course mth 158 It seems that along with this r2 asst my send file still contains r1 which I previously sent. I must be unclear on why it is there. Can I simply delete it from my paste? Please let me know how to make it right.
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21:21:24 query 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 --> (2x-3)/y (2*-2-3)/3 (-4-3)/3 -7/3 -2.333
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21:21:39 ** 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
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21:23:23 query 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 --> (4x)-(5y) (4*3)-(5*-2) 12--10 12+10 22
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21:26:18 ** 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 see that I failed to use the absolute value rule
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21:28:23 query 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 is the only one
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21:28:31 ** 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
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21:32:39 query R.2.76 \ 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 --> -.0625
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21:33:01 ** order of operations implies exponentiation before multiplication; the - in front of the 4 is not part of the 4 but is an implicit multiplication by -1. Thus only 4 is raised to the -2 power. -4^(-2) Since a^-b = 1 / (a^b), we have 4^-2 = 1 / (4)^2 = 1 / 16. The - in front then gives us -4^(-2) = - ( 1/ 16) = -1/16. If the intent was to take -4 to the -2 power the expression would have been written (-4)^(-2).**
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RESPONSE --> ok
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21:37:03 query 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 --> (3^-2*5^3)/(3^2*5) (.111*125)/(9*5) 13.889/45 .309
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21:37:20 ** (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
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22:09:54 query 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 5^-3*x^6/6^-3*y6 1/125x^6/1/216y^6 216y^6/125x^6
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22:11:31 [ 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 inverted the variables when they should have stayed the same.
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22:18:26 query 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 --> (-8x^3)^-2 1/64x^6
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22:21:19 ** 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 came up with the alterative answer but am unsure how to come up with it the other way(s) Am I right in the way I done it?
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22:28:51 query 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^2y/xy^2
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22:33:09 ** (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 do not understand I thought we had to have positive exponents. I can see the x^-3 but how y^-1 knowing x^-3 I would have assumed y^3
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22:46:58 query 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 --> 4x^-2(yz)^-1/(-5)^2x^4y^2z^-5 4x^-2y^-1z^-1/25x^4y^2z^-5 4z^6/25x^2y
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22:50:32 ** 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 believe I failed to group the x,y,z's correctly. I subtracted the x,y,z's out.
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22:51:22 query R.2.122 (was R.4.72). Express 0.00421 in scientific notation.
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RESPONSE --> 4.21*10^-3
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22:51:40 ** 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 --> ok
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22:51:56 query R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation.
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RESPONSE --> 9700
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22:52:02 ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **
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RESPONSE --> ok
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22:55:18 query R.2.152 \ 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 --> /x-98.6/>=1.5 /97-98.6/>=1.5 false = 1.6 /100-98.6/>=1.5 true = 1.4
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22:55:24 ** 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
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Љ{㗷 assignment #002 002. `query 2 College Algebra 09-03-2008