course Mth151 ͮJy}Ҫ
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13:27:00 `q001. Note that there are 7 questions in this assignment. Sketch three points A, B and C forming an equilateral triangle on a piece of paper, with point A at the lower left-hand corner, point B at the lower right-hand corner and point C at the top. Sketch the segments AB and AC. Now double the lengths of AB and AC, and place a point at each of the endpoints of these segments. Connect these new endpoints to form a new equilateral triangle. Two sides of this triangle will have three points marked while the new side will only have its two endpoints marked. Fix that by marking that middle point, so all three sides of your new triangle are marked the same. How many marked points were there in the original triangle, and how many are there in the new triangle?
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RESPONSE --> There are three marked points in the original triangle (A, B, C) and five marked points in the new triangle. confidence assessment: 0
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13:27:55 The original triangle had the three points A, B and C. When you extended the two sides you marked the new endpoints, then you marked the point in the middle of the third side. So you've got 6 points marked. Click on 'Next Picture' to see the construction. The original points A, B and C are shown in red. The line segments from A to B and from A to C have been extended in green and points marked at the ends of these segments. The new endpoints have been connected to form the third side of a larger triangle, and an equally spaced point has been constructed at the midpoint of that side. Your figure should contain the three original points, plus the three points added when the new side was completed.
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RESPONSE --> I did not count the point in the center between the double length of AC and AB (the new triangle) self critique assessment: 2
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13:31:02 `q002. Extend the two sides that meet at A by distances equal to the distance original lengths AC and AB and mark the endpoints of the newly extended segments. Each of the newly extended sides will have 4 marked points. Now connect the new endpoints to form a new right triangle. Mark points along the new side at the same intervals that occur on the other two sides. How many marked points are on your new triangle, and how many in the whole figure?
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RESPONSE --> 5 points on the new triangle and 11 in the whole figure. confidence assessment: 0
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13:31:39 You added the two new endpoints when you extended the sides. You then should have marked two new points on the new third side, so that each side contains 4 points including its endpoints. Your figure will now contain 10 marked points.
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RESPONSE --> I added an extra unnecessary point to get 5/11. self critique assessment: 2
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13:32:26 `q003. Continue the process for another step-extend each side by a distance equal to the original point-to-point distance. How many points do you have in the new triangle?
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RESPONSE --> 14 points in the new triangle. confidence assessment: 0
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13:32:50 You will add an endpoint to each newly extended side, so each of the new sides will contain 5 points. You will then have to add 3 equally spaced points to the new side, giving you a total of 13 points on the new triangle. In addition there are two marked points inside the triangle, for a total of 15 points. Click on 'Next Picture' to see the construction. The line segments along two sides of the triangle have again been extended and points marked at the ends of these segments. The new endpoints have been connected to form the third side of a larger triangle, and equally spaced points have been constructed along that side.
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RESPONSE --> Here I've forgotten a point. self critique assessment: 0
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13:33:25 `q004. Continue the process for one more step. How many points do you have in the new triangle?
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RESPONSE --> 21 points in the new triangle. confidence assessment: 1
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13:34:39 You will add an endpoint to each newly extended side, so each of the new sides will contain 6 points. You will then have to add 4 equally spaced points to the new side, giving you a total of 15 points on the new triangle. There are also 5 marked points inside the triangle for a total of 21 marked points.
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RESPONSE --> The triangle grows larger diagonally by a sequence of 1, 2, 3, 4, 5, so the next step is logically to add 6 points to the previous total of 15 to arrive at 21 points. self critique assessment: 2
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13:35:52 `q005. The sequence of marked points is 3, 6, 10, 15, 21. What do expect will be the next number in this sequence?
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RESPONSE --> 28 should be the next number in the sequence given that the pattern seems to be going (3, 4, 5, 6, ...) so 7 should be added to 21 to obtain 28. confidence assessment: 2
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13:37:39 `q006. How can you tell, in terms of the process you used to construct these triangles, that the next number should be 7 greater?
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RESPONSE --> The same way that the triangle appears to grow diagonally in a pattern of 1, 2, 3, 4, 5 for the marked points, the sequence of numbers also grows by 3, 4, 5, 6. confidence assessment: 1
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13:38:52 `q007. How do you know this sequence will continue in this manner?
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RESPONSE --> The sequence will continue in this manner because the triangle will cease to be an equal triangle if it does not. confidence assessment: 1
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13:39:38 Each time you extend the triangle, each side increases by 1. All the new marked points are on the new side, so the total number of marked points will increase by 1 more than with the previous extension.
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RESPONSE --> By extending the triangle each side increases by 1, therefore keeping the equality of the newly formed 'big' triangle. self critique assessment: 2
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13:49:56 `q001. There are seven questions in this set. See if you can figure out a strategy for quickly adding the numbers 1 + 2 + 3 + ... + 100, and give your result if you are successful. Don't spend more than a few minutes on your attempt.
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RESPONSE --> 100,000. If you add 99+1 and 98+2 etc. it should equal out to 100,000 by giving a one to one correspondence. confidence assessment: 0
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13:50:59 These numbers can be paired as follows: 1 with 100, 2 with 99, 3 with 98, etc.. There are 100 number so there are clearly 50 pairs. Each pair adds up to the same thing, 101. So there are 50 pairs each adding up to 101. The resulting sum is therefore total = 50 * 101 = 5050.
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RESPONSE --> I did the 1-1 correspondence incorrectly. self critique assessment: 2
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13:53:02 `q002. See if you can use a similar strategy to add up the numbers 1 + 2 + ... + 2000.
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RESPONSE --> 101000. 100 goes into 2000 20 times take the product of 5050 and multiply it by 20. confidence assessment: 0
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13:53:35 Pairing 1 with 2000, 2 with 1999, 3 with 1998, etc., and noting that there are 2000 numbers we see that there are 1000 pairs each adding up to 2001. So the sum is 1000 * 2001 = 2,001,000.
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RESPONSE --> I understand this now. self critique assessment: 1
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13:56:10 We can pair 1 with 501, 2 with 500, 3 with 499, etc., and each pair will have up to 502. However there are 501 numbers, so not all of the numbers can be paired. The number in the 'middle' will be left out. However it is easy enough to figure out what that number is, since it has to be halfway between 1 and 501. The number must be the average of 1 and 501, or (1 + 501) / 2 = 502 / 2 = 266. Since the other 500 numbers are all paired, we have 250 pairs each adding up to 502, plus 266 left over in the middle. The sum is 250 * 502 + 266 = 125,500 + 266 = 125,751. Note that the 266 is half of 502, so it's half of a pair, and that we could therefore say that we effectively have 250 pairs and 1/2 pair, or 250.5 pairs. 250.5 is half of 501, so we can still calculate the number of pairs by dividing the total number of number, 501, by 2. The total sum is then found by multiplying this number of pairs by the sum 502 of each pair: 250.5 * 502 = 125,766.
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RESPONSE --> 250 and a half pairs times the number 502 equals out to be 125,766 confidence assessment: 2
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13:56:37 `q004. Use this strategy to add the numbers 1 + 2 + ... + 1533.
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RESPONSE --> confidence assessment:
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13:58:00 Pairing the numbers, 1 with 1533, 2 with 1532, etc., we get pairs which each adult to 1534. There are 1533 numbers so there are 1533 / 2 = 766.5 pairs. We thus have a total of 1534 * 766.5, whatever that multiplies out to (you've got a calculator, and I've only got my unreliable head).
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RESPONSE --> 1533 numbers and divide them into pairs. This results into 766.5 pairs. 1,175044.5 self critique assessment: 1
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14:00:10 `q005. Use a similar strategy to add the numbers 55 + 56 + ... + 945.
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RESPONSE --> Take the difference of 945 and 55 which is 890 and divide that by 2 two make pairs which is 445 and find the product of that and 945 which is 420545. confidence assessment: 1
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14:00:39 We can pair up 55 and 945, 56 and 944, etc., obtaining 1000 for each pair. There are 945 - 55 + 1 = 891 numbers in the sum (we have to add 1 because 945 - 55 = 890 tells us how many 1-unit 'jumps' there are between 55 and 945--from 55 to 56, from 56 to 57, etc.. The first 'jump' ends up at 56 and the last 'jump' ends up at 945, so every number except 55 is the end of one of the 890 'jumps'. But 55 is included in the numbers to be summed, so we have 890 + 1 = 891 numbers in the sum). If we have 891 numbers in the sum, we have 891/2 = 445.5 pairs, each adding up to 1000. So we have a total of 445.5 * 1000 = 445,500.
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RESPONSE --> self critique assessment: 0
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14:01:27 Pairing 4 with 900, 8 with 896, etc., we get pairs adding up to 904. The difference between 4 and 900 is 896. The numbers 'jump' by 4, so there are 896 / 4 = 224 'jumps'. None of these 'jumps' ends at the first number so there are 224 + 1 = 225 numbers. Thus we have 225 / 2 = 112.5 pairs each adding up to 904, and our total is 112.5 * 904.
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RESPONSE --> 112.5 pairs times 904. self critique assessment: 0
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14:03:04 We can pair 1 and n, 2 and n-1, 3 and n-2, etc., in each case obtaining a sum of n + 1. There are n numbers so there are n/2 pairs, each totaling n + 1. Thus the total is n/2 * (n+1).
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RESPONSE --> n/2 * (n+1) if we have a one to one correspondence of 1 and n, 2 and n-1, 3 and n-2 self critique assessment: 0
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