R7 Homework

course mth158

9-22-09 @10:00pm

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

007. `* 7

* 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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Your solution:

This is equation for number 10 in the book.

(x^2 + 4x + 4) / (x^2 – 4)

(x+2) (x+2) / (x-2) (x+2) The x+2 cancel each other out

= (x+2) / (x-2)

confidence rating: 3

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Given Solution:

* * ** 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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Self-critique (if necessary):

Im not sure if my answer is correct or not because I did number 10 in the book and it was not the same as question 10 on this query.

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Self-critique Rating:

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Question:

* 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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Your solution:

= (x – 2) / 2x(2) * 2x (6) / (x-2) (x-2)

= 3 / (x-2)

confidence rating: 3

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Given Solution:

[ ( 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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Self-critique (if necessary): ok

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Self-critique Rating: ok

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Question:

* 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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Your solution:

= (3x – 1) / (3x + 2)

confidence rating: 3

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Given Solution:

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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Self-critique (if necessary):

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Self-critique Rating: ok

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Question:

* 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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Your solution:

= (x – 1) / (x^3) * (x^2 + 1) / (x^2 + 1) + (x)/ (x^2 + 1) * (x^3) / (x^3)

= (x^3 – x^2 + x – 1) / (x^3)(x^2 + 1) + (x^4) / (x^3)(x^2 + 1)

= (x^4 + x^3 – x^2 + x – 1) / (x^3)(x^2 + 1)

confidence rating: 3

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Given Solution:

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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Self-critique (if necessary):

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Self-critique Rating: ok

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Question:

* 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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Your solution:

In my book #58 is x-3, x^2+3x, x^3-9x

= x(x-3)(x+3)

confidence rating: 0

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Given Solution:

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:

• LCM = x(x-3)(x+3)(x^2+3)

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Self-critique (if necessary):

Since the question in this query is different than the question in the book I don’t know if my answer is correct.

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Self-critique Rating:

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Question:

* R.7.64 (was R.7.60). Show how you found and simplified the difference 3x / (x-1) - (x - 4) / (x^2 - 2x + 1).

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Your solution:

= (3x) / (x – 1) - (x – 4) / (x – 1)(x – 1)

= (3x) / (x – 1) * (x – 1) / (x – 1) - (x – 4) / (x – 1)(x – 1)

= [3x(x – 1) – x – 4] / [(x – 1)^2]

= (3x^2 – 3x – x – 4) / [(x – 1)^2]

= (3x^2 – 4x – 4) / [(x – 1)^2]

confidence rating: 3

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Given Solution:

* * ** Starting with

3x / (x-1) - (x-4) / (x^2 - 2x +1)

we factor the denominator of the second term to obtain (x - 1)^2. Since the first denominator (x - 1) is already a factor of the second, our common denominator is (x - 1)^2.

To express the given expression in terms of the common denominator we then 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.

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Self-critique (if necessary):

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Self-critique Rating: ok

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 grab onto 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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&#Your work looks very good. Let me know if you have any questions. &#