course Mth 151 Prof.Smith,I have taken test 1, test 2 (which you already graded), and test 3. I haven't seen my grades for test 1&3(hopefully I did ok), so I am working towards test 4.
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09:36:35 `q001. There are 5 questions in this set. From lectures and textbook you will learn about some of the counting systems used by past cultures. Various systems enabled people to count objects and to do basic arithmetic, but the base-10 place value system almost universally used today has significant advantages over all these systems. The key to the base-10 place value system is that each digit in a number tells us how many times a corresponding power of 10 is to be counted. For example the number 347 tells us that we have seven 1's, 4 ten's and 3 one-hundred's, so 347 means 3 * 100 + 4 * 10 + 7 * 1. Since 10^2 = 100, 10^1 = 10 and 10^0 = 1, this is also written as 3 * 10^2 + 4 * 10^1 + 7 * 10^0. How would we write 836 in terms of powers of 10?
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RESPONSE --> 8*100+3*10+6*1= 836 confidence assessment: 3
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09:37:10 836 means 8 * 100 + 3 * 10 + 6 * 1, or 8 * 10^2 + 3 * 10^1 + 6 * 10^0.
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RESPONSE --> so my answer was correct self critique assessment: 3
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09:38:12 `q002. How would we write 34,907 in terms of powers of 10?
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RESPONSE --> 34*1000+9*100+0*10+7*1 confidence assessment: 3
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09:40:13 34,907 means 3 * 10,000 + 4 * 1000 + 9 * 100 + 0 * 10 + 7 * 1, or 3 * 10^4 + 4 * 10^3 + 9 * 10^2 + 0 * 10 + 7 * 1.
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RESPONSE --> so we separate all the digits I took the first two and raised to the 1000 power. but it should have been 3*10,000+4*1,000+9*100+0*10+7*1 self critique assessment: 3
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09:46:27 `q003. How would we write .00326 in terms of powers of 10?
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RESPONSE --> .0*10,000+0*1,000+3*100+2*10+6*1 confidence assessment: 3
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09:53:23 First we note that .1 = 1/10 = 1/10^1 = 10^-1, .01 = 1/100 = 1/10^2 = 10^-2, .001 = 1/1000 = 1/10^3 = 10^-3, etc.. Thus .00326 means 0 * .1 + 0 * .01 + 3 * .001 + 2 * .0001 + 6 * .00001 = 0 * 10^-1 + 0 * 10^-2 + 3 * 10^-3 + 2 * 10^-4 + 6 * 10^-5 .
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RESPONSE --> decimal points are used on each number self critique assessment: 3
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09:57:10 `q004. How would we add 3 * 10^2 + 5 * 10^1 + 7 * 10^0 to 5 * 10^2 + 4 * 10^1 + 2 * 10^0?
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RESPONSE --> 357 542 -------- 899 confidence assessment: 3
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09:59:10 We would write the sum as (3 * 10^2 + 5 * 10^1 + 7 * 10^0) + (5 * 10^2 + 4 * 10^1 + 2 * 10^0) , which we would then rearrange as (3 * 10^2 + 5 * 10^2) + ( 5 * 10^1 + 4 * 10^1) + ( 7 * 10^0 + 2 * 10^0), which gives us 8 * 10^2 + 9 * 10^1 + 9 * 10^0. This result would then be written as 899.
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RESPONSE --> we rearrange, add and then translate self critique assessment: 3
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10:09:54 `q005. How would we add 4 * 10^2 + 7 * 10^1 + 8 * 10^0 to 5 * 10^2 + 6 * 10^1 + 4 * 10^0?
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RESPONSE --> 4*100+7*10+8*1 + 5*100+6*10+4*1 ________________ 9*100+13*10+12*1 or carry over 4*100+7*10+8*1 + 5*100+6*10+4*1 ________________ 1*10,000+0*100+4*10+2*1 which means: 1042 confidence assessment: 3
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10:13:03 We would write the sum as (4 * 10^2 + 7 * 10^1 + 8 * 10^0) + (5 * 10^2 + 6 * 10^1 + 4 * 10^0) , which we would then rearrange as (4 * 10^2 + 5 * 10^2) + ( 7 * 10^1 + 6 * 10^1) + ( 8 * 10^0 + 4 * 10^0), which gives us 9 * 10^2 + 13 * 10^1 + 12 * 10^0. Since 12 * 10^0 = (2 + 10 ) * 10^0 = 2 * 10^0 + 10^1, we have 9 * 10^2 + 13 * 10^1 + 1 * 10^1 + 2 * 10^0 = 9 * 10^2 + 14 * 10^1 + 2 * 10^0. Since 14 * 10^1 = 10 * 10^1 + 4 * 10^1 = 10^2 + 4 * 10^1, we have 9 * 10^2 + 1 * 10^2 + 4 * 10^1 + 2 * 10^0 = 10^10^2 + 4 * 10^1 + 2 * 10^0. Since 10*10^2 = 10^3, we rewrite this as 1 * 10^3 + 0 * 10^2 + 4 * 10^1 + 2 * 10^0. This number would be expressed as 1042.
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RESPONSE --> rearrange, add and translate self critique assessment: 3
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