course MTH 158

assignment #021

021. `query 21

College Algebra

07-07-2008

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assignment #021

021. `query 21

College Algebra

07-07-2008

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10:55:29

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

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RESPONSE -->

So y = k / sqrt(x)

4 = k / sqrt(9)

4 = k/3

k = 12

y = 12 / sqrt(x)

confidence assessment: 2

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10:55:43

** 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 -->

ok

self critique assessment: 3

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11:02:05

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 -->

z = k x^3 y^2

1 = k 2^3 * 3^2

1 = 72 k

k = -71

z = -71 x^3 y^2

confidence assessment: 2

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11:04:01

** 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 -->

I see that I wrongly multiplied x^3 and y^2 instead of adding, as well as subtracting the number by k instead of dividing.

self critique assessment: 2

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11:08:30

2.7.20 (was 2.6.20). Period varies directly with sqrt(length), const 2 pi / sqrt(32)

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RESPONSE -->

T = 2 pi / sqrt(32) * sqrt (1)

I don't know if I need to simplify anything additionally..

confidence assessment: 2

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11:09:46

** The equation is

T = k sqrt(L), with k = 2 pi / sqrt(32). So we have

T = 2 pi / sqrt(32) * sqrt(L). **

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RESPONSE -->

I meant to type sqrt(L) instead of sqrt(1)... when I wrote the problem down I put a lower-case L instead of an upper case L, and it looked like I'd written a 1...

self critique assessment: 3

your meaning was clear; no problem.

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11:10:55

**** What equation relates period and length? ****

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RESPONSE -->

Probably combined variation

confidence assessment: 2

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11:34:53

2.7.34 (was 2.6.30). Resistance dir with lgth inversely with sq of diam. 432 ft, 4 mm diam has res 1.24 ohms. **** What is the length of a wire with resistance 1.44 ohms and diameter 3 mm? Give the details of your solution.

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RESPONSE -->

r = kL/d^2

1.24 = k*432 / 4^2

1.24 = 432k / 16

1.24 = 27k

k = 1.24/27

The formula is r = .046 L / d^2

For the length of a wire with resistance 1.44 ohms and a diameter of 3mm, we substitute...

1.44 = .046 L / 3^2

1.44 = .046 L / 9

1.44 = .05 L

.046/9 is more like .005

L = 28.8

Since that isn't nearly near the length in the previous problem, I'm guessing that you would add that to the previous length?? So

r = k 450.8 / d^2

I really don't know...

confidence assessment: 1

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11:35:44

** We have

R = k * L / D^2. Substituting we obtain

1.24 = k * 432 / 4^2 so that

k = 1.24 * 4^2 / 432 = .046 approx.

Thus

R = .046 * L / D^2.

Now if R = 1.44 and d = 3 we find L as follows:

First solve the equation for L to get

L = R * D^2 / (.046). Then substitute to get

L = 1.44 * 3^2 / .046 = 280 approx.

The wire should be about 280 ft long. **

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RESPONSE -->

Ok.. why and how did you change 2.8 to 280?

self critique assessment: 2

Neither 2.8 nor 280 occurred until the last step, so 2.8 was not changed to 280.

280 is simply the solution to the equation.

Your equation would also have yielded 280. See the note I inserted into your solution.

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This looks good. See my notes. Let me know if you have any questions. &#