course Mth 158 So I am a lttle confused on my evaluating the exponents. I will go back over the text and now work problems. ?????R???????assignment #002
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19:42:23 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 --> x = - 2 and y = 3 (2x - 3)/y= replace the x with - 2 ans the y with 3, then evaluate (2*-2-3)/3= First multiply 2 by -2 which gives you - 4 (-4-3)/3= Then subtract -4 from 3 which gives you - 7 -7/3= - (7/3) confidence assessment: 3
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19:47:56 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 --> substitute x= 3 and y= - 2 |4x| - |5y| = |4(3)| - |5(-2)| = |12| - |-10| = 12-10 = 2 confidence assessment: 3
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19:48:11 ** 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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19:49:54 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 --> {x|x (can not = 0)} confidence assessment: 3
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19:52:46 ** 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: 2
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20:03:11 ** (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 --> The Laws of Exponents a^ - b = 1/a^b which gets rid of the negative exponent (-4)^ - 2 1/(-4)^2 1/16 (3^(-2)*5^3)/(3^2*5) Group factors with like bases a^b/a^c=a^(b-c) 3^(-2-2) * 5(3-1) 3^-4 * 5^2 a^ - b = 1/a^b which gets rid of the (1/3^4) * 5^2 negative exponent (1/81) * 25 25/81 self critique assessment: 2
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20:23:06 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/ (6y^ - 2)]^ - 3 By the laws of exponents first distribute the ^ - 3 through out the expression in the brackets as so [(5)^ - 3 * (x^ - 2)^ - 3 / (6^ - 3 * (y^ - 2)^ - 3)] = with using the law (a^m)^n=a^mn [5^ - 3 * x^(-2 * - 3)/ (6^ - 3 * y^(- 2* - 3))] [5^ - 3 * x^6 / (6^ - 3 * y^6)] now to get rid of the negative, use the recipical 6^3 * x^6 / 5^3 * y^6 216x^6/125y^6 confidence assessment: 2
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20:30:00 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 --> (- 8 x^3) ^ - 2 use the distribution of exponent -2 throughout the epression within the parenthesis as so: (-8^ - 2) * (x^3)^ - 2 then get rid of the negative exponent by recipical as so: (1/-8^2) * x^ - 6 (1/-8^2) * (1/x^6) and simplify (1/-64) * (1/x^6) 1/-64x^6 = -(1/64x^6) confidence assessment: 2
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20:31:24 ** 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 --> self critique assessment:
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20:31:44 ** 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: 2
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20:36:01 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) (y/x^2) / (x y^2) x^2/xy2 confidence assessment: 2
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20:51:27 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 --> OK how is x^ - 1 come about in the expression = x^-2 * y * x^-1 * y^-2 anyways... 4x^ - 2 (y z )^ - 1/ [(-5)^2 x^4 y^2 z^-5] 4x^ - 2 (y^ -1 * z^ - 1) / (-10)x^4 y^2 z^ - 5] 4z^5/ (y^1 * z^1)(-10x^4 y^2) 4z^(5-1) / -10 x^4 y^(2-1) 4z^4 / -10x^4y confidence assessment: 1
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20:54:51 ** 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 --> oooops! 5^2 is 25 I used 10 ok so I am a little confused. self critique assessment: 2
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20:55:40 ** 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 --> 4.21 * 10^ - 3 self critique assessment: 3
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20:56:09 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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20:59:48 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 --> | T - 98.6 | >1.5 substitue T for 97 and 100 as so: | 97 - 98.6 | =| -1.6 |= 1.6>1.5 |100 - 98.6 | =| 1.4 |= 1.4<1.5 confidence assessment: 3
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