course Mth 158 \קrx|bΕ|
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11:23:42 R.2.46 (was R.2.36) Evaluate for x = -2, and y = 3 the expression (2x - 3) / y and explan how you got your result.
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RESPONSE --> ( 2 ( - 2 ) - 3 ) / 3 ( - 4 - 3 ) / 3 - 7 / 3 Following the order of operations I did the parentheses first, with muliplication before addition. Then made sure the signs were correct. confidence assessment: 3
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11:24:23 ** Starting with (2x-3)/y we substitute x=-2 and y=3 to get (2*(-2) - 3)/3 = (-4-3)/3= -7/3. **
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RESPONSE --> I know how to plug in numbers into an equation. self critique assessment: 3
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11:27:42 R.2. 55 (was R.2.45) Evaluate for x = 3 and y = -2: | |4x| - |5y| | and explan how you got your result.
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RESPONSE --> I I4xI - I5yI I I I4(3)I - I5(-2)I I I I7I - I-10I I I I7I - I10I I I 7 - 10 I I -3 I 3 confidence assessment: 3
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11:28:40 ** Starting with | | 4x |- | 5y | | we substitute x=3 and y=-2 to get | | 4*3 | - | 5*-2 | | = | | 12 | - | -10 | | = | 12-10 | = | 2 | = 2. **
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RESPONSE --> I multiplied wrong, but I did have it correct in my notebook. self critique assessment: 3
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11:29:20 R.2.64 (was R.2.54) Explain what values, if any, must not be present in the domain of the expression (-9x^2 - x + 1) / (x^3 + x)
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RESPONSE --> all of them can be used. confidence assessment: 1
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11:30:43 ** The denominator of this expression cannot be zero, since division by zero is undefined. Since x^3 + x factors into (x^2 + 1) ( x ) we see that x^3 + x = 0 only if x^2 + 1 = 0 or x = 0. Since x^2 cannot be negative x^2 + 1 cannot be 0, so x = 0 is indeed the only value for which x^3 + x = 0. **
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RESPONSE --> I did not understad these problems, but I know that the denominator cannot be zero. self critique assessment: 3
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11:32:05 R.2.73 (was R.4.6). What is (-4)^-2 and how did you use the laws of exponents to get your result?
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RESPONSE --> -0.004 This problem can be answered by moving the decimal to the left two places. confidence assessment: 3
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11:33:54 **Since a^-b = 1 / (a^b), we have (-4)^-2 = 1 / (-4)^2 = 1 / 16. **
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RESPONSE --> I did this problem wrong because I was simply just moved the decimal over because I thought that was what I was suppose to do. self critique assessment: 3
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11:36:36 Extra Problem. What is (3^-2 * 5^3) / (3^2 * 5) and how did you use the laws of exponents to get your result?
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RESPONSE --> (9 * 125 ) / ( 9 * 5) 1125 / 45 25 I don't know how the exponent law helped me, I just simply applied the exponents and solved the equation. confidence assessment: 2
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11:37:12 ** (3^(-2)*5^3)/(3^2*5). Grouping factors with like bases we have 3^(-2)/3^2 * 5^3 / 5. Using the fact that a^b / a^c = a^(b-c) we get 3^(-2 -2) * 5^(3-1), which gives us 3^-4 * 5^2. Using a^(-b) = 1 / a^b we get (1/3^4) * 5^2. Simplifying we have (1/81) * 25 = 25/81. **
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RESPONSE --> I messed up around the negitive exponent. self critique assessment: 2
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11:41:18 R.2.94. Express [ 5 x^-2 / (6 y^-2) ] ^ -3 with only positive exponents and explain how you used the laws of exponents to get your result.
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RESPONSE --> I don't understand how to get this answer. It wasn't in the problems to do, and Icannot figure it out. confidence assessment: 3
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11:42:27 [ 5 x^-2 / (6 y^-2) ] ^ -3 = (5 x^-2)^-3 / (6 y^-2)^-3, since (a/b)^c = a^c / b^c. This simplifies to 5^-3 (x^-2)^-3 / [ 6^-3 (y^-2)^-3 ] since (ab)^c = a^c b^c. Then since (a^b)^c = a^(bc) we have 5^-3 x^6 / [ 6^-3 y^6 ] . We rearrange this to get the result 6^3 x^6 / (5^3 y^6), since a^-b = 1 / a^b.
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RESPONSE --> I understand that you must rearrange the problem to get the answer that we need. self critique assessment: 3
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11:43:36 Extra Problem. Express (-8 x^3) ^ -2 with only positive exponents and explain how you used the laws of exponents to get your result.
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RESPONSE --> I still don't understand about negitive exponents. confidence assessment: 3
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11:44:36 ** ERRONEOUS STUDENT SOLUTION: (-8x^3)^-2 -1/(-8^2 * x^3+2) 1/64x^5 INSTRUCTOR COMMENT:1/64x^5 means 1 / 64 * x^5 = x^5 / 64. This is not what you meant but it is the only correct interpretation of what you wrote. Also it's not x^3 * x^2, which would be x^5, but (x^3)^2. There are several ways to get the solution. Two ways are shown below. They make more sense if you write them out in standard notation. ONE CORRECT SOLUTION: (-8x^3)^-2 = (-8)^-2*(x^3)^-2 = 1 / (-8)^2 * 1 / (x^3)^2 = 1/64 * 1/x^6 = 1 / (64 x^5). Alternatively (-8 x^3)^-2 = 1 / [ (-8 x^3)^2] = 1 / [ (-8)^2 (x^3)^2 ] = 1 / ( 64 x^6 ). **
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RESPONSE --> I see how we got this answer and I still don't get negitive exponents. self critique assessment: 3
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11:47:47 R.2.90 (was R.4.36). Express (x^-2 y) / (x y^2) with only positive exponents and explain how you used the laws of exponents to get your result.
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RESPONSE --> (x^-2 y) / ( x y^2 ) (x^-2/x) * (y/y^2) xy/ x^2y^2 confidence assessment: 1
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11:48:26 ** (1/x^2 * y) / (x * y^2) = (1/x^2 * y) * 1 / (x * y^2) = y * 1 / ( x^2 * x * y^2) = y / (x^3 y^2) = 1 / (x^3 y). Alternatively, or as a check, you could use exponents on term as follows: (x^-2y)/(xy^2) = x^-2 * y * x^-1 * y^-2 = x^(-2 - 1) * y^(1 - 2) = x^-3 y^-1 = 1 / (x^3 y).**
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RESPONSE --> I still don't understand this negitive exponents. self critique assessment: 3
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11:49:00 Extra Problem. . Express 4 x^-2 (y z)^-1 / [ (-5)^2 x^4 y^2 z^-5 ] with only positive exponents and explain how you used the laws of exponents to get your result.
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RESPONSE --> I don't know. confidence assessment: 3
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11:50:05 ** Starting with 4x^-2(yz)^-1/ [ (-5)^2 x^4 y^2 z^-5] Squaring the -5 and using the fact that (yz)^-1 = y^1 * z^-1: 4x^-2 * y^-1 * z^-1/ [25 * x^4 * y^2 * z^-5} Grouping the numbers, and the x, the y and the z expression: (4/25) * (x^-2/x^4) * (y^-1/y^2) * (z^-1/z^-5) Simplifying by the laws of exponents: (4/25) * x^(-2-4) * y^(-1-2) * z^(-1+5) Simplifying further: (4/25) * x^-6 * y^-3 * z^4 Writing with positive exponents: 4z^4/ (25x^6 * y^3 ) **
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RESPONSE --> I'm still having problems with these exponents. self critique assessment: 3
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11:50:25 R.2.122 (was R.4.72). Express 0.00421 in scientific notation.
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RESPONSE --> 4.21 * 10^3 confidence assessment:
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11:50:54 ** 0.00421 in scientific notation is 4.21*10^-3. This is expressed on many calculators as 4.21 E-4. **
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RESPONSE --> I triedd to go backto write in the negitive exponent on 10. self critique assessment: 3
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11:51:20 R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation.
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RESPONSE --> 9,700 confidence assessment: 3
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11:51:45 ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **
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RESPONSE --> I understand the scientific notation. self critique assessment: 3
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11:54:13 R.2.150 (was R.2.78) If an unhealthy temperature is one for which | T - 98.6 | > 1.5, then how do you show that T = 97 and T = 100 are unhealthy?
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RESPONSE --> I 97 - 98.6 I > 1.5 I -1.6 I > 1.5 1.6 > 1.5 unhealthy temperature I 100 - 98.6 I > 1.5 I 1.4 I < 1.5 healthy temperature confidence assessment: 3
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11:55:01 ** You can show that T=97 is unhealthy by substituting 97 for T to get | -1.6| > 1.5, equivalent to the true statement 1.6>1.5. But you can't show that T=100 is unhealthy, when you sustitute for T then it becomes | 100 - 98.6 | > 1.5, or | 1.4 | > 1.5, giving us 1.4>1.5, which is an untrue statement. **
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RESPONSE --> This is why on the second problem I turned the greater than sign aound. self critique assessment: 3
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煳{GNǍk枷 assignment #002 002. `query 2 College Algebra 06-01-2007
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12:03:50 R.2.46 (was R.2.36) Evaluate for x = -2, and y = 3 the expression (2x - 3) / y and explan how you got your result.
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RESPONSE --> ( 2 (-2) - 3) / 3 (-4 -3) / 3 -7/3 confidence assessment: 2
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12:04:28 ** Starting with (2x-3)/y we substitute x=-2 and y=3 to get (2*(-2) - 3)/3 = (-4-3)/3= -7/3. **
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RESPONSE --> I got this answer and understood how to get it. self critique assessment: 3
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12:06:54 R.2. 55 (was R.2.45) Evaluate for x = 3 and y = -2: | |4x| - |5y| | and explan how you got your result.
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RESPONSE --> I I4(3)I - I5(-2)I I I I12I - I-10I I I 12 - 10 I 2 confidence assessment: 3
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12:07:35 ** Starting with | | 4x |- | 5y | | we substitute x=3 and y=-2 to get | | 4*3 | - | 5*-2 | | = | | 12 | - | -10 | | = | 12-10 | = | 2 | = 2. **
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RESPONSE --> I got this answer and know how to do this kind of problem. self critique assessment: 3
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12:08:03 R.2.64 (was R.2.54) Explain what values, if any, must not be present in the domain of the expression (-9x^2 - x + 1) / (x^3 + x)
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RESPONSE --> any, except for 0 confidence assessment: 3
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12:08:18 ** The denominator of this expression cannot be zero, since division by zero is undefined. Since x^3 + x factors into (x^2 + 1) ( x ) we see that x^3 + x = 0 only if x^2 + 1 = 0 or x = 0. Since x^2 cannot be negative x^2 + 1 cannot be 0, so x = 0 is indeed the only value for which x^3 + x = 0. **
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RESPONSE --> I got this. self critique assessment: 3
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12:09:17 R.2.73 (was R.4.6). What is (-4)^-2 and how did you use the laws of exponents to get your result?
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RESPONSE --> 4^2 16 confidence assessment: 0
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12:09:49 **Since a^-b = 1 / (a^b), we have (-4)^-2 = 1 / (-4)^2 = 1 / 16. **
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RESPONSE --> I didn't put the answer over a 1 because of the negitive sign. self critique assessment: 3
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12:11:01 Extra Problem. What is (3^-2 * 5^3) / (3^2 * 5) and how did you use the laws of exponents to get your result?
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RESPONSE --> I don't understand these questions (I've done this once, but it didn't save on my computer.) confidence assessment: 3
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12:11:34 ** (3^(-2)*5^3)/(3^2*5). Grouping factors with like bases we have 3^(-2)/3^2 * 5^3 / 5. Using the fact that a^b / a^c = a^(b-c) we get 3^(-2 -2) * 5^(3-1), which gives us 3^-4 * 5^2. Using a^(-b) = 1 / a^b we get (1/3^4) * 5^2. Simplifying we have (1/81) * 25 = 25/81. **
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RESPONSE --> I see how to get the answer,but don't really understand how. self critique assessment: 3
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12:12:25 R.2.94. Express [ 5 x^-2 / (6 y^-2) ] ^ -3 with only positive exponents and explain how you used the laws of exponents to get your result.
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RESPONSE --> 6^3 x^6 / 5^3 y^6 confidence assessment: 0
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12:12:43 [ 5 x^-2 / (6 y^-2) ] ^ -3 = (5 x^-2)^-3 / (6 y^-2)^-3, since (a/b)^c = a^c / b^c. This simplifies to 5^-3 (x^-2)^-3 / [ 6^-3 (y^-2)^-3 ] since (ab)^c = a^c b^c. Then since (a^b)^c = a^(bc) we have 5^-3 x^6 / [ 6^-3 y^6 ] . We rearrange this to get the result 6^3 x^6 / (5^3 y^6), since a^-b = 1 / a^b.
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RESPONSE --> I still don't understand. self critique assessment: 3
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12:13:05 Extra Problem. Express (-8 x^3) ^ -2 with only positive exponents and explain how you used the laws of exponents to get your result.
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RESPONSE --> 1 / 64x^5 confidence assessment: 2
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12:13:58 ** ERRONEOUS STUDENT SOLUTION: (-8x^3)^-2 -1/(-8^2 * x^3+2) 1/64x^5 INSTRUCTOR COMMENT:1/64x^5 means 1 / 64 * x^5 = x^5 / 64. This is not what you meant but it is the only correct interpretation of what you wrote. Also it's not x^3 * x^2, which would be x^5, but (x^3)^2. There are several ways to get the solution. Two ways are shown below. They make more sense if you write them out in standard notation. ONE CORRECT SOLUTION: (-8x^3)^-2 = (-8)^-2*(x^3)^-2 = 1 / (-8)^2 * 1 / (x^3)^2 = 1/64 * 1/x^6 = 1 / (64 x^5). Alternatively (-8 x^3)^-2 = 1 / [ (-8 x^3)^2] = 1 / [ (-8)^2 (x^3)^2 ] = 1 / ( 64 x^6 ). **
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RESPONSE --> I get how to make exponents negitive. self critique assessment: 3
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12:14:37 R.2.90 (was R.4.36). Express (x^-2 y) / (x y^2) with only positive exponents and explain how you used the laws of exponents to get your result.
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RESPONSE --> 1/ x^3 y confidence assessment: 2
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12:15:35 ** (1/x^2 * y) / (x * y^2) = (1/x^2 * y) * 1 / (x * y^2) = y * 1 / ( x^2 * x * y^2) = y / (x^3 y^2) = 1 / (x^3 y). Alternatively, or as a check, you could use exponents on term as follows: (x^-2y)/(xy^2) = x^-2 * y * x^-1 * y^-2 = x^(-2 - 1) * y^(1 - 2) = x^-3 y^-1 = 1 / (x^3 y).**
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RESPONSE --> With the exponents of making it negitive, you put it over a one. self critique assessment: 3
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12:16:08 Extra Problem. . Express 4 x^-2 (y z)^-1 / [ (-5)^2 x^4 y^2 z^-5 ] with only positive exponents and explain how you used the laws of exponents to get your result.
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RESPONSE --> 4z^4 / 26x^6 * y^3 confidence assessment: 2
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12:16:55 ** Starting with 4x^-2(yz)^-1/ [ (-5)^2 x^4 y^2 z^-5] Squaring the -5 and using the fact that (yz)^-1 = y^1 * z^-1: 4x^-2 * y^-1 * z^-1/ [25 * x^4 * y^2 * z^-5} Grouping the numbers, and the x, the y and the z expression: (4/25) * (x^-2/x^4) * (y^-1/y^2) * (z^-1/z^-5) Simplifying by the laws of exponents: (4/25) * x^(-2-4) * y^(-1-2) * z^(-1+5) Simplifying further: (4/25) * x^-6 * y^-3 * z^4 Writing with positive exponents: 4z^4/ (25x^6 * y^3 ) **
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RESPONSE --> This again is a negitive exponent problem. self critique assessment: 3
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12:17:29 R.2.122 (was R.4.72). Express 0.00421 in scientific notation.
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RESPONSE --> 4.21 * 10 ^-3 confidence assessment: 3
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12:18:12 ** 0.00421 in scientific notation is 4.21*10^-3. This is expressed on many calculators as 4.21 E-4. **
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RESPONSE --> I understand scientific notation. self critique assessment: 3
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12:18:30 R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation.
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RESPONSE --> 9700 confidence assessment: 3
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12:19:01 ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **
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RESPONSE --> this is scientific noation and I understand this process. self critique assessment: 3
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12:21:44 R.2.150 (was R.2.78) If an unhealthy temperature is one for which | T - 98.6 | > 1.5, then how do you show that T = 97 and T = 100 are unhealthy?
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RESPONSE --> I 97 - 98.6 I > 1.5 I-1.6I > 1.5 1.6 > 1.5 UNHEALTHY TEMPERATURE I 100 - 98.6 I > 1.5 I 1.4 I > 1.5 1.4 < 1.5 HEALTHY TEMPERATURE confidence assessment: 3
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12:22:32 ** You can show that T=97 is unhealthy by substituting 97 for T to get | -1.6| > 1.5, equivalent to the true statement 1.6>1.5. But you can't show that T=100 is unhealthy, when you sustitute for T then it becomes | 100 - 98.6 | > 1.5, or | 1.4 | > 1.5, giving us 1.4>1.5, which is an untrue statement. **
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RESPONSE --> This is why I changed to sign on the second one to make it correct. self critique assessment: 3
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