37_Cal_Solution of Calculus_6e

37_Cal_Solution of Calculus_6e - y (5) = 1 (x 0); 2...

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y ( 5) = 1 2 ( x 0); alternatively, 2 y +10= x. C01S0M.022: The equation 3 x 2 y = 4 of the other line may be written in the form y = 3 2 x 2, revealing that it and L have slope 3 2 . Hence an equation of L is y ( 3) = 3 2 ( x 2); that is, y = 3 2 x 6 . C01S0M.023: The equation y 2 x = 10 may be written in the form y =2 x + 10, showing that it has slope 2. Hence the perpendicular line L has slope 1 2 . Therefore an equation of L is y 7= 1 2 ( x ( 3)); that is, x +2 y =11 . C01S0M.024: The segment S joining (1 , 5) and (3 , 1) has slope ( 1 ( 5)) / (3 1) = 2 and midpoint (2 , 3), and hence L has slope 1 2 and passes through (2 , 3). So an equation of L is y ( 3) = 1 2 ( x 2); that is, x +2 y = 4 . C01S0M.025: The graph of y = f ( x )=2 2 x x 2 is a parabola opening downward. The only such graph is shown in Fig. 1.MP.6. C01S0M.026:
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This note was uploaded on 09/14/2011 for the course MATH 101 taught by Professor Cheng during the Spring '11 term at Ecole Hôtelière de Lausanne.

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