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M111_Assignment01Both_fall_2007

Course: MATH 111, Fall 2009
School: UMBC
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College Okanagan Math 111 (71) Fall 2007 Assignment One Problems & Solutions Instructor: Clint Lee Due: Tuesday October 16 Instructions. Do all parts of all 10 questions. Show all signicant steps in your work. You will not receive any marks for just an answer, even if it is correct. Use complete English language sentences to answer questions where a written answer or explanation is required. The number of...

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College Okanagan Math 111 (71) Fall 2007 Assignment One Problems & Solutions Instructor: Clint Lee Due: Tuesday October 16 Instructions. Do all parts of all 10 questions. Show all signicant steps in your work. You will not receive any marks for just an answer, even if it is correct. Use complete English language sentences to answer questions where a written answer or explanation is required. The number of points for each question is given in the left margin, total 60. The questions in this assignment consist of all of the questions on Term Test One plus additional questions. The questions that appeared on Term Test One are labelled with a number in square brackets, giving the corresponding number of the question on the test. 1. [2] Consider the linear equation 3x + 4y = 12. (a) [1a] Put the linear equation above into standard form. Solution Solving for y gives 3 4y = 3x + 12 y = 4 x + 3 This is the standard form of the equation. [2] (b) [1b] Determine the x and y-intercepts of the line with the linear equation above. Solution Reading off from the standard form gives the y-intercept as (0, 3). Setting y = 0 in the original equation gives 3x = 12 x = 4 So the x-intercept is (4, 0). [2] (c) [1c] Determine the slope of the line with the linear equation above. Solution 3 The slope of the line is m = 4 . [2] (d) Graph the linear equation above. Put at least one value for x and y on each axis. Solution The graph is shown at the right. y 5 3x +4y=12 5 x Math 111 (71) Fall 2007 Assignment One Problems & Solutions Page 2 2. [2] A towing company charges a xed amount in addition to a per-kilometer charge to tow a car. The total charge y in dollars is related to the towing distance x in kilometers by the equation y = 30 + 2x. (a) [2a] Sketch the graph of this equation for 0 x 25. Solution The graph is shown at the right. y 100 80 60 40 20 5 10 15 20 25 x y=30+2x [1] (b) [2b] What is the charge of towing a car for 15 kilometers? Solution The charge for towing a car 15 kilometers is y = 30 + 2(15) = 60 dollars. [2] (c) [2c] If the company charged a customer $80 to tow his car, what was the corresponding towing distance? Solution Solve the equation 80 = 30 + 2x for x to give x = 25 kilometers. 3. [3] Consider the system of linear inequalities y 5x + 5x + 9y y x + x 0, y 0 Solution The boundary line 5x + y = 100, call it line I, has x and y-intercepts at (20, 0) and (0, 100). Note that the grid given does not include y = 100, so use another point on the line, say (12, 40). The boundary line 5x + 9y = 180, call it line II, has x and y-intercepts at (0, 20) and (36, 0). The boundary line x + y = 5, call it line III, has x and y-intercepts at (0, 5) and (5, 0). Writing each of the inequalities in standard form shows that the feasible set lies below lines I and II, but above line III. The two inequalities x 0 and y 0 tell us that the feasible set is only in the rst quadrant, to the right of the y-axis and above the xaxis. The feasible set is shown at the right. 100 180 5 (a) [3a] Graph the feasible set determined by the system. y 30 I D C A B Problem 3 E II III 30 x continued . . . Math 111 (71) Fall 2007 Assignment One Problems & Solutions Page 3 . . . Problem 3 [5] continued (b) [3b] Find all of the coordinates of the vertices of the feasible set. Solution All but vertex C is an x or y-intercept of one of the lines. So we can nd their coordinates without further work. They are A(5, 0), B(20, 0), D (0, 20), and E(0, 5). To nd the coordinates of the vertex C we solve the system of equations 5x 5x + y = 100 + 9y = 180 Subtracting the rst equation from the second gives 8y = 80 y = 10. Substituting y = 10 into either of the equations give 5x = 90 x = 18. So the coordinates of vertex C are (18, 10). 4. Suppose that the supply and demand equations for a new CD at Sam Goodies are q = 3p 12 and q = 2p + 23 respectively, where p is the unit price of the CD and q is the quantity. [3] (a) [4a] What are the supply and demand when the price is $10? Is $10 too high a price to charge for the CD? Why or why not? Solution When the price is $10, p = 10 and the supply equation gives q = 30 12 = 18 and the demand equation gives q = 20 + 23 = 3. This price is too high because demand is lower than supply. [3] (b) [4b] Find the equilibrium price and the corresponding number of units. Solution The equilibrium price is when the supply and demand equations give the same level of demand. This is the case when 3p 12 = 2p + 23 5p = 35 p = 7 The equilibrium price is $7 and at this price both the supply and demand equations give q = 3(7) 12 = 2(7) + 23 = 9 [3] (c) Find the p-intercepts of the supply and demand lines and give an economic interpretation of both intercepts. You do not need to draw the graphs. Solution The p-intercept of the supply line is given by 3p 12 = 0 p = 4. This is the price below which a supplier will not produce any of this product. The p-intercept of the demand line is given by 2p + 23 = 0 p = 11.5. This is the price above which nobody will buy this product. Math 111 (71) Fall 2007 Assignment One Problems & Solutions Page 4 [4] 5. Find the value of a for which the three points (1, 3), (6, 5), and ( a, 1) all lie on the same straight line. Solution The slope of the line joining (1, 3) and (6, 5) is m= 2 53 = 61 5 For the point ( a, 1) to be on the same line as the other two points, the slope of the line joining it to either of the two points must equal this slope. Thus, the of value a is given by 6 2 5 (1) = = 30 = 12 2a a = 9 6a 6a 5 [5] 6. [5] Solve the system of linear equations below using Gaussian elimination. If there is no solution, state so; if there are innitely many solutions, give the general solution and two particular solutions. x 3y 5z = 1 3x + 7y + 9z = 1 x + 4z = 5 Solution The augmented matrix is 1 3 1 1 3 5 7 9 1 0 4 5 Applying Gaussian elimination to this matrix gives 1 3 1 1 [2][2]+3[1] [2] 2 [2] 1 1 3 5 3 5 1 1 1 3 5 1 [3][3]+(1)[1] [3] 2 [3] 7 9 1 4 0 1 3 2 0 2 6 0 4 5 0 3 9 6 0 1 3 2 1 0 0 [3][3]+(1)[2] [1][1]+3[2] 0 1 0 4 3 0 5 2 0 The system has an innite number of solutions given by the system x y + 4z = 5 + 3z = 2 The leading variables are x and y and free variable is z. Expressing x and y in terms of z gives x = 5 4z and y = 2 3z. Two possible particular solutions can be obtained by substituting any two values of z into these. Choosing z = 0 gives x = 5, y = 2, z = 0. Choosing z = 2 gives x = 3, y = 4, z = 2. Math 111 (71) Fall 2007 Assignment One Problems & Solutions Page 5 [3] 7. (a) [7] A marketing company wishes to conduct a $15,000,000 advertising campaign in three media radio, television, and newspaper. The company wants to spend twice as much money in television advertising as in radio and newspaper advertising combined. Also, the company wants to spend a total of $13,000,000 in radio and television combined. Let x, y and z represent the amount in millions of dollars spent in radio, television, and newspaper, respectively. Set up a system of linear equations for x, y, and z. Solution There are three conditions on the variables x, y and z. The rst is that the total amount spent is to be $15,000,000. This gives the equation x + y + y = 15. The second is that the amount spent on television is twice as much as that spent on radio and newspaper combined. This gives y = 2( x + z) = 2x + 2z 2x y + 2z = 0. The third is that the amount spent on radio and television combined is $13,000,000. This gives x + y = 13. The system of equation for x, y, and z is z = 15 x + y + 2x y + 2z = 0 x + y = 13 [4] (b) Solve the system of equations above to determine how much the company should spend in each medium. Solution The augmented matrix is 1 2 1 1 1 1 1 2 0 15 0 13 [2] 1 [2] 3 15 [3](1)[3] 30 2 Applying Gaussian elimination to this matrix gives 1 2 1 1 1 1 2 1 0 [2][2]+(2)[1] 15 1 [3][3]+(1)[1] 0 0 13 0 1 [1][1]+(1)[2] 0 0 1 3 0 0 1 0 0 0 1 1 0 1 3 [1][1]+(1)[3] 10 2 1 0 0 1 1 0 1 0 1 15 10 2 The system has a unique solution with x = 3, y = 10, and z = 2. Math 111 (71) Fall 2007 Assignment One Problems & Solutions Page 6 [4] 8. [7] In a certain provincial legislature the percentage of legislators voting for or against lowering the legal drinking age by various party afliations is summarized by this matrix. For 0.6 A = 0.3 1.0 Against 0.4 Liberal 0.7 New Democrat 0.0 Independent There are 60 Liberals, 30 New Democrats, and 10 Independents in the legislature. Use a matrix calculation to determine how many legislators voted for lowering the drinking age and how many voted against lowering the drinking age. Solution Represent the membership of the legislature using a 1 3 matrix M= Lib 60 New Dem 30 Ind 10 The...

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time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] 0.000 199.955 107.913 -73.390 42.589 -70.954 42.482 2.296 -72.172 42.536 67.669 -60.000 191.190 103.183 -73.371 42.3
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] 0.000 198.858 107.321 -73.317 42.832 -71.027 42.240 11.517 -72.172 42.536 67.669 -60.000 189.528 102.286 -73.427 42.577 -71.357 41
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] 0.000 199.871 107.868 -73.385 42.627 -70.959 42.444 3.883 -72.172 42.536 67.669 -60.000 192.080 103.663 -73.508 42.1
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] 0.000 199.374 107.600 -73.355 42.744 -70.989 42.327 8.531 -72.172 42.536 67.669 -60.000 197.466 106.570 -73.442 42.5
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] 0.000 199.813 107.837 -73.382 42.646 -70.962 42.425 4.668 -72.172 42.536 67.669 -60.000 195.121 105.304 -73.490 42.2
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] 0.000 197.858 106.781 -71.373 41.869 -72.971 43.203 15.756 -72.172 42.536 67.669 -60.000 243.575 131.454 -72.951 43.4
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] 0.000 198.526 107.142 -72.779 41.763 -71.565 43.308 13.081 -72.172 42.536 67.669 -60.000 196.105 105.835 -73.046 41.9
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] 0.000 197.911 106.810 -73.197 43.006 -71.147 42.065 15.560 -72.172 42.536 67.669 -60.000 191.635 103.423 -73.197 42.9
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] 0.000 197.884 106.795 -72.961 41.862 -71.383 43.209 15.661 -72.172 42.536 67.669 -60.000 194.533 104.987 -72.963 41.9
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] 0.000 198.289 107.014 -73.260 42.145 -71.084 42.927 14.090 -72.172 42.536 67.669 -60.000 197.839 106.771 -73.220 42.1
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] 0.000 197.824 106.763 -72.985 43.193 -71.359 41.878 15.879 -72.172 42.536 67.669 -60.000 202.041 109.039 -72.995 43.2
Harvard - MM - 545
<?xml version="1.0" encoding="UTF-8"?><Error><Code>NoSuchKey</Code><Message>The specified key does not exist.</Message><Key>1b4ab5fee493cead478cc9dca0883b865af0f908.txt</Key><RequestId>CD4F4F8B1291BE80</RequestId><HostId>Jv7vEylFxDCBNjuVHRUVfeIcjGhq
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] 0.000 198.060 106.891 -71.441 41.827 -72.903 43.245 14.996 -72.172 42.536 67.669 -60.000 197.090 106.367 -72.865 43.2
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] 0.000 199.736 107.795 -71.931 41.655 -72.413 43.416 5.545 -72.172 42.536 67.669 -60.000 217.002 117.113 -72.242 41.6
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] 0.000 198.045 106.882 -72.908 41.829 -71.436 43.242 15.056 -72.172 42.536 67.669 -60.000 198.172 106.951 -72.601 41.7
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] 0.000 199.868 107.866 -73.385 42.628 -70.959 42.444 3.916 -72.172 42.536 67.669 -60.000 215.709 116.415 -73.519 42.3
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 197.691 106.691 -73.088 43.115 -71.256 41.957 16.353 -72.172 42.536 67.669 4V8 1B9
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 198.058 106.889 -73.226 42.971 -71.118 42.1 15.005 -72.172 42.536 67.669 DDH OWD
Harvard - MM - 545
<?xml version="1.0" encoding="UTF-8"?><Error><Code>NoSuchKey</Code><Message>The specified key does not exist.</Message><Key>2bf3fc852dccab2ba56b827d8f3430b815398e5a.txt</Key><RequestId>46B9FDA8F435AF8B</RequestId><HostId>dWLK/+sJIDmZidhqz2bm/MEMBxbZ
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 199.374 107.6 -73.355 42.744 -70.989 42.327 8.534 -72.172 42.536 67.669 DDH BOS
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 198.156 106.942 -73.242 42.12 -71.102 42.951 14.623 -72.172 42.536 67.669 GBR LWM
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 198.596 107.18 -73.294 42.874 -71.05 42.197 12.765 -72.172 42.536 67.669 DDH OWD
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 199.006 107.401 -73.329 42.808 -71.015 42.264 10.748 -72.172 42.536 67.669 DDH BOS
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 198.54 107.15 -73.288 42.883 -71.056 42.188 13.016 -72.172 42.536 67.669 DDH OWD ORE GDM
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 198.12 106.923 -73.236 42.959 -71.108 42.113 14.766 -72.172 42.536 67.669 DDH OWD
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 198.04 106.879 -73.223 42.975 -71.121 42.096 15.076 -72.172 42.536 67.669 DDH OWD ORE GDM
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 200 107.937 -73.392 42.539 -70.952 42.532 0.148 -72.172 42.536 67.669 PSF BVY
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 198.618 107.192 -73.296 42.201 -71.048 42.87 12.667 -72.172 42.536 67.669 GBR LWM
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 199.073 107.437 -73.334 42.275 -71.01 42.797 10.381 -72.172 42.536 67.669 GBR LWM
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 199.985 107.929 -73.392 42.504 -70.952 42.567 1.338 -72.172 42.536 67.669 PSF BVY ORE ORE
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 198.237 106.986 -73.253 42.135 -71.091 42.936 14.3 -72.172 42.536 67.669 GBR LWM ORE GDM
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 198.865 107.325 -73.318 42.831 -71.026 42.241 11.485 -72.172 42.536 67.669 DDH OWD
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 197.865 106.785 -73.186 43.019 -71.158 42.053 15.728 -72.172 42.536 67.669 DDH 1B9
Harvard - MM - 545
<?xml version="1.0" encoding="UTF-8"?><Error><Code>NoSuchKey</Code><Message>The specified key does not exist.</Message><Key>caca94b9412c801a4ac162c0224e9ec0565acacc.txt</Key><RequestId>776FEE47E4AB9B75</RequestId><HostId>gKu9m4x0MqFA5IevLuAbE5qo1zDN
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 199.789 107.823 -73.381 42.418 -70.963 42.653 4.96 -72.172 42.536 67.669 PSF BVY ORE ORE
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 199.029 107.413 -73.331 42.804 -71.013 42.268 10.624 -72.172 42.536 67.669 DDH BOS
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 199.883 107.874 -73.386 42.622 -70.958 42.449 3.688 -72.172 42.536 67.669 PSF BOS ORE ORE
Harvard - MM - 545
<?xml version="1.0" encoding="UTF-8"?><Error><Code>NoSuchKey</Code><Message>The specified key does not exist.</Message><Key>943b6b2a28f57a16e8cf45f11a2b415258498429.txt</Key><RequestId>A88BD4E3894E1E6C</RequestId><HostId>ttK43XX543rDwqGfKWk5I8keBmxZ
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 199.641 107.744 -71.891 41.662 -72.453 43.409 6.462 -72.172 42.536 67.669 5B3 CNH
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 199.964 107.918 -72.084 41.639 -72.26 43.433 2.054 -72.172 42.536 67.669 IJD CNH
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 199.839 107.851 -72.359 41.648 -71.985 43.424 4.332 -72.172 42.536 67.669 IJD CNH ORE GDM
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 199.096 107.45 -71.714 41.706 -72.63 43.365 10.251 -72.172 42.536 67.669 5B3 VSF ORE GDM
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 199.89 107.878 -72.326 41.644 -72.018 43.427 3.572 -72.172 42.536 67.669 IJD CNH ORE GDM
Harvard - MM - 545
time[min] xsec[km] xsec[nmi] x1.1st y1.1st x2.1st y2.1st xsec2[nmi] ctr.x ctr.y dpease[nmi] arpt1.1st arpt2.1st arpt1.2nd arpt2.2nd0 197.889 106.798 -73.192 42.06 -71.152 43.012 15.641 -72.172 42.536 67.669 GBR MHT GDM ORE