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Plugging these numbers into the formula yields d = [(xp xq)2 + ( yp yq)2]1/2 = {[1 ( 4)]2 + ( 6 5)2}1/2 = [32 + ( 11)2]1/2 = (9 + 121)1/2 = 1301/2 Using a calculator and rounding off to three decimal places, we get d 11402 8 Let s call the endpoints of our line segment L by the names P and Q, such that P = ( 4,5) and Q = ( 5, 3) The coordinate values of these points are xp = 4 yp = 5 xq = 5 yq = 3 Using the formula to find the midpoint (xm,ym), we obtain (xm,ym) = [(xp + xq)/2,( yp + yq)/2] = {[ 4 + ( 5)]/2,[5 + ( 3)]/2} = ( 9/2,2/2) = ( 9/2,1) In decimal form, the ordered pair is exactly (xm,ym) = ( 45,1) 9 Let s call the endpoints of M by the names P and Q, such that P = ( 5, 3) and Q = (1, 6) The coordinates are xp = 5 yp = 3 xq = 1 yq = 6
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Plugging these numbers into the midpoint formula, we get (xm,ym) = [(xp + xq)/2,( yp + yq)/2] = {( 5 + 1)/2,[ 3 + ( 6)]/2} = ( 4/2, 9/2) = ( 2, 9/2) When we express this ordered pair in decimal form, we have exactly (xm,ym) = ( 2, 45) 10 We can call the endpoints of N by the names P and Q, such that P = (1, 6) and Q = ( 4,5) This time, we have xp = 1 yp = 6 xq = 4 yq = 5 When we put these values into the formula for the midpoint, we come up with (xm,ym) = [(xp + xq)/2,( yp + yq)/2] = {[1 + ( 4)]/2,( 6 + 5)/2} = ( 3/2, 1/2) In decimal form, this is exactly (xm,ym) = ( 15, 05)
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1 There are 2p radians in a full circle of 360 If we assume that p 314159, then a full circle has a radian measure of 2p 2 314159 628318 To get the radian measure in 1 , we divide 2p by 360 That gives us 1 628318/360 00175 rounded off to four decimal places
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2 If we go 7/8 of the way around a circle counterclockwise, we rotate thorough an angle of 7/8 2p, or 7p /4 3 An angle of 120 is 1/3 of a circular rotation, because 120 is 1/3 of 360 If we go 1/3 of the way around a circle counterclockwise, that s an angle of 1/3 2p, or 2p /3 4 Imagine that we travel over the earth in a great circle (the shortest path between two points on the surface of a sphere, as measured on that surface) for 1000 /p km If the earth s circumference is 40,000 km and the planet is a perfectly smooth sphere, then our distance traveled is (1,000 /p )/40,000 = 1000/(40,000p) = 1/(40p) of a complete circumnavigation If we travel exactly once around the earth along a great circle, we go through an angle of 2p The angular separation, in radians, of two points located 1000 /p km apart on the surface is therefore [1/(40p)] 2p = (2p)/(40p) = 2/40 = 1/20 5 Figure A-1 shows the graphs of y = 2 sin x (solid curve) and y = sin x (dashed curve) The graph of y = 2 sin x resembles the graph of y = sin x, but the amplitude is doubled 6 Figure A-2 shows the graphs of y = sin 2x (solid curve) and y = sin x (dashed curve) The graph of y = sin 2x resembles the graph of y = sin x, but the frequency is doubled
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