Assign 21

course Mth158

I can't remember if I have submitted this assignment or not.

f̐Ēҙصassignment #021

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.

021. `query 21

College Algebra

03-17-2007

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15:00:41

2.7.8 (was 2.6.6). y inv with sqrt(x), y = 4 when x = 9.

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RESPONSE -->

y=k/x

y=4 x=9

y=k/sqrt(x)

4=k/sqrt(9)

4=k/3

12=k

y=12/x

confidence assessment: 3

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15:00:53

** The inverse proportionality to the square root gives us y = k / sqrt(x).

y = 4 when x = 9 gives us

4 = k / sqrt(9) or

4 = k / 3 so that

k = 4 * 3 = 12.

The equation is therefore

y = 12 / sqrt(x). **

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RESPONSE -->

whooah!

self critique assessment: 2

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15:04:01

2.7.12 (was 2.6.10). z directly with sum of cube of x and square of y; z=1 and x=2 and y=3.

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RESPONSE -->

y=kx

z=k(x^3+y^2)

1=k(2^3+3^2)

1=k(8+9)

1=17k

1/17=17k/17

1/17=k

z=1/17(x^3+y^2)

confidence assessment: 3

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15:04:14

** The proportionality is

z = k (x^3 + y^2).

If x = 2, y = 3 and z = 1 we have

1 = k ( 2^3 + 3^2) or

17 k = 1 so that

k = 1/17.

The proportionality is therefore

z = (x^3 + y^2) / 17. **

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RESPONSE -->

good good!

self critique assessment: 2

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I think I've seen this, but I don't think it's been posted.

In any case, great work.