course Mth 151] ەvxNǔassignment #007
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00:17:09 `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 --> Were 3 Now 6 confidence assessment: 3
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00:17:20 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 --> self critique assessment: 3
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00:18:42 `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 --> Not sure if I understand what they are asking for but I think it is 8 and 9 confidence assessment: 1
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00:19:04 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 --> ok I kind of get it but not entirely self critique assessment: 3
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00:19:54 `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 --> 16? confidence assessment: 1
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00:20:14 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 --> I was close. I'm not sure how I got 16 and they got 15 self critique assessment: 2
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00:20:43 `q004. Continue the process for one more step. How many points do you have in the new triangle?
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RESPONSE --> 21 confidence assessment: 2
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00:20:50 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 --> Got it! self critique assessment: 3
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00:21:02 `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 confidence assessment: 3
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00:21:08 The differences between these numbers are 3, 4, 5, 6. The next difference, according to this pattern, should be 7, which would make the next number 28.
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RESPONSE --> Got it self critique assessment: 3
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00:21:35 `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 --> Because the number keeps going up by one more each time the triangle gets bigger confidence assessment: 3
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00:21:44 When you extend the triangle again, you will add two new endpoints and each side will now have 7 points. The 7 points on the new triangle will be all of the new points.
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RESPONSE --> Yep self critique assessment: 3
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00:22:07 `q007. How do you know this sequence will continue in this manner?
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RESPONSE --> Triangles are geometrical figures that always have the same proportions confidence assessment: 2
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00:22:16 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 --> ok self critique assessment: 3
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00:26:46 `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 --> I'm honestly not sure at all. confidence assessment: 3
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00:27:05 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 --> Ah ok self critique assessment: 3
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00:27:36 `q002. See if you can use a similar strategy to add up the numbers 1 + 2 + ... + 2000.
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RESPONSE --> 2001 * 1000 = 2001000 confidence assessment: 3
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00:27:42 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 --> got it self critique assessment: 3
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00:28:35 `q003. See if you can devise a strategy to add up the numbers 1 + 2 + ... + 501.
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RESPONSE --> 501* 250 + 501=125751 confidence assessment: 2
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00:29:02 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 --> got it slightly different then that but got it confidence assessment: 3
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00:30:32 `q004. Use this strategy to add the numbers 1 + 2 + ... + 1533.
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RESPONSE --> 1533*766+1533=1175811 confidence assessment: 2
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00:30:54 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 --> Haha ok :-) I got the same thing just my way self critique assessment: 3
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00:31:34 `q005. Use a similar strategy to add the numbers 55 + 56 + ... + 945.
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RESPONSE --> 1000*445=445000 confidence assessment: 2
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00:31:50 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 --> Ah I did my math wrong self critique assessment: 3
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00:32:38 `q006. Devise a strategy to add the numbers 4 + 8 + 12 + 16 + ... + 900.
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RESPONSE --> 448.5*904=405444 confidence assessment: 2
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00:32:59 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 --> definitely did that math wrong but i'm not sure where... self critique assessment: 3
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00:33:32 `q007. What expression would stand for the sum 1 + 2 + 3 + ... + n, where n is some whole number?
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RESPONSE --> (1+2+3+n) I guess confidence assessment: 1
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00:33:42 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 --> huh? self critique assessment: 3
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