#$&* course MTH 152 4/17 6 020. ``q Query 20*********************************************
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Given Solution: `aSTUDENT SOLUTION: The ray MO, designated by the letters MO with a single arrow over the top, originates at the point M, passes through the point O and continues forever. The ray OM, designated by the letters OM with a single arrow over the top, originates at the point O, passes through the point M and continues forever. The two rays have in common the line segment OM, which would be designated by the letters OM with a 'bar' over the top. This would be the intersection of the two rays. The union of the two rays would form the line OM, which continues forever in both directions and is designated by the letters OM with a 'double arrow' over the top (a double arrow looks something like <-->). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query 9.1.54 lines SR and TP intersect at Q, where Q lies between S and R, and also between T and P. What are the names of the pairs of vertical angles for this figure? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Vertical angles are created when angles face opposite each other as a result of two crossing lines with the same vertex point. In this problem, two vertical angles are created as a result of the crossing of lines SQP and RQT. The vertical angles formed are SQT and PQR. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The point Q lies between S and R on the first line, and between T and P on the second. The angles formed by these two intersecting lines, running clockwise around the figure, are SQT, SQP, PQR and RQT. A pair of vertical angles consists two alternate angles from this list. The only possibilities are SQT and PQR SQP and RQT and these are the possible pairs of vertical angles. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query 9.1.60 Angles 5x - 129 deg and 2x - 21 deg are vertical angles. What is the value of x and how did you obtain it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Vertical pairs are always equal to each other, so the values of the vertical pairs in this problem can be set up to equal each other. If (5x-129) deg=(2x-21) deg, then we can solve the amount of the vertical pairs by solving the equation like any other normal equation. If we subtract 2x from each side of the equation then we are left with 3x-129=-21. We can then add 129 to -21, giving us an equation of 3x=108. When we divide each side by 3, we are left with the answer to the equation. X is equal to 36 deg. When we substitute 36 for x in each side of the equation, we find that each vertical angle is equal to 51 deg. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since the angles are vertical angles, they are equal to each other, therefore, I set them up to be equal to each other and then solved. To check myself, I then substituted my answer in for x on both sides of the equation to make sure they were equal. Starting with 5x - 129 deg = 2x - 21 deg subtract 2x from both sides to get 3x - 129 deg = -21 deg. Add 129 deg to both sides to get 3x = 108 deg. Divide both sides by 3 to get x = 36 deg. To check substitute 36 deg in for x in the equation and simplify, getting 51 deg = 51 deg. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query 9.1.72 The complement of an angle is 10 deg less that 1/5 of its supplement. What is the measure of the angle and how did you get it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In this particular problem, the complement of the angle is 10 degrees less than 1/5 of the angles supplement. If x is equal to the degree of the angle, then our supplement can be listed as 180 deg-x. In the same way, our complement can be listed as 90 deg-x. If the complement of the angle is 10 degrees less than 1/5 the supplement of the angle, then we have an overall equation of 1/5(180 deg-x)-10 deg=90 deg-x. We can solve this equation like any other normal equation. For our first step, we can multiply each side of the equation by 5. This leaves us with 450deg-5x=180 deg-x-50 deg. Next we can add x-450 to each side of the equation, giving us an equation of -4x=-320. When we divide 320 by 4, we find that x is equal to 80. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Let x be the degree measure of the angle. Then the supplement is 180 deg - x; 10 deg less than 1/5 the supplement is 1/5(180 deg - x) - 10 deg. The complement is 90 deg - x. So the equation is 90 deg - x = 1/5(180 deg - x) - 10 deg. Multiplying both sides by 5 we get 450 deg - 5 x = 180 deg - x - 50 deg so that 450 deg - 5 x = 130 deg - x. Adding x - 450 deg to both sides we get -4x = -320 deg so that x = 80 deg. Checking against the conditions of the problem: The complement of 80 deg is 10 deg. The supplement of 80 deg is 100 deg; 1/5 the supplement is 1/5 * 100 deg = 20 deg, so the complement 10 deg is 10 deg less than the supplement 20 deg. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!