course I have looked through the orientation again and cannot find my posted assignments. Any suggestions? Thanks for your help.Obviously, I found the assignment in the send file and I hope I have sent the correct problems. 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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17:15:13 ** 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 --> ok self critique assessment: 3
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17:17:16 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 --> ( 4 *3 ) - (5 *2 ) = 2 confidence assessment: 3
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17:17:31 ** 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 --> ok self critique assessment: 3
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17:18:20 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 --> zero is excluded from the domain. confidence assessment: 3
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17:18:49 ** 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 --> ok self critique assessment: 3
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17:21:37 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 --> 1/ -4 * -4 = 1/ 16 A^ -n = 1/a^n confidence assessment: 3
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17:21:50 **Since a^-b = 1 / (a^b), we have (-4)^-2 = 1 / (-4)^2 = 1 / 16. **
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RESPONSE --> ok self critique assessment: 3
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17:26:32 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 --> 25/81 I subtracted the exponents and multiplied the resulting values. confidence assessment: 2
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17:26:49 ** (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 --> ok self critique assessment: 2
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17:50:18 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 --> (5x^-2)^ -3 / (6y^ -2)^ -3 =( 5^ -3) / ( 6^ -3) *(y^ -3) = (x^6/ 5^3)/ [ (y^6)/ (6^3) ] = ( x^6/ 5^3) *[(6^3)/y^6) =6^3x^6 / 5^3 y^6 confidence assessment: 2
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17:50:41 [ 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 --> ok self critique assessment: 3
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17:54:39 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/ (-8x^3)^2 = 1/ 64x^6 confidence assessment: 2
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17:55:45 ** 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 --> ok self critique assessment: 3
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18:00:11 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 -1) * y^(1-2) = x^ -3 * y^ -1=1/x^3y confidence assessment: 2
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18:00:28 ** (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 --> ok self critique assessment: 3
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18:41:21 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 --> [ (4 * 1/x^2)] * [ 1 / (yz)^ -1] / [ ( 25x^4) / y^5] = [4/ (x^y^2)]* [y^5 / ( 25x^4)]= [4 *y^5] /[25x^6y^2] confidence assessment: 2
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18:45:02 ** 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 --> ok self critique assessment: 2
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18:46:00 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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18:46:14 ** 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 --> ok self critique assessment: 3
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18:46:45 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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18:47:11 ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **
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RESPONSE --> ok self critique assessment: 3
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18:51:17 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 --> 97- 98.6= 1.6 1.6 is greater than 1.5 thus you are healthy. 100-98.6=1.4, which inplies you are not healthy. confidence assessment: 3
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18:51:31 ** 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 --> ok self critique assessment: 3
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