assignment 2

course

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

17:12:01 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 --> (2*-2 - 3)/3= multiply 2 and -2 first (-4-3)/3 =subtract -4 and -3 -7/3

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17:12:08 ** 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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17:17:33 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 --> | |4*3| - |5*-2| | = | |12| - |-10|| = |12 - 10 | = 2

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17:17:42 ** 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

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17:22:55 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 --> X cannot equal 0. since anything multiplied or divided with 0 always equals 0.

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17:23:17 ** 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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17:26:40 query 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 = 1/16 since a ^ -n = 1/a^n

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17:26:47 **Since a^-b = 1 / (a^b), we have (-4)^-2 = 1 / (-4)^2 = 1 / 16. **

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RESPONSE --> ok

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17:36:32 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) = 3^ (-2-2) * 5^ (3-1) = 3^ -4 * 5^ 2= 1/3^4 *25 = 1/81 * 25 = 25/81

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17:37:00 ** (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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17:49:03 [ 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 --> [ 5 x^-2 / (6 y^-2) ] ^ -3= 5^-3x^6 / 6^-3 y^6 = 1/5^3*x^6 / 1/6^3*y^6 = 6^3x^6 / 5^3y^6

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17:53:52 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 --> (-8 x^3) ^ -2 = -8^-2 * x^-6 = 1/-8^2 * 1/x^6 = 1/64 * 1/x^6 = 1/64x^6

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17:54:21 ** 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

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17:57:02 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 --> (x^-2 y) / (x y^2) = x^(-2-1) * y^(1-2) = x^-3 * y^-1 = 1/x^3y^1

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17:57:14 ** (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

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19:13:19 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 --> 4 x^-2 (y z)^-1 / [ (-5)^2 x^4 y^2 z^-5 ]= 4x^-2 y^-1 z^-1 / 25 x^4 y^2 Z^-5 = 4/25 * x^(-2-4) * y^(-1-2) * z^(-1+5) = 4/25 * x^-6 * y^-3 * z^4 = 4/25 * 1/x^6 * 1/y^3 * z^4= 4z^4/25x^6y^3

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19:13:24 ** 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

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19:14:10 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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19:14:16 ** 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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19:15:02 query R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation.

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RESPONSE --> 9.7 * 10^3 = 9700 move decimal to right 3 places

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19:15:06 ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **

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RESPONSE --> ok

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19:17:48 query 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 | 97 - 98.6 | > 1.5 1.6>1.5 (Unhealthy) | T - 98.6 | > 1.5 | 100 - 98.6 | > 1.5 1.4>1.5 (Healthy)

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19:17:56 ** 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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You did very well on these questions and seem to understand everything thoroughly. Let me know if you have questions.