Assignment R2

course MTH 158

C餈ܭassignment #002

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002. `query 2

College Algebra

06-09-2008

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17:56:14

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 -->

Replace x and y with their numbers...

2(-2) -3 / 3 = -4 -3 / 3 = -7 / 3

confidence assessment: 3

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17:56:27

** 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 -->

understood

self critique assessment: 3

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18:01:20

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 -->

First, evaluate x for 3 and y for -2

| |4(3)| - |5(-2)| | = | |12| - |-10| |

change |-10| to the actual value 10 and take it out of the lines.

| 12 - 10 |

And subtract

| 2 | = 2

confidence assessment: 3

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18:01:31

** 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 -->

understood

self critique assessment: 3

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18:02:14

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 = 0

confidence assessment: 3

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18:02:35

** 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 -->

self critique assessment:

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18:03:39

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 -->

Since it has the negative exponent, it becomes

1 / (-4)^2.

Now multiply -4 times itself, which makes 16.

confidence assessment: 3

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18:03:57

**Since a^-b = 1 / (a^b), we have

(-4)^-2 = 1 / (-4)^2 = 1 / 16. **

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RESPONSE -->

understood

self critique assessment: 2

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18:15:48

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 -->

First solve positive exponents...

3^-2 * 5^3 / 3^2 * 5 = 3^-2 * 125 / 9 * 5

Multiply denominator....

3^-2 * 125 / 45

Now split them up and fix negative exponent...

3^-2 / 1 * 125/45 = 1/9 * 25/9 = 25/81

confidence assessment: 3

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18:15:56

** (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 -->

self critique assessment: 3

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18:42:04

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 -->

First rearrange...

[5x^-2]^-3 / [6y^-2]^-3 since [a/b]^2 = a^2 / b^2

then rearrange some more...

5^-3 (x^-2)^-3 / 6^-3 (y^-2)^-3 since (ab)^2 = a^2 * b^2

multiply negative exponents on xs and ys...

5^-3 * x^6 / 6^-3 * y^6

And flip parts of fraction still with negative exponents...

6^3 * x^6 / 5^3 * y^6 = 216x^6 / 125y^6

confidence assessment: 2

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18:42:23

[ 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 -->

understood

self critique assessment: 2

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18:48:10

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 -->

First rearrange...

-8^-2 (x^3)^-2 since (ab)^2 = a^2 * b^2

now solve the x

-8^-2 x^-6

and turn it into a fraction

1 / -8^2 * x^6 = 1 / 64x^6

confidence assessment: 2

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18:49:03

** 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 -->

Wasn't that my answer?

self critique assessment: 1

That's what you meant, but you didn't group your denominator.

1 / 64x^6 means divide 1 by 64, then multiply the result by x^6. It's clear what you intended; be careful about the grouping.

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18:55:28

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 -->

First rearrange...

x^-2 / x * y / y^2 since ab / ba = a/b * b/a

now flip the fraction with the negative exponent

x / x^2 * y / y^2

and simplify

1/xy

confidence assessment: 2

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18:56:15

** (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 -->

not understood

self critique assessment: 1

&#

Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions (to which I will respond).

&#

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18:59:52

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 -->

no idea

confidence assessment: 0

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19:05:07

** 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 -->

understood

self critique assessment: 2

&#

You did not respond to the question, and should therefore have responded with a very detailed self-critique to the given solution, addressing what you do and do not understand about every of the solution. You should analyze the given solution in this manner, phrase by phrase.

&#

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19:05:57

R.2.122 (was R.4.72). Express 0.00421 in scientific notation.

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RESPONSE -->

4.21 * 10^-3

confidence assessment: 2

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19:06:10

** 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 -->

understood

self critique assessment: 3

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19:06:31

R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation.

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RESPONSE -->

9700

confidence assessment: 3

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19:06:40

** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **

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RESPONSE -->

understood

self critique assessment: 3

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19:11:07

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 -->

To show that T = 97 is unhealthy:

| T - 98.6 | > 1.5

To show that T = 100 is unhealthy:

| T - 98.6 | > 1.3

confidence assessment: 2

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19:12:01

** 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 -->

understood

self critique assessment: 2

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Good work on most questions. See my notes on self-critique. You are welcome to make a copy of this document, insert specific questions and/or self-critiques, mark your insertions with &&&& and submit it.