course Mth 158 kqGůg{֠assignment #001 yܩTFbxCΧ assignment #002 002. `query 2 College Algebra 01-28-2007
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19:38:24 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 --> For the problem I insert -2 into the equation for x and 3 into the equation for y. The new problem now looks like this: {2*(-2)-3} / 3. I first multiply 2 by -2 and I got -4 so now my problem looks like (-4 - 3) / 3. Then I subtracted 3 from -4 and got -7. Now my problem reads -7/3. My answer is -7/3 or -2.3 repeating. confidence assessment: 2
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19:39:26 ** 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 messed up and hit enter response too many times. self critique assessment:
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19:47:30 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 --> 0 cannot be in the expression because you cannot divide by 0. confidence assessment: 1
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19:47:57 01-28-2007 19:47:57 ** 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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NOTES -------> ** 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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19:50:26 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 --> Using the laws of exponents I put the equation into this form 1/-4^2 My answer for the problem 1/-4^2 is 1/-16 confidence assessment: 2
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19:50:45 01-28-2007 19:50:45 **Since a^-b = 1 / (a^b), we have (-4)^-2 = 1 / (-4)^2 = 1 / 16. **
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NOTES -------> **Since a^-b = 1 / (a^b), we have (-4)^-2 = 1 / (-4)^2 = 1 / 16. **
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19:59:35 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 used the espression so that all exponents are positive. My answer is 1/1^4 confidence assessment: 2
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19:59:52 01-28-2007 19:59:52 ** (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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NOTES -------> ** (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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20:27:05 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 first put [5 x^-2 / (6 y^-2)]^-3. I then multiplied the negitive exponent of -2 by -3 and got 6. This made my equation look like this: 5x^6 / 6^-3(y^-2)^-3. Then I multiplied it on out to get 5x^6/1/216y^6. I then multiplied 216 by 5 and got 1080 and put it ofver y^6. My answer is 1080x^6/y^6. confidence assessment: 2
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20:27:22 01-28-2007 20:27:22 [ 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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NOTES -------> [ 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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21:35:42 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 first multiply the 3 by -2 and get -6. this makes my equation -8x^-6. I then divide 1 by 8x^6, so my answer is 1/-8x^6. confidence assessment: 2
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21:36:13 01-28-2007 21:36:13 ** 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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NOTES -------> ** 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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22:25:13 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 --> I multiplied X^-2/x*y/y2 and that equaled x^-2-1*y^1-2. I then got x^-3*1/y. Whenever an exponent is 0 or negative I assumed that the base is not 0. So I got a negative esponent for x so I put x^0 * 1/y and my answer is x^0/y. confidence assessment: 2
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22:25:40 01-28-2007 22:25:40 ** (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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NOTES -------> ** (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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22:36:16 R.2.122 (was R.4.72). Express 0.00421 in scientific notation.
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RESPONSE --> Ok I will use the problem and save it in my notes so I can study it and the solution to it. The scientific notation for 0.00421 is 4.21*10^-3 confidence assessment: 3
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22:38:15 R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation.
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RESPONSE --> The decimal notation for 9.7*10^3 is 9700.0 confidence assessment: 3
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22:38:30 ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **
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RESPONSE --> ok self critique assessment:
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22:50:16 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 --> 98.6 - 1.5 equals 97.1 which is less than 98.6 which is unhealthy because your temperature is not high enough. The equation would read 98.6 - x = unhealthy body temperature. For x you would replace it with 1.5 and your unhealthy body temperature would be 97. For a body temperature above 98.6 like in the equation 98.6 + 1.5 = 100 which is a high body temperature it would be unhealthy. Both body temperatures exceed 1.5. T = 97 T = 100 [T-98.6]>1.5 [T-98.6]>1.5 [97-98.6]>1.5 [100-98.6]>1.5 [-1.6]>1.5 [1.4]>1.5 untrue statement 1.6 > 1.5 confidence assessment: 3
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22:50:36 01-28-2007 22:50:36 ** 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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NOTES -------> ** 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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