) (6 points) The function f(1) == (x2 - 2x + 1)(x + 3) wa has vertical asymptotes at i) x = -3,0 ii) = -3,0,1 iii) x = 0,1 iv) not listed d) (6 points] The solution to the initial value problem v'(x) = 473 – cos(x) +e", y(0) = 3 is i) y(x) = x + sin(x) + e* + 2 ii) y(I) = x4 – sin(x) +et+2 iii) y(x) = 2* – sin(1) + et + 3 iv) not listed

1 , mark 1. If u = (1, -1, 3) and v = (2,1, -1). Find u.v. A: -2 B: -1 C: 0 D:1 E: 2

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