Assignment 2

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.

......!!!!!!!!...................................

10:44:08 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.

......!!!!!!!!...................................

RESPONSE --> (2(-2) - 3) / 3 = ((-4) - 3) / 3 = -7 / 3 First I replaced the x and y values with the appropriate numbers. Then I multiplied (2(-2)) to get (-4). Then I subtracted 3 from -4 to get -7. The most reduced fraction possible is (-7 / 3).

.................................................

......!!!!!!!!...................................

10:44:20 ** Starting with (2x-3)/y we substitute x=-2 and y=3 to get (2*(-2) - 3)/3 = (-4-3)/3= -7/3. **

......!!!!!!!!...................................

RESPONSE --> Ok.

.................................................

......!!!!!!!!...................................

10:49:44 query R.2. 55 (was R.2.45) Evaluate for x = 3 and y = -2: | |4x| - |5y| | and explan how you got your result.

......!!!!!!!!...................................

RESPONSE --> | |4(3)| - |5(-2)| | = | |12| - |10| | = |2| = 2 I substituted the x and y values with the appropriate number. I then multiplied (4(3)) and (5(-2)), to get (12) and (10). The absolute symbols invalidate the negative of (-2), thus (10) is positive. Then I subtracted (10) from (12) to get 2.

.................................................

......!!!!!!!!...................................

10:50:38 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)

......!!!!!!!!...................................

RESPONSE --> Ok.

.................................................

����������Ӏj|��b�t�z_����L�{�� assignment #002 ��f�zۣ��O������������ College Algebra 06-01-2006

......!!!!!!!!...................................

10:54:42 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.

......!!!!!!!!...................................

RESPONSE --> (2(-2) - 3) / 3 = (-4 - 3) / 3 = -7 / 3 I subsituted the appropriate values for x and y. Then I multiplied (2(-2) to get (-4), after which I subtracted (3) from it. This gave me (-7), and as (-7) cannot be divided by (3) into a integer, I left it as a fraction.

.................................................

......!!!!!!!!...................................

10:54:48 ** Starting with (2x-3)/y we substitute x=-2 and y=3 to get (2*(-2) - 3)/3 = (-4-3)/3= -7/3. **

......!!!!!!!!...................................

RESPONSE --> Ok.

.................................................

......!!!!!!!!...................................

10:57:24 query R.2. 55 (was R.2.45) Evaluate for x = 3 and y = -2: | |4x| - |5y| | and explan how you got your result.

......!!!!!!!!...................................

RESPONSE --> | |4(3)| - |5(-2)| | = | |12| - |10| | = |2| = 2 I substituted the appropriate values for x and y. Then i multiplied (4 * 3) to get (12) and (5 * 2) to get (10). The absolute symbols invalidate the negative in (-2), so it becomes (2), making (10) positive. Then I subtracted (10) from (12) to get (2).

.................................................

......!!!!!!!!...................................

10:57:42 ** Starting with | | 4x |- | 5y | | we substitute x=3 and y=-2 to get | | 4*3 | - | 5*-2 | | = | | 12 | - | -10 | | = | 12-10 | = | 2 | = 2. **

......!!!!!!!!...................................

RESPONSE --> Ok, no need to write so much, apparently.

Show your thinking, explain your reasoning, but for your own sake keep it reasonably brief. No problem for me either way--I read quickly.

.................................................

......!!!!!!!!...................................

11:01:20 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)

......!!!!!!!!...................................

RESPONSE --> {x | x is not equal to 0 or -1} A zero will make the denominator equal to zero (0^3 + 0 = 0), and that makes the fraction undefined. A (-1) has the same effect ((-1)^3 + -1) = 0).

.................................................

......!!!!!!!!...................................

11:04:01 ** 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. **

......!!!!!!!!...................................

RESPONSE --> Ok, so [(-1)^2 + 1) = 2], then (2 + (-1) = 1), therefore (-1) can be in the domain. I didn't realize it could be factored

.................................................

......!!!!!!!!...................................

11:08:10 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?

......!!!!!!!!...................................

RESPONSE --> The law of [(a^m)^n = a^mn] can be used to answer this question. (-4) is raised to the first power, and that product is raised by the 2cd power. (-4) * (-4) = 16.

.................................................

......!!!!!!!!...................................

11:09:50 **Since a^-b = 1 / (a^b), we have (-4)^-2 = 1 / (-4)^2 = 1 / 16. **

......!!!!!!!!...................................

RESPONSE --> The problem in the book is written as (-4)^2, unlike the program, which has it as (-4)^-2, which gives a different answer.

.................................................

......!!!!!!!!...................................

11:19:24 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?

......!!!!!!!!...................................

RESPONSE --> (3^-2 * 5^3) / (3^2 * 5) = [(1 / 3^2) * 5^3] / (3^2 * 5) = [(1 / 9) * 125] / (9 * 5) = [(125 / 9)] / (45) = 25 / 81 The first law used was [(a^-n) = (1 / a^n) if a is not 0]. This changed (3^-2) to (1 / 3^2). Then I evaluated all the exponents, so : (1 / 3^2) became (1 / 9), (5^3) became (125), (3^2) became (9). Then I followed the rest of the order of operations to arrive at (25 / 81).

.................................................

......!!!!!!!!...................................

11:22:15 ** (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. **

......!!!!!!!!...................................

RESPONSE --> That's a different way of doing what I did. In this method, you group the (3) and the (5) bases. The second law used, [a^b / a^c = a^(b-c)], was not in the book, but perhaps it was in the CD. I'll have to look that over when I get home. We still get the same answer, but my guess is that this is the proper way to do it.

.................................................

......!!!!!!!!...................................

11:36:38 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.

......!!!!!!!!...................................

RESPONSE --> most likely wrong, but . . . [(5x^-2) / (6y^-2]^-3 = [(1 / 5x^2) / (1 / 6y^2)]^-3 =

Right idea, but the 5 is not grouped with the x^-2 and the 6 is not grouped with the y^-2.

[(5x^-2) / (6y^-2]^-3 = [5 / x^2) / (6 / y^2)]^-3.

Due to the above, this should read

[(5^-3) / ((x^2)^-3)] / [(6^-3) / ((y^2)^-3)] = [(1 / 5^3) / (1 / x^6)] / [(1 / 6^3) / (1 / y^6)].

Division by a fraction is the same as multiplication by the reciprocal:

[(1 / 5^3) / (1 / x^6)] / [(1 / 6^3) / (1 / y^6)] =

[(1 / 5^3) * (x^6 / 1)] / [(1 / 6^3) * (y^6 / 1)] = (x^6 / 5^3) / (y^6 / 6^3) = (x^6 / 5^3) / (6^3 / y^6) = 6^3 x^6 / (5^3 y^6).

You could of course write 5^3 and 6^3 as 125 and 216, respectively.

.................................................

......!!!!!!!!...................................

11:39:25 [ 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.

......!!!!!!!!...................................

RESPONSE --> Yeah, I was wrong, and I don't understand any of the answer. Apparently, I used the right laws (or two of the right laws), just in the wrong order, and I will need help with this.

.................................................

......!!!!!!!!...................................

11:45:00 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.

......!!!!!!!!...................................

RESPONSE --> (-8x^3)^-2 = (-8^-2)(x^3)^-2 = (-8^-2)(x^-6) = (1 / (-8^2)x^6) = (1 / 64x^6) first law: [(ab)^n = a^n b^n] second law: [(a^m)^n = a^mn] third law: [a^-n = (1 / a^n)]

.................................................

......!!!!!!!!...................................

11:48:59 ** 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 ). **

......!!!!!!!!...................................

RESPONSE --> I must remember to put parenthesis around figures that need to be treated as a single term. I see where I made a mistake using the alternative solution, the closest to mine. I should have used the law of (a^-n) = (1 / a^n) first, and then do everything else, instead of using it last in this case. It's looking more and more like I need help with exponents.

.................................................

......!!!!!!!!...................................

11:58:05 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.

......!!!!!!!!...................................

RESPONSE --> (x^-2 y) / (x y^2) = [(x^-2) / (x)] * [(y) / (y^2)] = (x / x^2) * (y / y^2) =

The (x) should stay in the denominator. Just because you move the x^-2 to the denominator doesn't mean that the x needs to move to the numerator. The x^-2 and the x are independent factors; what happens to one doesn't affect the other.

That would give you

(1 / (x^2 ( x) ) * (y / y^2), which gives you

(1 / x^3 ) * (y / y^2),

etc.

.................................................

......!!!!!!!!...................................

11:59:11 ** (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).**

......!!!!!!!!...................................

RESPONSE --> I don't understand, I just don't understand.

Actually you do. You're getting 90% of your steps right, and you're applying the laws systematically and intelligently. Just need to watch out for a few pitfalls.

.................................................

......!!!!!!!!...................................

12:01:41 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.

......!!!!!!!!...................................

RESPONSE --> I don't know. I've never been that great with exponents, and this program doesn't make it an easier to understand them or what the problem is.

Hopefully my responses will help.

.................................................

......!!!!!!!!...................................

12:02:13 ** 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 ) **

......!!!!!!!!...................................

RESPONSE --> See previous response, and you'll know what I'll say.

.................................................

......!!!!!!!!...................................

12:02:56 query R.2.122 (was R.4.72). Express 0.00421 in scientific notation.

......!!!!!!!!...................................

RESPONSE --> 4.21 * 10^-3

.................................................

......!!!!!!!!...................................

12:03:04 ** 0.00421 in scientific notation is 4.21*10^-3. This is expressed on many calculators as 4.21 E-4. **

......!!!!!!!!...................................

RESPONSE --> Ok.

.................................................

......!!!!!!!!...................................

12:03:26 query R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation.

......!!!!!!!!...................................

RESPONSE --> 9700.0

.................................................

......!!!!!!!!...................................

12:03:36 ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **

......!!!!!!!!...................................

RESPONSE --> ok.

.................................................

......!!!!!!!!...................................

12:06:04 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?

......!!!!!!!!...................................

RESPONSE --> |97 - 98.6| > 1.5 = |1.6| > 1.5 = A temperature of 97 is unhealthly because it is more than 1.5 degrees away from the norm. |100 - 98.6| > 1.5 = |1.4| > 1.5 = A temperature of 100 is healthy because it is less than 1.5 degrees away from the norm.

.................................................

......!!!!!!!!...................................

12:06:13 ** 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. **

......!!!!!!!!...................................

RESPONSE --> Ok.

.................................................

"

Your work on this assignment is actually pretty good. See my notes and let me know if you have further questions or need further clarification.