query r2

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.

The program writes everything done on a given day to the same SEND file. You should simply delete in the manner you describe. Be sure to delete it after pasting it into the form so you don't accidentally change in the SEND file itself. That way if you happen to make an error and delete the wrong thing, you still have the original file and can easily correct the error.

I went ahead and deleted the first file this time. Thanks for heads-up.

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assignment #002

002. `query 2

College Algebra

09-03-2008

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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?

You should have grouped your denominator. You didn't show the steps you followed in your solution, so I can tell how you arrived at your answer. However except for the grouping your answer is correct, so I suspect you know how to simplify this expression.

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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

y / y^2 = 1 / y, which is the same as y^(-1).

Alternatively, y * y^(-2) = y^1 * y^(-2) = y^(1 - 2) = y^(-1), or 1 / y.

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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.

You also left out some important signs of grouping, which appears to have caused some errors.

For example

4x^-2y^-1z^-1/25x^4y^2z^-5 = 4 x^2 y z^(-6) / 25, consistent with your answer, but the expression is

4x^-2y^-1z^-1/ (25x^4y^2z^-5), which gives a very different result.

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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

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