section R2

Good. You have a few errors but your procedures are sound. See my notes, in boldface like this one, and be sure you see the notes about grouping your expressions correctly.

If anything is not clear let me know, and include as many specifics as possible.

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20:19:58 ** 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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20:24:10 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 --> I 4(3)I - I 5(-2) I = I12I - I-10I = 12-10=2 Not sure how to type in absolute value signs but you get the idea.

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20:24:25 ** 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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20:27:47 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 --> (-9x^2 -x +1) / (x^3 +x) = -9x^2 - x -1 / x(x^2+1) denom factors to x(x+1)(x+1) -9x^2 -x +1 / x(x+1)(x+1) x=0 and x=-1 cannot be in domain

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20:30:42 ** 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 --> Why isn't x=-1 included? Isn't x^2 + 1 a perfect square?

The given solution is incorrect, but x^2 + 1 is not the perfect square of any real-valued binomial. (x+1)^2 would be a perfect square, but (x + 1)^2 = x^2 + 2x + 1, not x^2 + 1.

The correct solution is as follows:

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.

It appears that you are using an older version of the query program. The version currently posted on the Sup Study ... pages contains this solution as a correction to the previous erroneous solution.

Better go to the Sup Study > Course Documents > Downloads > College Algebra and download the newer version. The full name of the program is query_ca_qa_format_thru_30_for_new_edition_0705.

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20:31:59 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 = 16 -4 * -4 = 16

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20:33:35 **Since a^-b = 1 / (a^b), we have (-4)^-2 = 1 / (-4)^2 = 1 / 16. **

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RESPONSE --> The exponent in that problem isn't negative.

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20:38:28 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 * 5^2 / 3^2 *5 = 3^-2/3^2 * 5^2/5 = 3^-2-2 * 5^2-1 3^-4 * 5^2 = a^-n = 1/a^-n 5^2 = 25 and 3^4 = 81 25/81

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20:39:27 ** (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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20:51:35 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 --> using the laws (a^m)^n = (a)^m*n and a^-n= 1/a^n (5x^-3)^-3 / (6y^-2)^-3 = 125x^6 / 6^-3y^6 = 125x^6 / 1/216y^6 = 216y^6/125x^6

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20:56:24 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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. So I need to flip the 6 into a postive in the numerator. Is that correct?

[ 5 x^-2 / (6 y^-2) ] ^ -3 = 5^-3 (x^-2)^-3 / ( 6^(-3) (y^-2)^(-3) ) = (1/125) x^6 / ( 1/216 * y^6) = (216/125) * x^6 / y^6, or if you prefer (216 x^6) / (125 y^6).

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21:03:37 ** 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 was responding to the previous question. I hadn't even gotten to this problem yet. Apparently I clicked too soon. I am trying to backtrack to the question posed here.

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21:06:26 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^-2y / xy^2 = x^(-2-1) / x * y^(1-2) / y = x^-3 * y^-1 using a^-n=1/a^n = 1/x^3y

The answer should be expressed as 1 / ( x^3 y).

By order of operations, 1/x^3y means (1 / x^3) * y, which is not what you intended. I could see what you meant, though, and on paper I believe your solution would have been correct.

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21:07:37 ** (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 --> yes

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21:17:39 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/x^4 * y^-1/y^2 * z^-1/z^5 * (-5)^2 subtracting exponents first you have: 4x(-2-4) * y(-1-2) *z (-1--5) * (-5)^2 = 4x^-6 * y^-3 * z^4 * -5^2 = 1/4x^6 * 1/y^3 * 25z^4 = 25z^4/4x^6y^3

The 4 never gets into the denominator. It isn't linked with the x in any way.

And the 25 never gets into the numerator.

Again I can tell what you mean, but be careful about grouping numerators and especially denominators. Except for the 4 and the 25 in the wrong places, you did very well.

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21:22:46 ** 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 --> In my response 25 was num and 4 the denom but our variables and powers are the same. Are both answers correct?

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21:23:14 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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21:24:17 ** 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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21:24:33 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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21:24:40 ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **

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RESPONSE --> ok

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21:28:53 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 --> for t=97 | 97-98.6| greater than or equal to 1.5 |-1.6| = 1.6 is greater than 1.5 for t=100 |100-98.6| greater than 1.5 |1.4} = 1.4 is less than 1.5 not an unhealthy temp.

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21:29:10 ** 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 --> right

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