Assignment 7

course MTH 158

I'm really having problems with this lesson, is there any extra resources to practice with?

鸙昜弘闖~邶裞驂臐㛡埼基瀡ssignment #007

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007. `query 7

College Algebra

09-19-2007

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17:58:21

R.7.10 (was R.7.6). Show how you reduced (x^2 + 4 x + 4) / (x^4 - 16) to lowest terms.

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RESPONSE -->

First factor the numerator:

x^2 + 2x + 2x +4

x(x+2) + 2(x+2)

(x+2)(x+2)

then the denominator:

(x^2-4)(x^2-4)

(x+2)(x-2)(x+2)(x-2)

cancel out common factors, leaving

1/(x-2)^2

confidence assessment: 2

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17:58:31

09-19-2007 17:58:31

** We factor the denominator to get first (x^2-4)(x^2+4), then (x-2)(x+2)(x^2+4). The numerator factors as (x+2)^2. So the fraction is

(x+2)(x+2)/[(x-2)(x+2)(x^2+4)], which reduces to

(x+2)/[(x-2)(x^2+4)]. **

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NOTES ------->

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18:03:00

R.7.28 (was R.7.24). Show how you simplified[ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ].

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RESPONSE -->

first factor:

x^2 - 4x + 4

x^2 - 2x -2x +4

x (x-2) -2 (x-2)

(x-2)(x-2)

Then flip the reciprocal:

[(x-2)/4x] * [12x/(x-2)(x-2)]

cancel out like terms:

3x/(x-2)

confidence assessment: 2

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18:03:21

** [ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ] =

(x-2) * / 4x * 12 x / (x^2 - 4x + 4) =

(x-2) * 12 x / [ 4x ( x^2 - 4x + 4) ] =

12 x (x-2) / [4x ( x-2) ( x-2) ] =

3/(x - 2) **

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RESPONSE -->

I forgot to cancel out one of the x's.....

self critique assessment: 3

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18:04:40

R.7.40 (was R.7.36). Show how you found and simplified the sum (2x - 5) / (3x + 2) + ( x + 4) / (3x + 2).

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RESPONSE -->

There is already a common denominator, so add the numerarator and simplify:

(3x-1)/(3x+2)

confidence assessment: 3

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18:04:51

** We have two like terms so we write

(2x-5)/(3x+2) + (x+4)/(3x+2) = [(2x-5)+(x+4)]/(3x+2). Simplifying the numerator we have

(3x-1)/(3x+2). **

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RESPONSE -->

self critique assessment: 3

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18:13:50

R.7.52 (was R.7.48). Show how you found and simplified the expression(x - 1) / x^3 + x / (x^2 + 1).

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RESPONSE -->

first to find LCD:

[(x-1)/x^3] * [(x^2+1)/(x^2+1)]

and

[x/(x^2+1)] * [x^3/x^3]

giving you:

[(x-1)(x^2+1)]/x^3(x^2+1)] + [x(x^3)/x^3(x^2+1)]

we now have a LCD, x^3(x^2 +1), add numerator, after factoring

(x^5 + 1x -1)/x^3(x^2+1)

confidence assessment: 1

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18:13:58

09-19-2007 18:13:58

** Starting with (x-1)/x^3 + x/(x^2+1) we multiply the first term by (x^2 + 1) / (x^2 + 1) and the second by x^3 / x^3 to get a common denominator:

[(x-1)/(x^3) * (x^2+1)/(x^2+1)]+[(x)/(x^2+1) * (x^3)/(x^3)], which simplifies to

(x-1)(x^2+1)/[ (x^3)(x^2+1)] + x^4/ [(x^3)(x^2+1)]. Since the denominator is common to both we combine numerators:

(x^3+x-x^2-1+x^4) / ) / [ (x^3)(x^2+1)] . We finally simplify to get

(x^4 +x^3 - x^2+x-1) / ) / [ (x^3)(x^2+1)] **

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NOTES ------->

Your procedure was good; you didn't show the details of your steps after getting [(x-1)(x^2+1)]/x^3(x^2+1)] + [x(x^3)/x^3(x^2+1)], but chances are any error would be easy to correct based on the given solution.

You didn't self-critique, however, so I can't tell exactly what went wrong.

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18:17:03

R.7.58 (was R.7.54). How did you find the LCM of x - 3, x^3 + 3x and x^3 - 9x, and what is your result?

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RESPONSE -->

i do not know

confidence assessment: 0

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18:17:06

09-19-2007 18:17:06

** x-3, x^3+3x and x^3-9x factor into

x-3, x(x^2+3) and x(x^2-9) then into

(x-3) , x(x^2+3) , x(x-3)(x+3).

The factors x-3, x, x^2 + 3 and x + 3 'cover' all the factors of the three polynomials, and all are needed to do so. The LCM is therefore:

x(x-3)(x+3)(x^2+3) **

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NOTES ------->

&#

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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18:21:46

R.7.64 (was R.7.60). Show how you found and simplified the difference3x / (x-1) - (x - 4) / (x^2 - 2x + 1).

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RESPONSE -->

first factor out the second denominator:

(x-1)(x-1)

then multiply the first by (x-1) to get a common denominator of (x-1)^2

combine like terms of the numerarator, leaving

[3x^2-4x-4]/(x-1)^2

confidence assessment: 2

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18:21:58

** Starting with 3x / (x-1) - (x-4) / (x^2 - 2x +1) we factor the denominator of the second term to obtain (x - 1)^2. To get a common denominator we multiply the first expression by (x-1) / (x-1) to get

3x(x-1)/(x-1)^2 - (x-4)/(x-1)^2, which gives us

(3x^2-3x-x-4) / (x-1)^2 = (3x^2 - 4x - 4) / (x-1)^2.

DRV**

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RESPONSE -->

self critique assessment: 3

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18:22:23

09-19-2007 18:22:23

QUESTION FROM STUDENT: On the practice test I'm having problems with problem #5 I don't know where to start or how to set it up. I'm probably missing something simple and will probably feel stupid by seeing the solution. Could you help with this problem.

A retailer is offering 35% off the purchase price of any pair of shoes during its annual charity sale. The sale price of the shoes pictured in the advertisement is $44.85. Find the original price of the shoes by solving the equation p-.35p = 44.85 for p.

INSTRUCTOR RESPONSE: It's very easy to get ahold of the wrong idea on a problem and then have trouble shaking it, or to just fail to look at it the right way. Nothing stupid about it, just human nature.

See if the following makes sense. If not let me know.

p - .35 p = 44.85. Since p - .35 p = 1 p - .35 p = (1 - .35) p = .65 p we have

.65 p = 44.85. Multiplying both sides by 1/.65 we get

p = 44.85 / .65 = etc. (you can do the division on your calculator); you'll get something near $67).

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NOTES ------->

................................................."

You could have used detailed self-critique on a couple of these problems, but I don't see any big overall problems.

Submit copies of the problems you didn't completely get and include detailed self-critiques. I'm pretty sure I can help straighten out the few details you might be missing.

&#

In your self-critique you need a phrase-by-phrase analysis of the given solution, detailing everything you do and do not understand.
Deconstruct the given solution and explain in detail what you do and do not understand about every part.

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Assignment 7

Your procedure was good; you didn't show the details of your steps after getting [(x-1)(x^2+1)]/x^3(x^2+1)] + [x(x^3)/x^3(x^2+1)], but chances are any error would be easy to correct based on the given solution. You didn't self-critique, however, so I can't tell exactly what went wrong.

&#

You could have used detailed self-critique on a couple of these problems, but I don't see any big overall problems. Submit copies of the problems you didn't completely get and include detailed self-critiques. I'm pretty sure I can help straighten out the few details you might be missing.

&#

In your self-critique you need a phrase-by-phrase analysis of the given solution, detailing everything you do and do not understand. Deconstruct the given solution and explain in detail what you do and do not understand about every part.

Your procedure was good; you didn't show the details of your steps after getting [(x-1)(x^2+1)]/x^3(x^2+1)] + [x(x^3)/x^3(x^2+1)], but chances are any error would be easy to correct based on the given solution.

You didn't self-critique, however, so I can't tell exactly what went wrong.

You could have used detailed self-critique on a couple of these problems, but I don't see any big overall problems.

Submit copies of the problems you didn't completely get and include detailed self-critiques. I'm pretty sure I can help straighten out the few details you might be missing.

In your self-critique you need a phrase-by-phrase analysis of the given solution, detailing everything you do and do not understand.
Deconstruct the given solution and explain in detail what you do and do not understand about every part.