# barcode in ssrs 2008 y = x + 3 = 3 + 3 = 0 in Visual Studio .NET Drawer Quick Response Code in Visual Studio .NET y = x + 3 = 3 + 3 = 0

y = x + 3 = 3 + 3 = 0
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The two intersection points are shown in Fig. 5-4. Notice that we could have solved 13x 2 + 54x + 45 = 0 by factoring the left side, (13x + 15)(x + 3) = 0 However, such a factorization is sometimes dif cult to discover, whereas the quadratic formula can be applied automatically.
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5.3 Find the perpendicular distance from the point P(2, 3) to the line L :
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3y + x = 3.
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Let the perpendicular from P to L hit L at the point X (see Fig. 5-5). If we can nd the coordinates of X, then the distance PX can be computed by the distance formula (2.1). But X is the intersection of line L with line M through P perpendicular to L . The slope-intercept equation of L is 1 y = x+1 3
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Fig. 5-4
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Fig. 5-5
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which shows that the slope of L is 1 . Therefore, by Theorem 4.2, the slope of M is 1/ 1 = 3 so that a point-slope 3 3 equation of M is y 3 = 3(x 2) Solving for y, we obtain the slope-intercept equation of M , y 3 = 3x 6 Now solve the system L: 3y + x = 3 M: y = 3x 3 By the second equation, substitute 3x 3 for y in the rst equation, 3(3x 3) + x = 3 9x 9 + x = 3 10x = 12 6 12 = x= 10 5 or y = 3x 3
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INTERSECTIONS OF GRAPHS
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By the equation for M , when x = 6 , 5 y=3 Thus, point X has coordinates ( 6 , 3 ), and 5 5 PX = 2 6 2 3 2 + 3 = 5 5 4 2 + 5 12 2 = 5 16 144 + = 25 25 160 25 6 5 3= 18 3 3= 5 5
160 = = 25 Use the approximation
4 10 16 10 = = 0.8 10 5 5
10 3.16 (the symbol means is approximately equal to ) to obtain PX 0.8(3.16) 2.53
Supplementary Problems
5.4 Find the intersections of the following pairs of graphs, and sketch the graphs. (a) The lines L : x 2y = 2 and M : 3x + 4y = 6 (b) The lines L : 4x + 5y = 10 and M : 5x + 4y = 8 (c) The line x + y = 8 and the circle (x 2)2 + (y 1)2 = 25 (d) The line y = 8x 6 and the parabola y = 2x 2 (e) The parabolas y = x 2 and x = y2 (f) The parabola x = y2 and the circle x 2 + y2 = 6 (g) The circles x 2 + y2 = 1 and (x 1)2 + y2 = 1 (h) The line y = x 3 and the hyperbola xy = 4 (i) The circle of radius 3 centered at the origin and the line through the origin with slope 2 3 (j) The lines 2x y = 1 and 4x 2y = 3 5.5 GC Solve Problem 5.4 parts (a) (j) by using a graphing calculator. 5.6 (a) Using the method employed in Problem 5.3, show that a formula for the distance from the point P(x1 , y1 ) to the line L : Ax + By + C = 0 is |Ax1 + By1 + C| A2 + B2 (b) Find the distance from the point (0, 3) to the line x 2y = 2. 5.7 Let x represent the number of million pounds of mutton that farmers offer for sale per week. Let y represent the number of dollars per pound that buyers are willing to pay for mutton. Assume that the supply equation for mutton is y = 0.02x + 0.25 that is, 0.02x + 0.25 is the price per pound at which farmers are willing to sell x million pounds of mutton per week. Assume also that the demand equation for mutton is y = 0.025x + 2.5
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