Read e-book online 101 short cuts in math anyone can do PDF

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To obtain the slope and y-intercept, we transform the equation into its slopeintercept form. To do this we solve for y. 2x − 3 y = 6 y= 2 x−2 3 2 3 slope: m = ; y-intercept: (0, – 2) 57. To obtain the slope and y-intercept, we transform the equation into its slopeintercept form. To do this we solve for y. x+ y =1 y = – x +1 slope: m = – 1; y-intercept: (0, 1) 59. x = – 4 The slope is not defined; there is no yintercept. y=5 slope: m = 0; y-intercept: (0, 5) 63. To obtain the slope and y-intercept, we transform the equation into its slopeintercept form.

We write the solution set either as {x | x < 3 or x > 4} or as all x in the interval (– ∞, 3) or (4, ∞). 5 27 79. First we solve the equation 4 x 2 + 9 = 6 x and use the solutions to separate the real number line. 4x 2 + 9 = 6x 4x 2 − 6x + 9 = 0 This equation has no real solutions. Its discriminant, b 2 – 4ac = 36 – 144 = – 108, is negative. The value of 4 x 2 − 6 x + 9 either is always positive or always negative. To see 2 which is true, we test x = 0. Since 4 ( 0 ) − 6 ( 0 ) + 9 = 9 is positive, we conclude that expression is always positive, and the inequality 4 x 2 + 9 < 6 x has no solution.

4 5 point (0, 0), which is the y-intercept. So, we use the slope-intercept form of the line: y = mx + b y − 0 = 3 (x − (− 4 )) 45. We are given the slope m = y= 47. Since the slope is undefined, the line is vertical. The equation of the vertical line containing the point (1, 4) is: x =1 0 − (−1) 1 = 2−0 2 y = mx + b 1 x −1 2 2y = x − 2 y= x − 2y = 2 49. Since the slope = 0, the line is horizontal. The equation of the horizontal line containing the point (1, 4) is: y=4 51. 9 53. To obtain the slope and y-intercept, we transform the equation into its slopeintercept form.

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