course phy 201
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˂Qǽ Student Name: assignment #007
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14:27:19 `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 --> 3 on the original triangle 6 on the new triangle
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14:27:40 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 --> ok
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14:37:57 `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 --> 4 points on the new triangle. 10 points on the whole figure.
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14:38:01 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
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14:40:14 `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 --> 5 points in the new triangle, 15 for the whole triangle.
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14:40:23 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 --> ok
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14:42:24 `q004. Continue the process for one more step. How many points do you have in the new triangle?
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RESPONSE --> 6 points for the new triangle, 21 for the whole triangle.
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14:42:32 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 --> ok
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14:43:40 `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
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14:43:45 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 --> ok
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14:45:14 `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 difference of the numbers goes in the pattern of 3,4,5,6,.. so adding the difference of 7 to the last number 21= 28.
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14:45:25 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 --> ok
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14:47:14 `q007. How do you know this sequence will continue in this manner?
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RESPONSE --> By drawing out the trangle, the next number of points to make the triangle, would be 8,then 9, than 10....
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14:47:21 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
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aPBƚJ Student Name: assignment #008
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19:23:29 `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
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19:29:29 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 started working it out like this, but then I thought it was wrong.
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19:31:50 `q002. See if you can use a similar strategy to add up the numbers 1 + 2 + ... + 2000.
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RESPONSE --> There are 1000 pairs adding to 2001. 1000 * 2001= 2001000
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19:31:57 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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19:34:44 `q003. See if you can devise a strategy to add up the numbers 1 + 2 + ... + 501.
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RESPONSE --> There are 250.5 pairs adding to 502. 250.5 * 502=125, 751.
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19:36:50 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, but when I muliply, the 250.5 * 502= 125,751? Am I doing something wrong?
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19:40:12 `q004. Use this strategy to add the numbers 1 + 2 + ... + 1533.
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RESPONSE --> There are 766.5 pairs totaling 1534. 766.5 * 1534= 1,175,811.
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19:40:23 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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19:42:58 `q005. Use a similar strategy to add the numbers 55 + 56 + ... + 945.
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RESPONSE --> there are 472.5 pairs totaling 1000. 472.5 * 1000= 472,500
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19:45:25 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 --> ok. = 445,500.
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19:52:42 `q006. Devise a strategy to add the numbers 4 + 8 + 12 + 16 + ... + 900.
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RESPONSE --> 900* 4+1=846 #'s in the sum. 900/4 = 225 pairs. 904 * 225= 203,400.?????
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19:55:50 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 --> ok= 101,700
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19:57:47 `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+...+5,6,7...
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19:59:00 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
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ؗIfHЅ{ assignment #007 űpZ Liberal Arts Mathematics I 02-12-2006
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14:06:02 Query 1.2.6 seq 2, 51, 220, 575, 1230, 2317 ... by successive differences
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RESPONSE --> 2, 57, 220, 575, 1230, 2317,...,3992 55, 163, 355, 655, 1087, 1675 108, 192, 300, 432, 588 84, 108, 132, 156 24, 24, 24
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14:06:15 ** If the sequence is 2, 57, 220, 575, 1230, 2317, ... then we have: 2, 57, 220, 575, 1230, 2317, # 3992 55, 163, 355, 655, 1087, # 1675 108, 192, 300, 432, # 588 84, 108, 132, # 156 24, 24, The final results, after the # signs, are obtained by adding the number in the row just below, in the following order: Line (4) becomes 132+24=156 Line (3) becomes 432+156=588 Line (2) becomes 1087+588=1675 Line (1) becomes 2317+1675=3992 The next term is 3992. **
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RESPONSE --> ok
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14:08:43 1.2.18 1^2 + 1 = 2^2 - 2; 2^2 + 2 = 3^2 - 3; 3^2 + 3 = etc.
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RESPONSE --> = 4^2 + 4 = 5^2 - 5 = 16 + 4 = 25 - 5 = 20 = 20 yes, true statement.
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14:08:49 ** The next equation in the sequence would be 4^2 + 4 = 5^2 - 5 The verification is as follows: 4^2 + 4 = 5^2 - 5 simplifies to give you 16 + 4 = 25 - 5 or 20 = 20 **
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RESPONSE --> ok
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14:12:35 1.2.30 state in words (1 + 2 + ... + n ) ^ 2 = 1^3 + 2^3 + ... + n^3
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RESPONSE --> The couning numbers to the second(^2), are = to the numbers to the Third (^3) power.
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14:13:35 ** the equation says that the square of the sum of the first n counting numbers is equal to the sum of their cubes **
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RESPONSE --> ok
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14:18:56 1.2.36 1 st triangular # div by 3, remainder; then 2d etc. Pattern.
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RESPONSE --> 1/3= .3333... 3/3= 1 6/3=2 10/3= 3.3 15/3= 5 21/3= 7 28/3= 9.3333... 36/3=12 the pattern is .33....., then two whole numbers, then whole numbers with repeating .333...., then two whole numbers, then two whole numbers with repeating .3333....
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14:19:43 ** The triangular numbers are 1, 3, 6, 10, 15, 21, . . . . We divide these by 3 and get the sequence of remainders. When you divide 1 by 3 you get 0 with remainder 1 (3 goes into 1 zero times with 1 left over). 3 divided by 3 gives you 1 with remainder 0. 6 divded by 3 is 2 with remainder 0. 10 divided by 3 is 3 with remainder 1. Therefore the remainders are 1,0,0,1,0,0. It turns out that the sequence continues as a string of 1,0,0 's. At this point that is an inductive pattern, but remmeber that the sequence of triangular numbers continues by adding successively larger and larger numbers to the members of the sequence. Since the number added always increases by 1, and since every third number added is a multiple of 3, is isn't too difficult to see how the sequence of remainders comes about and to see why it continues as it does. COMMON ERROR: .3333333,1,2,3.3333333,etc. INSTRUCTOR CORRECTION: You need the remainders, not the decimal equivalents. When you divide 1 by 3 you get 0 with remainder 1 (3 goes into 1 zero times with 1 left over). 3 divided by 3 gives you 1 with remainder 0. 6 divded by 3 is 2 with remainder 0. 10 divided by 3 is 3 with remainder 1. Therefore the remainders are 1,0,0,1,0,0 and the sequence continues as a string of 1,0,0 's. COMMON ERROR: 1/3, 1, 2, 3 1/3 CORRECTION: These are the quotients. You need the remainders. If you get 1/3 that means the remainder is 1; same if you get 3 1/3. If you just getting whole number (like 1 or 2 in your calculations) the remainder is 0. In other words, when you divide 1 by 3 you get 0 with remainder 1 (3 goes into 1 zero times with 1 left over). 3 divided by 3 gives you 1 with remainder 0. 6 divded by 3 is 2 with remainder 0. 10 divided by 3 is 3 with remainder 1. The remainders form a sequence 1,0,0,1,0,0 and the sequence continues as a string of 1,0,0 's. **
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RESPONSE --> ok
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14:22:06 1.2.48 use formula to find the 12 th octagonal number. Explain in detail how you used the formula to find this number.
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RESPONSE --> O 12= 12(6* 12 - 4)/ 2 = 12*68/2 = 816/2= 408
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14:22:14 ** The pattern to formulas for triangular, square, pentagonal, hexagonal, heptagonal and octagonal numbers is as follows: Triangular numbers: n / 2 * [ n + 1 ] note that this is the same as Gauss' formula Square numbers: n / 2 * [ 2n + 0 ] or just n^2 Pentagonal #'s: n / 2 * [ 3n - 1 ] Hexagonal #'s: n / 2 * [ 4n - 2 ] Heptagonal #'s: n / 2 * [ 5n - 3 ] Octagonal #'s: n / 2 * [ 6n - 4 ] The coefficient of n in the bracketed term starts with 1 and increases by 1 each time, and the +1 in the first bracketed term decreases by 1 each time. You will need to know these formulas for the test. The last formula is for octagonal numbers. To get n = 12 octangonal number use n/2 * [ 6n - 4 ] to get 12 / 2 * [ 6 * 12 - 4 ] = 6 * 68 = 408. **
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RESPONSE --> ok
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қGˤ\àz^wH assignment #008 űpZ Liberal Arts Mathematics I 02-12-2006
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18:06:19 1.3.6 9 and 11 yr old hosses; sum of ages 122. How many 9- and 11-year-old horses are there?
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RESPONSE --> There are 5, 9-year-old horses and 7, 11-year-old horses, for a total of 45 + 77= 122.
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18:10:01 ** If there was one 11-year-old horse the sum of the remaining ages would have to be 122 - 11 = 111, which isn't divisible by 9. If there were two 11-year-old horses the sum of the remaining ages would have to be 122 - 2 * 11 = 100, which isn't divisible by 9. If there were three 11-year-old horses the sum of the remaining ages would have to be 122 - 3 * 11 = 89, which isn't divisible by 9. If there were four 11-year-old horses the sum of the remaining ages would have to be 122 - 4 * 11 = 78, which isn't divisible by 9. If there were five 11-year-old horses the sum of the remaining ages would have to be 122 - 5 * 11 = 67, which isn't divisible by 9. The pattern is 122 - 11 = 111, not divisible by 9 122 - 2 * 11 = 100, not divisible by 9 122 - 3 * 11 = 89, not divisible by 9 122 - 4 * 11 = 78, not divisible by 9 122 - 5 * 11 = 67, not divisible by 9 122 - 6 * 11 = 56, not divisible by 9 122 - 7 * 11 = 45, which is finally divisible by 9. Since 45 / 9 = 5, we have 5 horses age 9 and 7 horses age 11. **
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RESPONSE --> ok, I got the right answer, but I made to list of each year-old horse, from one horse to 12 horses. Then I used trial and error to add the two and find the two that together had the sum of 122.
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18:18:16 Query 1.3.10 divide clock into segments each with same total
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RESPONSE --> I'm not clear on what the question if asking for, but I'm going to answer it like the question in my book where it asks you to divide it into three equal segments with two straight lines. The lines to divide the clock into two equal segments, with two straight lines: one line going from the 12-4 and another line going from the 8 to the first line. The three regions are: 12-4, 4-8, 8-12.
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18:21:06 ** The total of all numbers on the clock is 78. So the numbers in the three sections have to each add up to 1/3 * 78 = 26. This works if we can divide the clock into sections including 11, 12, 1, 2; 3, 4, 9, 10; 5, 6, 7, 8. The numbers in each section add up to 26. To divide the clock into such sections the lines would be horizontal, the first from just beneath 11 to just beneath 2 and the second from just above 5 to just above 8. Horizontal lines are the trick. You might have to draw this to see how it works. **
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RESPONSE --> ok, I understand know.
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18:49:05 Query 1.3.18 M-F 32 acorns each day, half of all acorns eaten, 35 acorns left after Friday
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RESPONSE --> on monday, they started with 128 acorns. Mon. had 128+ 32=160, ate 1/2= 80. Tues. had 80+ 32= 112, ate 1/2= 56. Wed. had 56+ 32= 88, ate 1/2= 44. Thur. had 44+ 32= 76, ate 1/2= 38. Fri. had 38 + 32= 70, ate 1/2 =35. Then on Sat morning they had 35 acorns left! I started from the end and worked my way through the problem, then went back through it to see if it made since.
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18:49:33 ** You have to work this one backwards. If they were left with 35 on Friday they had 70 at the beginning (after bringing in the 32) on Friday, so they had 70 - 32 = 38 at the end on Thursday. So after bringing in the 32 they had 2 * 38 = 76 at the beginning of Thursday, which means they had 76 - 32 = 44 before the 32 were added. So they had 44 Wednesday night ... etc. **
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RESPONSE --> ok
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18:59:32 Query 1.3.30 Frog in well, 4 ft jump, 3 ft back.
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RESPONSE --> On the 16th day the frog will reach the top.
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19:00:52 ** COMMON ERROR: 20 days CORRECTION: The frog reaches the 20-foot mark before 20 days. On the first day the frog jumps to 4 ft then slides back to 1 ft. On the second day the frog therefore jumps to 5 ft before sliding back to 2 ft. On the third day the frog jumps to 6 ft, on the fourth to 7 ft., etc. Continuing the pattern, on the 17th day jumps to 20 feet and hops away. The maximum height is always 3 feet more than the number of the day, and when max height is the top of the well the frog will go on its way. **
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RESPONSE --> I got this concept, but when it reached the top, I didn't drop it down,for night and then make it to the seventh day.
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19:08:03 Query 1.3.48 How many ways to pay 15 cents?
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RESPONSE --> 6 ways. 1 dime, 5 pennies. 15 pennies. 1 dime, 1 nickel. 3 nickels. 2 nickels, 5 pennies. 1 nickel, 10 pennies.
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19:09:04 ** To illustrate one possible reasoning process, you can reason this out in such a way as to be completely sure as follows: The number of pennies must be 0, 5, 10 or 15. If you don't use any pennies you have to use a dime and a nickle. If you use exactly 5 pennies then the other 10 cents comes from either a dime or two nickles. If you use exactly 10 pennies you have to use a nickle. Or you can use 15 pennies. Listing these ways: 1 dime, 1 nickel 1 dime, 5 pennies 2 nickels, 5 pennies 3 nickels 15 pennies 1 nickel 10 pennies **
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RESPONSE --> ok
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19:17:14 Query 1.3.52 Given 8 coins, how do you find the unbalanced one in 3 weighings
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RESPONSE --> Take the 8 coins and divide them into two groups of 4. For the first weighing,put one pair of 4 againist the other pair of 4. For the 2nd weighing, take the lighter pair of 4, and divide them into two pairs of two. Weigh them. For the 3rd weighing, take the lighter pair of two and divide them into one againist one. Weigh them and the lighter one is the bad coin.
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19:17:30 ** Divide the coins into two piles of 4. One pile will tip the balance. Divide that pile into piles of 2. One pile will tip the balance. Weigh the 2 remaining coins. You'll be able to see which coin is heavier. **
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RESPONSE --> ok
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