Query 29

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course MTH 151

05/05/13 1:50 am

029. `query 29

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Question: `q7.3.18 (1/3) / 6 = 1/18. Is this ratio equation valid or not and how did you determine your answer?

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Your solution:

(1/3) / 6 = 1/18

(1/3 ) / 6 = 1 / 18

1/3 * 18 = 6

1 * 6 = 6

This ratio equation would be Valid!

confidence rating #$&*: 3

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Given Solution:

`a**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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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q7.3.20 z/8 = 49/56. Solve this proportionality for z.

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Your solution:

Z = 49

8 56

8*49=392/56= 7

Z = 7

confidence rating #$&*: 3

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Given Solution:

`a**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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Self-critique (if necessary):  OK

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Self-critique Rating: OK

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Question: `q7.3.42 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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Your solution:

.45/8 = .056

.49/16 = .030

1.59/50 = .031

16 oz box would be your best bet!

confidence rating #$&*: 3

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Given Solution:

`a** 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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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q7.3.45 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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Your solution:

X would = 1 simply by cross multiplying and dividing!

confidence rating #$&*: 3

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Given Solution:

`a** 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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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `qIf 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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Your solution:

Im not going to lie the fractions in the ratios threw me off on this question I struggled with it and ended up looking at the given solution 

confidence rating #$&*: 0

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Given Solution:

`a** 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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Self-critique (if necessary):

I got it! I am so happy that we are finally to equations. I actually feel like my brain is working on more than just word problems!

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Self-critique Rating: OK

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Question: `q7.3.72. 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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Your solution:

4 = 9

75 x

Illumination at 9 feet would be 168.75 feet

confidence rating #$&*: 2

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Given Solution:

`a**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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Self-critique (if necessary):

Totally did that one wrong but looking at the solution I can see where I went wrong and hopefully can avoid it on the test!

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Self-critique Rating: OK

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Question: `q7.3.66 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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Your solution:

L =27 if w = 10 w = 18 L= ?

L = k/w

27 = K/10

K= 27*10

K= 270

L= 270/w

w=18

L=270/18

L= 15

confidence rating #$&*: 3

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Given Solution:

`a**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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Self-critique (if necessary):

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Self-critique rating:

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Question: `q7.3.66 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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Your solution:

L =27 if w = 10 w = 18 L= ?

L = k/w

27 = K/10

K= 27*10

K= 270

L= 270/w

w=18

L=270/18

L= 15

confidence rating #$&*: 3

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Given Solution:

`a**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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Self-critique (if necessary):

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Self-critique rating:

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