21

course Mth 158

xRw{cwassignment #021

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v慬nؤ˺Zۋ

College Algebra

07-18-2006

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22:26:43

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

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RESPONSE -->

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

Inverse variation of y= k / x or.. y = k / sqrt(x)

if y ='s 4 when x='s 9 then substituting into the equation you get..

4 = k / sqrt(9)

sqrt (9) is 3 so that..

4 = k / 3

k = 4 * 3 = 12

y = 12 / sqrt(x)

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22:27:17

** 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

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22:28:03

** 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 -->

ok

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22:33:21

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

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RESPONSE -->

Varies directly.. y=kx

T varies directly with the square root of its length l (in feet) so that..

T = k sqrt(L)

where k = 2 pi / sqrt(32).

putting it into the formula we have..

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

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22:34:01

** 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

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22:42:47

query 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 -->

Resistance dir with lgth inversely with sq of diam...

R = k * L / D^2

1.24 = k * 432 / 4^2

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

R = .046 * L / D^2

R = 1.44 d = 3

looking for L..

L = R * D^2 / (.046)

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

making the wire around 280 feet

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22:43:06

** 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

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Good work. Let me know if you have questions.