7_3

course Mth 151

???y~????}???assignment #029

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029. `query 29

College Algebra

08-02-2007

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21:26:38

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 -->

The ratio is valid.

(1/3) / 6 = 1/18

Cross multiply:

1/3 * 18 = 6

6 * 1 = 6

confidence assessment: 3

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21:26:50

**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

self critique assessment: 3

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21:28:28

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

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RESPONSE -->

z/8 = 49/56

Cross multiply:

56z = 8 * 49

56z = 392

/56 /56

z = 7

7/8 = 49/56

confidence assessment: 3

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21:28:32

**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

self critique assessment: 3

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21:32:12

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 -->

8 oz. for $ .45

45/8 = 5.625 cents per ounce

16 oz. for $. 49

49/16 = 3.0625 cents per ounce

50 oz. for $1.59

159/50 = 3.18 cents per ounce

16 oz for $ . 49 is the best value

confidence assessment: 3

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21:32:24

** 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

self critique assessment: 3

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21:34:11

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 -->

x/3 = 2/6

Cross multiply:

6x = 6

x = 1

confidence assessment: 3

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21:34:19

** 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 -->

OK

self critique assessment: 3

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21:39:46

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 -->

z = k/x

9 = k/(2/3)

Cross multiply:

9 (2/3) = k

6 = k

z = 6/ (5/4)

Cross multiply:

5/4z = 6

z = 6/ (5/4)

z= 6 * (4/5)

z= 24/5

confidence assessment: 2

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21:40:01

** 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 -->

OK

self critique assessment: 3

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21:49:30

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 -->

y = k/x

75 = k/4^2

75 = k/16

Cross multiply:

75 *16 = k

1200 = k

y = 1200/ 9^2

y = 1200/81

y = 14.8 ft candles

confidence assessment: 2

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21:49:37

**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 -->

OK

self critique assessment: 3

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21:53:15

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 -->

y = k/x

so L = k/w because L is inversely proportional to width

27 = k/10

cross multiply

27 * 10 = k

270 = k

L = 270 /18

L = 15

confidence assessment: 3

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21:53:21

**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

self critique assessment: 3

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