Assignment 21

course mth 158

???H?????????v?assignment #021021. `query 21

College Algebra

07-23-2008

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01:46:52

**** query 2.5.8 / 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

4=k/9

36=k

y=36/4

y=9

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01:47:10

** 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 -->

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01:49:41

query 2.5.12 / 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(8+9)

1=17k

k=1/17

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01:49:52

** 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 -->

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01:55:36

query 2.5.20 / 2.7.20 (was 2.6.20). Period varies directly with sqrt(length), const 2 pi / sqrt(32)

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RESPONSE -->

P=k(sqrt L)

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01:56:53

** 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 -->

ok I didn't know that you were saying that k=2 pi/ sqrt(32)

I see

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

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01:57:37

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

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RESPONSE -->

p=kl

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02:02:56

query 2.5.42 / 2.7.42 (was 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 -->

1.24=k 432/16

1.24=27k

.046=kor 1.24/27=k

1.44=.046 L/3^2

12.96=.046L

L=281.74

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02:03:07

** 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 -->

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I think you understand the process. If there's anything you don't understand, you're welcome to submit a copy with self-critiques inserted and marked &&&&.