course MTH 151 әyn֜assignment #007
......!!!!!!!!...................................
15:29:24 Query 1.2.6 seq 2, 51, 220, 575, 1230, 2317 ... by successive differences
......!!!!!!!!...................................
RESPONSE --> The book has 57 instead of 51 so that is what I used. 57-2=55 220-57=163 575-220=355 1230-575=655 2317-1230=1087 163-55=108 355-163=192 655-355=300 1087-655=432 192-108=84 300-192=108 432-300=132 108-84=24 321-108=24 then work backwards 132+24=156 432+156=588 1087+588=1675 2317+1675=3992 so the next number would be 3992 confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:29:39 ** 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. **
......!!!!!!!!...................................
RESPONSE --> correct self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:35:26 1.2.18 1^2 + 1 = 2^2 - 2; 2^2 + 2 = 3^2 - 3; 3^2 + 3 = etc.
......!!!!!!!!...................................
RESPONSE --> the next number in the sequence would be 4^2+4=5^2-5 because it is just increasing each number by 1. confidence assessment: 1
.................................................
......!!!!!!!!...................................
15:36:14 ** 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 **
......!!!!!!!!...................................
RESPONSE --> correct self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:39:05 1.2.30 state in words (1 + 2 + ... + n ) ^ 2 = 1^3 + 2^3 + ... + n^3
......!!!!!!!!...................................
RESPONSE --> the sum of the numbers in the first set squared will equal the sum of the second set cubed. confidence assessment: 1
.................................................
......!!!!!!!!...................................
15:39:36 ** the equation says that the square of the sum of the first n counting numbers is equal to the sum of their cubes **
......!!!!!!!!...................................
RESPONSE --> correct self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:51:27 1.2.36 1 st triangular # div by 3, remainder; then 2d etc. Pattern.
......!!!!!!!!...................................
RESPONSE --> 1/3=0 with 1 remainder 3/3=1 with 0 remainder 6/3=2 with 0 remainder 10/3=3 with 1 remainder 15/3=5 with 0 remainder 21/3=7 with 0 remainder the pattern is 1,0,0,1,0,0 with remainders confidence assessment: 1
.................................................
......!!!!!!!!...................................
15:51:43 ** 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. **
......!!!!!!!!...................................
RESPONSE --> correct self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:56:31 1.2.48 use formula to find the 12 th octagonal number. Explain in detail how you used the formula to find this number.
......!!!!!!!!...................................
RESPONSE --> using the formula in the book it would 12(6*12-4)/2=408 confidence assessment: 1
.................................................
......!!!!!!!!...................................
15:57:46 ** 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. **
......!!!!!!!!...................................
RESPONSE --> correct but the formula in the book showed n(6n-4)/2 which is what I used. self critique assessment: 3
.................................................
Ϝzَ~̚٬ assignment #008 008. `Query 8 College Algebra 02-14-2009
......!!!!!!!!...................................
16:13:34 1.3.6 9 and 11 yr old hosses; sum of ages 122. How many 9- and 11-year-old horses are there?
......!!!!!!!!...................................
RESPONSE --> I tried different numbers but I came up with 7*11=77 then I subtracted 77 from 122 which equals 45 then divided that by 9 which equals 5. The other numbers could not be divided by 9. So the answer is 7 11 year olds and 5 9 year olds. confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:13:41 ** 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. **
......!!!!!!!!...................................
RESPONSE --> correct self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:24:45 Query 1.3.10 divide clock into segments each with same total
......!!!!!!!!...................................
RESPONSE --> 1+2+3+4+5+6+7+8+9+10+11+12=78/3=26 so each segment would have to equal 26 after calculating on paper for a few minutes I came up with, 1,2,11,12 3,4,9,10 5,6,7,8 confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:24:54 ** 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. **
......!!!!!!!!...................................
RESPONSE --> correct self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:27:54 Query 1.3.18 M-F 32 acorns each day, half of all acorns eaten, 35 acorns left after Friday
......!!!!!!!!...................................
RESPONSE --> on this one I did 32*5=160/2=80=45 confidence assessment: 0
.................................................
......!!!!!!!!...................................
16:28:28 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok I should have worked this one out backwards I understand. self critique assessment: 2
.................................................
......!!!!!!!!...................................
16:35:04 Query 1.3.30 Frog in well, 4 ft jump, 3 ft back.
......!!!!!!!!...................................
RESPONSE --> on the first day he jumps to 4 ft then back 3 to 1 ft the next day he jumps 4 ft to make it 5 ft then back down to 2 ft. by the time he reaches the 16th day he would be at 19 ft then the 17th day he can jump out. confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:35:31 ** 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. **
......!!!!!!!!...................................
RESPONSE --> correct, i started to say 20 at first too. self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:38:27 Query 1.3.48 How many ways to pay 15 cents?
......!!!!!!!!...................................
RESPONSE --> I came up with, 15 pennies 3 nickels 5 pennies and 2 nickels 10 pennies and 1 nickel 5 pennies and 1 dime 1 nickel and 1 dime 6 different ways. confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:38:39 ** 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 **
......!!!!!!!!...................................
RESPONSE --> correct self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:45:29 Query 1.3.52 Given 8 coins, how do you find the unbalanced one in 3 weighings
......!!!!!!!!...................................
RESPONSE --> I would put 4 coins and 4 on the other the one with the fake coin would be lighter and be higher on the scale. After that I would take those 4 coins and put 2 on each side. Again the lighter side would raise up and then divide those 2. The last weighing will determine the fake coin. confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:45:37 ** 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. **
......!!!!!!!!...................................
RESPONSE --> correct self critique assessment: 3
.................................................