Assignment 29

course Mth 151

ېyyd률assignment #029

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Liberal Arts Mathematics I

07-27-2006

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13:28:00

7.3.18 (1/3) / 6 = 1/18. Is this ratio equation valid or not and how did you determine your answer?

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RESPONSE -->

(1/3) / 6 = 1/18

1/3(18) = 6(1)

6 = 6

valid

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13:28:19

**If we multiply both sides by 6 * 18 we get

6 * 18 * (1/3 ) / 6 = 6 * 18 * (1 / 18) or

18 * 1/3 = 6. Note that the effect here is the same as that of 'cross-multiplying', but it's a good idea to remember that 'cross-multiplying' is really a shortcut way to think of multiplying both sides by the common denominator.

Since 18 * 1/3 = 18 / 3 = 6, the equation 18 * 1/3 = 6 is true, which verifies the original equality. **

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RESPONSE -->

ok

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13:30:35

7.3.20 z/8 = 49/56. Solve this proportionality for z.

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RESPONSE -->

z/8 = 49/56

49(8) = 56z

392 = 56z

z = 7

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13:30:39

**Multiply both sides by 8 * 56 to get

8 * 56 * z / 8 = 8 * 56 * 49 / 56. Simplify to get

56 * z = 8 * 49. Divide both sides by 56 to get

z = 8 * 49 / 56. Simplify to get

z = 7. **

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RESPONSE -->

ok

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13:31:58

7.3.42 8 oz .45; 16 oz. .49; 50 oz. 1.59`sb Which is the best value per unit for green beans and how did you obtain your result?

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RESPONSE -->

.45/8 = .056

.49/16 = .0306

1.59/50 = .0318

The 16 oz is the cheapest.

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13:33:17

** 45 cents / 8 oz = 5.63 cents / oz.

49 cents / 16 oz = 3.06 cents / oz.

159 cents / 50 oz = 3.18 cents / oz.

16 oz for .49 is the best value at 3.06 cents / oz. **

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RESPONSE -->

ok

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13:35:20

7.3.45 triangles 4/3, 2, x; 4, 6, 3. What is the value of x and how did you use an equation to find it?

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RESPONSE -->

I compared the two triangles in the following way:

(4/3)/4

x/3

2/6

I reduced the 2/6 to 1/3 and then thought that x/3 is the same as 1/3, so x = 1.

I don't think this way is quite right, but the triangle part confused me just a little.

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13:35:53

** the 4/3 corresponds to 4, 2 corresponds to 6, and x corresponds to 3.

The ratios of corresponding sides are all equal.

So 4/3 / 4 = 2 / 6 = x / 3.

Just using x / 3 = 2 / 6 we solve to get x = 1.

We would have obtained the same thing if we had used x / 3 = 4/3 / 4. **

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RESPONSE -->

Oh. I guess I was right.

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13:39:08

If z = 9 when x = 2/3 and z varies inversely as x, find z when x = 5/4. Show how you set up and used an equation of variation to solve this problem.

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RESPONSE -->

y = k/x^2

9 = k/(2/3)^2

9 = k/(4/9)

multiply both sides by 4/9

k = 4

y = 4/(5/4)^2

y = 4/(25/16)

multiply both sides by 25/16

y = 2.56 or 64/25

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13:41:33

** If z varies inversely as x then z = k / x.

Then we have

9 = k / ( 2/3). Multiplying both sides by 2/3 we get

2/3 * 9 = k so

k = 6.

Thus z = 6 / x. So when x = 5/4 we have

z = 6 / (5 /4 ) = 24 / 5 = 4.8. Note that the translations of other types of proportionality encountered in this chapter include:

z = k x^2: z varies as square of x.

z = k / x^2: z varies inversely as square of x.

z = k x: z is proportional to x. **

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RESPONSE -->

Don't you square the first x (2/3)?

In the problem as solved here, z varies inversely with x, and the given solution is correct.

If the problem has z varying inversely with x^2, you would then square the x.

It's possible that my problem statement here doesn't match that of the text. Either way, I think you'll understand.

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13:48:36

7.3.72. Illumination is inversely proportional to the square of the distance from the source. Illumination at 4 ft is 75 foot-candles. What is illumination at 9 feet?

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RESPONSE -->

I couldn't seem to figure out the problem using the inverse variations (y = k/x or y = k/x^2) so I tried working the problem using a proportion equation.

4/75 = 9/x

4x = 9(75)

4x = 675

x = 168.75

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13:53:39

**Set up the variation equation I = k / r^2, where I stands for illumination and r for distance (you might have used different letters). This represents the inverse proportionality of illumination with the square of distance.

Use I = 75 when r = 4 to get

75 = k / 4^2, which gives you

k = 75 * 4^2 = 75 * 16 = 1200.

Now rewrite the proportionality with this value of k: I = 1200 / r^2.

To get the illumination at distance 9 substitute 9 for r to get

I = 1200 / 9^2 = 1200 / 81 = 14.8 approx..

The illumination at distance 9 is about 14.8.

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RESPONSE -->

When I started to use the appropriate formulas (instead of cross multiplying) I mixed up where the numbers should go. My beginning equation was

4 = k/75^2. In general, I get confused on what x and y could stand for.

Now that I reread the problem in the book, I see why 75 was y and 4 is x^2.

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13:58:41

7.3.66 length inv prop width; L=27 if w=10; w = 18. L = ?

Explain how you set up and used a variation equation to obtain the length as a function of width, giving your value of k. Then explain how you used your equation to find the length for width 18

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RESPONSE -->

My question 66 isn't about the length inv prop width; L=27 if w=10; w = 18. L = ?.

I see now, number 64.

x - width

y - length

k - area

xy = k

10(27) = k

k = 270

y = k/x

y = 270/18

y = 15

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13:58:50

**Set up the variation equation L = k / w, which is the inverse proportion.

Use L = 27 when w = 10 to get

27 = k / 10, which gives you

k = 27 * 10 = 270.

Now we know that L = 270 / w.

So if w = 18 you get

L = 270 / 18 = 15. **

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RESPONSE -->

ok

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I believe you understand this. Let me know if you have questions on anything.