course mth 158 \קrx|bΕ|assignment #002
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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
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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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ƈ{ѫރ assignment #001 001. `query 1 College Algebra 06-01-2007
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12:37:45 R.1.14 (was R.1.6) Of the numbers in the set {-sqrt(2), pi + sqrt(2), 1 / 2 + 10.3} which are counting numbers, which are rational numbers, which are irrational numbers and which are real numbers?
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RESPONSE --> counting numbers: none rational numbers: 1/2 + 10.3 irrational numbers: -sprt(2); pi + sprt(2) real numbers: -sprt(2); pi + sprt(2); 1/2 + 10.3 confidence assessment: 3
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12:38:08 ** Counting numbers are the numbers 1, 2, 3, .... . None of the given numbers are counting numbers Rational numbers are numbers which can be expressed as the ratio of two integers. 1/2+10.3 are rational numbers. Irrational numbers are numbers which can be expressed as the ratio of two integers. {-sqrt(2)}, pi+sqrt(2) are irrational numbers.. **
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RESPONSE --> I undertand the different categories of numbers. self critique assessment: 3
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12:38:34 R.1.32 (was R.1.24) Write in symbols: The product of 2 and x is the product of 4 and 6
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RESPONSE --> 2 * x = 4 * 6 confidence assessment: 3
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12:38:53 ** The product of 2 and x is 2 * x and the product of 4 and 6 iw 4 * 6. To say that these are identical is to say that 2*x=4*6. **
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RESPONSE --> I know how to write words into symbols. self critique assessment: 3
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12:42:58 R.1.50 (was R.1.42) Explain how you evaluate the expression 2 - 5 * 4 - [ 6 * ( 3 - 4) ]
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RESPONSE --> 2 - 5 * 4 - [ 6 * ( 3 - 4) ] 2 - 5 * 4 - [ 6 * (-1) ] 2 - 5 * 4 - [-6] 2 - 5 * 4 + 6 2 - 20 + 6 -18 + 6 -12 confidence assessment: 3
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12:43:50 **Starting with 2-5*4-[6*(3-4)]. First you evaluate the innermost group to get 2-5*4-[6*-1] . Then multiply inside brackets to get 2-5*4+6. Then do the multiplication to get 2-20+6. Then add and subtract in order, obtaining -12. **
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RESPONSE --> I understand how to solve equations. self critique assessment: 3
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12:45:17 R.1.80 (was R.1.72) Explain how you use the distributive property to remove the parentheses from the express (x-2)(x-4).
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RESPONSE --> ( x - 2 ) ( x - 4 ) x ^2 - 4 x - 2 x + 8 x^2 - 6 x + 8 confidence assessment: 3
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12:47:35 ** COMMON ERROR: Using FOIL. FOIL is not the Distributive Law. FOIL works for binomial expressions. FOIL follows from the distributive law but is of extremely limited usefulness and the instructor does not recommend relying on FOIL. Starting with (x-2)(x-4) ; one application of the Distributive Property gives you x(x-4) - 2(x-4) . Applying the property to both of the other terms we get x^2 - 4x - (2x -8). Simplifying: x^2 - 4x - 2x + 8 or x^2 - 6x + 8. *
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RESPONSE --> I understand the FOIL system and how to solve problems using it. self critique assessment: 3
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12:49:36 R.1.86 (was R.1.78) Explain why (4+3) / (2+5) is not equal to 4/2 + 3/5.
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RESPONSE --> The firstequation is the product of the numbers and the second is adding two different equations together. confidence assessment: 2
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12:50:17
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RESPONSE --> I know the order of operations, but I was answering by just looking at the problems. confidence assessment: 3
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zztK߮F^а assignment #001 001. `query 1 College Algebra 06-01-2007
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12:53:06 R.1.14 (was R.1.6) Of the numbers in the set {-sqrt(2), pi + sqrt(2), 1 / 2 + 10.3} which are counting numbers, which are rational numbers, which are irrational numbers and which are real numbers?
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RESPONSE --> Counting: none Rational: 1 / 2 + 10.3; Irrational: -sqrt(2); pi + sqrt(2) Real: 1 / 2 + 10.3; -sqrt(2); pi + sqrt(2) confidence assessment: 3
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12:53:19 ** Counting numbers are the numbers 1, 2, 3, .... . None of the given numbers are counting numbers Rational numbers are numbers which can be expressed as the ratio of two integers. 1/2+10.3 are rational numbers. Irrational numbers are numbers which can be expressed as the ratio of two integers. {-sqrt(2)}, pi+sqrt(2) are irrational numbers.. **
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RESPONSE --> I know the types of numbers. self critique assessment: 3
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12:53:55 R.1.32 (was R.1.24) Write in symbols: The product of 2 and x is the product of 4 and 6
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RESPONSE --> 2 * x = 4 * 6 confidence assessment: 3
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12:54:10 ** The product of 2 and x is 2 * x and the product of 4 and 6 iw 4 * 6. To say that these are identical is to say that 2*x=4*6. **
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RESPONSE --> I understand this process. self critique assessment: 3
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12:56:13 R.1.50 (was R.1.42) Explain how you evaluate the expression 2 - 5 * 4 - [ 6 * ( 3 - 4) ]
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RESPONSE --> 2 - 5 * 4 - [ 6 * ( 3 - 4) ] 2 - 5 * 4 - [ 6 * ( -1 ) ] 2 - 5 * 4 - [ -6 ] 2 - 20 - [ -6 ] 2 - 20 + 6 -18 + 6 -12 confidence assessment: 3
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12:56:28 **Starting with 2-5*4-[6*(3-4)]. First you evaluate the innermost group to get 2-5*4-[6*-1] . Then multiply inside brackets to get 2-5*4+6. Then do the multiplication to get 2-20+6. Then add and subtract in order, obtaining -12. **
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RESPONSE --> I know how to solve equations. self critique assessment: 3
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12:58:58 R.1.80 (was R.1.72) Explain how you use the distributive property to remove the parentheses from the express (x-2)(x-4).
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RESPONSE --> (x-2)(x-4) x^2 - 4x - 2x + 8 x^2 - 6x + 8 confidence assessment: 3
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12:59:13 ** COMMON ERROR: Using FOIL. FOIL is not the Distributive Law. FOIL works for binomial expressions. FOIL follows from the distributive law but is of extremely limited usefulness and the instructor does not recommend relying on FOIL. Starting with (x-2)(x-4) ; one application of the Distributive Property gives you x(x-4) - 2(x-4) . Applying the property to both of the other terms we get x^2 - 4x - (2x -8). Simplifying: x^2 - 4x - 2x + 8 or x^2 - 6x + 8. *
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RESPONSE --> I understand how to use FOIL. self critique assessment: 3
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13:01:35 R.1.86 (was R.1.78) Explain why (4+3) / (2+5) is not equal to 4/2 + 3/5.
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RESPONSE --> The first equation is going to add the numbers together and then divided them. The second equation is adding the two problems together. confidence assessment: 3
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13:02:13 ** Good answer but at an even more fundamental level it comes down to order of operations. (4+3)/(2+5) means 7/7 which is equal to 1. By order of operations, in which multiplications and divisions precede additions and subtractions, 4/2+3/5 means (4/2) + (3/5), which gives us 2+3/5 = 2 3/5 **
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RESPONSE --> I was simply solving this how they look, I didn't think about the order of operations. self critique assessment: 3
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