course MTH 151 I apologize for submitting late work but chatper 1 has really gotten to me. I found that these types of problems are most difficult for me.
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19:11:10 `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 --> Ok, the only strategy I came up with is: 1) adding 1 - 9 = 45 2) adding 10 - 19 =145 3) adding 20 - 29 =245 4) adding 30 - 39 = 345 and so on all the way through 90 - 99 = 945. Then at the end, add the 100 which gives you the answer of 5050.
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19:11:52 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 --> OK, I should have remembered the problems in the previous chapter for this one.
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19:13:05 `q002. See if you can use a similar strategy to add up the numbers 1 + 2 + ... + 2000.
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RESPONSE --> By solving this problem like the first one, you should get the answer of 2,001,000.
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19:13:11 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 --> OK
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19:15:03 `q003. See if you can devise a strategy to add up the numbers 1 + 2 + ... + 501.
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RESPONSE --> I think you would solve this problem the same way as the first two only you would pair up 500 to 1, 499 to 2, and so on, and then add one to the answer. This answer should be 125251.
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19:16:11 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 --> OK, I missed this one but I understand the answer you have given here.
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19:22:26 `q004. Use this strategy to add the numbers 1 + 2 + ... + 1533.
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RESPONSE --> I think the solution to this one would be 1175811.
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19:23:05 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 --> OK
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19:31:03 `q005. Use a similar strategy to add the numbers 55 + 56 + ... + 945.
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RESPONSE --> With this problem, there is an equal number of pairs so the solution would be 445,000.
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20:24:35 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 --> Oh ok. I see how you got this. I didn't take into consideration that he unit jumps are different from the number of units.
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20:27:28 `q006. Devise a strategy to add the numbers 4 + 8 + 12 + 16 + ... + 900.
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RESPONSE --> OK, just a guess... 904 * (225 + 1) = 204304
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20:47:57 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 --> This one really got me. I didn't divide 225 by 2. I'll go back and try to study more on problems like these. They are seeming quite a bit more diffucult for me to get than the other chapters. I understand the answers when they are given but I am not so sure that I can figure them out completely just yet. I'll work on this.
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21:01:47 `q007. What expression would stand for the sum 1 + 2 + 3 + ... + n, where n is some whole number?
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RESPONSE --> I understand what this question is asking for but I'm not sure how to to put it in expression form. I know that if n=4, the sum would be 10. But I'm not sure about the expression.
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21:03:44 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 --> OK, I see. I can work this problem by the answer given here but I'm not so sure that I could have came up with this formula on my own.
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n{P?q???|??W??s?? Student Name: assignment #011
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21:32:34 `q001. . There are five questions in this set. Draw four points in a square pattern (i.e., if the points were properly connected, they should form a square). From each of the points, draw a straight line to each of the other points. How many lines did you draw?
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RESPONSE --> Well, I drew 12 individual lines although there are only 6 lines because they overlap each other.
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21:32:59 Each corner of the square will connected to each of the other three corners, so from each corner you would have drawn three lines. Since there are four corners, had you followed the instructions precisely you would have drawn 4 * 3 = 12 lines. However each of these lines will be identical with another line you would have drawn, since for any two corners you would be drawing a line from the first to the second then another overlapping line from the second to the first. Therefore you might have said that there are 6 lines.
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RESPONSE --> OK
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21:33:32 `q002. How many triangles are there in the figure you drew?
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RESPONSE --> There are 8 triangles in the figure I drew.
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21:33:38 You should have a total of 8 triangles. The diagonals divide the square up into 4 small triangles. Each diagonal also divides the square into 2 larger triangles. Since there are 2 diagonals there are 4 larger triangles. The 4 small triangles and the 4 larger triangles total 8 triangles.
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RESPONSE --> OK
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21:36:57 `q003. The lines you drew should form a square with its two diagonals. Label the corners of the square A, B, C and D, going in order around the square, and label the center where the diagonals cross E. Now list all possible combinations of 3 of the letters A, B, C, D, E (note: combinations don't care about order, so A D E is the same as D A E or E A D or any other combination of these same three letters, so list each possible combination only once. That is, if you list for example ADE you won't list DAE).
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RESPONSE --> There would be 4 three-letter combinations: ACD ABC ABD BCD
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21:38:54 The easiest way to list these sequences is alphabetically: ABC, ABD, ABE all start with AB; then ACD and ACE start with AC and ADE starts with AD. This is a list of all possible combinations containing A. We next list all possible remaining combinations containing B: BCD, BCE and BDE. Then we write down CDE, the only remaining combination containing C. We thus have the 10 combinations ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE.
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RESPONSE --> Oh my goodness! I left out point E altogether. This was just a simple mistake.
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21:40:24 `q004. Of the 10 combinations ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE, which form triangles on your figure?
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RESPONSE --> ABD,ABE,ACD,ADE,BCD,BCE,CDE
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21:40:58 ABC forms a large triangle from A to B to C and back to A. The same is true of ABD, ACD and BCD. These are the large triangles in the figure. ACE and BDE form straight lines, not triangles. ABE, ADE, BCE and CDE form small triangles. Thus of the 10 possible combinations of labeled points, we find the 4 large triangles and the 4 small triangles we saw earlier, in addition to 2 straight lines which do not form triangles. Since any triangle in the figure must be labeled by three of the five points A, B, C, D, E, we see that these are the only triangles that can be formed.
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RESPONSE --> OK.
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21:41:59 `q005. Sketch the same figure as before, but without the line segment from A to B. Now how may triangles are there?
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RESPONSE --> There would be 5 triangles
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21:42:07 Any possible triangle must still come from the list ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE. We again see that ACE and BDE form straight lines so do not count as triangles. Now ABC, ABD and ABE do not form triangles because the line segment AB is now missing. This leaves us the five triangles ACD, ADE, BCE, BCE and CDE.
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RESPONSE --> ok
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