#$&* course MTH 152 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aSTUDENT SOLUTION: The ray MO, designated by the letters MO with a single arrow over the top, originates at the point M, passes through the point O and continues forever. The ray OM, designated by the letters OM with a single arrow over the top, originates at the point O, passes through the point M and continues forever. The two rays have in common the line segment OM, which would be designated by the letters OM with a 'bar' over the top. This would be the intersection of the two rays. The union of the two rays would form the line OM, which continues forever in both directions and is designated by the letters OM with a 'double arrow' over the top (a double arrow looks something like <-->). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): - I noticed once I drew it out that this actually makes a shape, constituting a line that continues forever in both directions. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Query 9.1.54 lines SR and TP intersect at Q, where Q lies between S and R, and also between T and P. What are the names of the pairs of vertical angles for this figure? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - Assuming a generalization of the sum of both angles = 180, they are either acute (< 90^) or obtuse (>90^) whose combined sums will always =180 in Euclidean space confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): - Note that each end and point will designated each particular angle. - Am I wrong to generalize that TQR and RQT are essentially equivalent, and thus not counted due to being the same thing? ------------------------------------------------ Self-critique Rating: ok
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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): Okay. ------------------------------------------------ Self-critique Rating: ********************************************* 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: - I note that complements generally refer to 90 - the value of x, denoted as (90 - x) - And supplements generally refer to 180 - the value of x, denoted as (180 - x) - Since the complement is 10 degrees less than 1/5 of the supplement, one can write this as - (90 - x) = 10 - 1/5(180 - x) - Instead of distributing the fraction, I choose to multiply both sides of the equation (as its inverse, which is way easier to me) - 450 - 5x = 50 - 180 - x - 450 - 4x = 130 - -4x = -320 - 4x = 320 - X = 80 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Let x be the degree measure of the angle. Then the supplement is 180 deg - x; 10 deg less than 1/5 the supplement is 1/5(180 deg - x) - 10 deg. The complement is 90 deg - x. So the equation is 90 deg - x = 1/5(180 deg - x) - 10 deg. Multiplying both sides by 5 we get 450 deg - 5 x = 180 deg - x - 50 deg so that 450 deg - 5 x = 130 deg - x. Adding x - 450 deg to both sides we get -4x = -320 deg so that x = 80 deg. Checking against the conditions of the problem: The complement of 80 deg is 10 deg. The supplement of 80 deg is 100 deg; 1/5 the supplement is 1/5 * 100 deg = 20 deg, so the complement 10 deg is 10 deg less than the supplement 20 deg. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment. - I really miss algebra and geometry. Glad to see angles again! " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment. - I really miss algebra and geometry. Glad to see angles again! " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course MTH 152 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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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): - I noticed once I drew it out that this actually makes a shape, constituting a line that continues forever in both directions. ------------------------------------------------ Self-critique Rating: