course Mth 151 I cannot locate q_a_ 20. When I downloaded the programs the q_a_ files are divided up into q_a_ initials, q_a_ 1-19, q_a_ 21-26, and q_a_ 28-29. ?????ü????????assignment #020????~v???€??G?Liberal Arts Mathematics I
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16:38:55 query 4.3.6 number following base-six 555
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RESPONSE --> the number before would be 554 base six and I'm not for sure but I think it is 1000
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16:40:56 ** COMMON ERROR: 556. INSTRUCTOR COMMENT: The digit 6 does not exist in base six for the same reason as there is no single digit for 10 in base 10 (digits stop at 9; in general there is no digit corresponding to the base). CORRECT SOLUTION: 555{base 6} = 5 * 6^2 + 5 * 6^1 + 5 * 6^0. If you add 1 you get 5 * 6^2 + 5 * 6^1 + 5 * 6^0 + 1, which is equal to 5 * 6^2 + 5 * 6^1 + 6 * 6^0. But 6 * 6^0 is 6^1, so now you have 5 * 6^2 + 6 * 6^1 + 0 * 6^0. But 6 * 6^1 is 6^2 so you have 6 * 6^2 + 0 * 6^1 + 0 * 6^0. But 6 * 6^2 is 6^3 so the number is 6 * 6^3 + 0 * 6^2 + 0 * 6^1 + 0 * 6^0. So the number following 555{base 6} is 1000{base 6}. The idea is that we can use a number greater than 5 in base 6, so after 555 we go to 1000, just like in base 10 we go from 999 to 1000. **
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RESPONSE --> ok
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16:45:13 query 4.3.20 34432 base five
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RESPONSE --> 3 * 5^4 + 4 * 5^3 = 4 * 5^2 + 3 * 5^1 + 2 * 5^0 1875 +500 +100 +15 +2 2492 calculator shortcut: 3*5+4*5+4*5+3*5+2 = 2492
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16:45:33 **34432 base five means 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0. 5^2 = 625, 5^3 = 125, 5^2 = 25, 5^1 = 5 and 5^0 = 1 so 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0= 3 * 625 + 4 * 125 + 4 * 25 + 3 * 5 + 2 * 1 = 1875 + 500 + 100 + 15 + 2 = 2492. **
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RESPONSE --> ok
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16:47:59 Explain how you use the calculator shortcut to get the given number.
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RESPONSE --> This is how I understand it by reading the book. Start with the first number on the left and * it by the base. Then add the next digit, * by the base and keep repeating this process until you get to the last digit on the right. Only add this digit do not multiply by the base.
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16:48:43 ** 2 + 5 (3 + 5(4 + 5(4 + 3 * 5))) multiplies out to the given number because, for example, the 3 * 5 at the end will be multiplied by the three 5? to its left to give 3 * 5 * 5 * 5 * 5 = 3 * 5^4, which matches the 3 * 5^4 in the sum 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0. So we do 3 + 5, then 4 + the answer, then 5 * the answer, then 4 + the answer, then 5 * the answer, then 3 + the answer, then 5 * the answer, then 2 + the answer. **
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RESPONSE --> ok
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16:52:38 query 4.3.40 11028 decimal to base 4
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RESPONSE --> 4^0= 1 0 leaving 0 4^1= 4 1 0 4^2= 16 1 4 4^3= 64 0 20 4^4= 256 3 20 4^5= 1024 2 788 4^6=4096 2 2836 2230110 base four
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16:53:09 ** 4^0 = 1 4^1 = 4 4^2 = 16 4^3 = 64 4^4 = 256 4^5 = 1024 4^6 = 4096 (4*7 = 16386, which is larger than the given 11028) So to ?uild up?11028 we need 2 * 4^6 = 8192, leaving 2836. 2 * 4^5 = 2048, leaving 788. 3 * 4^4 = 768, leaving 20. 0 * 4^3, because we need only 20, which is less than 64. 1 * 4^2 = 16, leaving 4. 1 * 4^1 = 4, leaving 0.
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RESPONSE --> ok
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16:54:17 0 * 4^1. Thus our number is 2230110 base 4.
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RESPONSE --> I am confused about what problem this is or how to work it.
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16:55:07 query 4.3.51 DC in base 16 to binary
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RESPONSE --> 11011100 base 2
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16:55:47 ** C stands for decimal 12, which in binary is 1100. D stands for decimal 13, which in binary is 1101. Since our base 16 is the fourth power of 2, our number can be written 11011100, where the 1101 in the first four places stands for D and the 1100 in the last four places stands for C. Note that this method works only when one base is a power of the other.**
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RESPONSE --> ok
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17:00:57 Is a base-9 number always even, always odd, or sometimes even and sometimes odd if the number ends in an even number?
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RESPONSE --> Sometimes even and sometimes odd 1,2,3,4,5,6,7,8,10,11,12,13,14,15,16,17,18,20 It starts off odd, even and works until number 10 which becomes odd. Once you reach 20 it is back odd even again.
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17:07:06 ** You can investigate this question by trying a variety of examples. For example, 82nine = 8 * 9^1 + 2 * 9^0 = 72 + 2 = 74 in decimal. This is even because you are adding two even multiples of the odd numbers 9^1 = 9 and 9^0 = 1. You have to use an even multiple of 9^0 = 1 because the number ends with an even number. But you don't have to use an even multiple of 9^1 = 9. So we try something like 72nine = 7 * 9^1 + 2 * 9^0 = 63 + 2 = 65 in decimal and we see that the result is odd. The key is that in base nine, the powers of nine are always odd numbers. So when converting you get from a base-9 number with two even digits a number which is even * odd + even = even, while from a base-9 number with an odd digit followed by an even digit you get odd * odd + even = odd; many other possible combinations show the same thing but this is one of the simpler examples that shows why we get odd for some base-9 numbers, and even for others. For a couple of specifics, 70 in base 9 is 63 in base ten, and is odd. However 770 in base 9 is even. **
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RESPONSE --> This problem was confusing for me.
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