If your solution to stated problem does not match the given solution, you should self-critique per instructions at

 

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

 

 

Your solution, attempt at solution.  If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

 

029.  `query 29

 

 

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

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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. **

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

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

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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.  **

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

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?

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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. **

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

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?

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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.  **

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

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.

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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. **

 

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

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? 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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.

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

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

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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. **