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asp.net barcode reader free Intersection point P in VS .NET
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Create Bar Code In Java Using Barcode creation for Java Control to generate, create bar code image in Java applications. Draw Bar Code In Java Using Barcode maker for Java Control to generate, create bar code image in Java applications. horizontal axes, we know that its slope is 1 Because we ve been told that line L passes through the origin, we know that its yintercept is 0 From algebra, we remember that the slopeintercept form of the Cartesian equation for a straight line is y = mx + b where m is the slope and b is the yintercept Plugging in 1 for m and 0 for b, we find that the Cartesian equation for line L is y = x We ve been told that circle C is centered at the origin and has a radius of 3 units From algebra, we recall that the general form for the equation of a circle centered at the origin is x2 + y2 = r2 where r is the radius When we plug in either 3 or 3 for r, we find that the Cartesian equation for circle C is x2 + y2 = 9 9 Here s the system of Cartesian equations that we ve found, representing line L and circle C as shown in Figs 38 and A5: y = x and x 2 + y2 = 9 Let s replace y in the second equation by x, so we get x2 + ( x)2 = 9 Because ( x)2 = x2 for any real number x, we can rewrite the above equation as x2 + x2 = 9 which simplifies to 2x2 = 9 and further to x2 = 9/2 The solutions to this equation are x = (9/2)1/2 Draw UPCA Supplement 5 In ObjectiveC Using Barcode maker for iPhone Control to generate, create Universal Product Code version A image in iPhone applications. Barcode Generator In ObjectiveC Using Barcode creator for iPad Control to generate, create barcode image in iPad applications. 480 WorkedOut Solutions to Exercises: 19 or x = (9/2)1/2 To solve for y, we must plug in these values of x to either of the equations in our original system Let s use y = x When we put the first of these solutions into that equation, we obtain y = (9/2)1/2 which tells us that one of the points is (x,y) = [(9/2)1/2, (9/2)1/2] When we plug the second solution for x into the equation y = x, we get y = [ (9/2)1/2] = (9/2)1/2 so we know that the other point is (x,y) = [ (9/2)1/2,(9/2)1/2] By inspecting Fig A5, we can see that the points must be P = [ (9/2)1/2,(9/2)1/2] and Q = [(9/2)1/2, (9/2)1/2] 10 To get the Cartesian equivalents of the points we found when we solved Problems 6 and 7, we use the conversion formulas x = r cos q and y = r sin q The polar form of point P is (q,r) = (3p /4,3) In this case, we have x = 3 cos (3p /4) = 3 ( 21/2)/2 = (9/2)1/2 and y = 3 sin (3p /4) = 3 21/2/2 = (9/2)1/2 so the ordered pair is (x,y) = [ (9/2)1/2,(9/2)1/2] 4
The polar form of point Q is (q,r) = (7p /4,3) In this case, we have x = 3 cos (7p /4) = 3 21/2/2 = (9/2)1/2 and y = 3 sin (7p /4) = 3 ( 21/2)/2 = (9/2)1/2 so the ordered pair is (x,y) = [(9/2)1/2, (9/2)1/2] We have found that P = [ (9/2)1/2,(9/2)1/2] and Q = [(9/2)1/2, (9/2)1/2] These results agree with what we got when we solved Problem 9 They are the Cartesian coordinates of points P and Q as shown in Figs 38 and A5 4
1 Here are the two vectors we ve been told to work with: a = ( 3,6) and b = (2,5) In this situation, xa = 3, xb = 2, ya = 6, and yb = 5 The Cartesian sum a + b is a + b = [(xa + xb),( ya + yb)] = [( 3 + 2),(6 + 5)] = ( 1,11) Reversing the order of the sum, we get b + a = [(xb + xa),( yb + ya)] = [2 + ( 3),(5 + 6)] = ( 1,11) 482 WorkedOut Solutions to Exercises: 19 The Cartesian difference a  b is a  b = [(xa xb),( ya yb)] = [( 3 2),(6 5)] = ( 5,1) Reversing the order of the difference, we obtain b  a = [(xb xa),( yb ya)] = {[2 ( 3)],(5 6)} = [(2 + 3),(5 6)] = (5, 1) 2 Imagine that we have an arbitrary Cartesian vector a = (xa,ya) Its Cartesian negative is a = ( xa, ya) By definition, the Cartesian sum vector a + ( a) is a + ( a) = {[xa + ( xa)],[ya + ( ya)]} = [(xa xa),( ya ya)] = (0,0) = 0 Reversing the order of the sum, we get a + a = [( xa + xa),( ya + ya)] = {[xa + ( xa)],[ya + ( ya)]} = [(xa xa),( ya ya)] = (0,0) = 0 3 As with the solutions to Problems 1 and 2, demonstrating this fact is a mere exercise in arithmetic Nevertheless, we can get some practice in mathematical rigor by carefully working our way through each step in the process According to the formula for the Cartesian difference between two vectors from the chapter text, we have a  b = [(xa xb),( ya yb)] and b  a = [(xb xa),( yb ya)] Now let s look closely at the coordinates for these two vectors, and compare them The x coordinate of a  b is the real number xa xb, while the x coordinate of b  a is the real number xb xa From prealgebra, we remember that when we reverse the order of the difference between two numbers, we get the negative In this case, it means xb xa = (xa xb)

