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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

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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.

 

 

014.

 

 

Question:  `q001.    Note that this assignment has 7 questions

 

If a(n) = a(n-1) + 2^n, with a(0) = 3, then substitute in turn the values 1, 2, 3 and 4 into the equation to obtain the values a(1), a(2), a(3) and a(4).

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution: 

If we substitute n = 1 into a(n) = a(n-1) + 2^n we get

a(1) = a(1-1) + 2^1 or, since 1-1 = 0 and 2^1 = 2

a(1) = a(0) + 2.  Since we are given a(0) = 3 we now have

a(1) = 3 + 2 = 5.

 

If we substitute n = 2 into a(n) = a(n-1) + 2^n we get

a(2) = a(2-1) + 2^2 or, since 2-1 = 1 and 2^2 = 4

a(2) = a(1) + 4.  Since in the previous step we found that a(1) = 5 we now have

a(2) = 5 + 4 = 9.

 

If we substitute n = 3 into a(n) = a(n-1) + 2^n we get

a(3) = a(3-1) + 2^3 or, since 3-1 = 2 and 2^3 = 8

a(3) = a(2) + 8.  Since in the previous step we found that a(2) = 9 we now have

a(3) = 9  + 8 = 17.

 

If we substitute n = 4 into a(n) = a(n-1) + 2^n we get

a(4) = a(4-1) + 2^4 or, since 4-1 = 3 and 2^4 = 16

a(4) = a(3) + 16.  Since in the previous step we found that a(3) = 17 we now have

a(4) = 17 + 16 = 33.

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question:  `q002.  If a(n) = 2 * a(n-1) + n with a(0) = 3, then what are the values of a(1), a(2), a(3) and a(4)?

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution: 

If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get

a(1) = 2 * a(1-1) + 1 or since 1-1 = 0

a(1) = 2 * a(0) + 1.  Since we know that a(0) = 3 we have

a(1) = 2 * 3 + 1 = 7.

 

If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get

a(2) = 2 * a(2-1) + 2 or since 2-1 = 1

a(2) = 2 * a(1) + 2.  Since we know that a(0) = 3 we have

a(2) = 2 * 7 + 2 = 16.

 

If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get

a(3) = 2 * a(3-1) + 3 or since 3-1 = 2

a(3) = 2 * a(2) + 3.  Since we know that a(0) = 3 we have

a(3) = 2 * 16 + 3 = 35.

 

If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get

a(4) = 2 * a(4-1) + 4 or since 4-1 = 3

a(4) = 2 * a(3) + 4.  Since we know that a(0) = 3 we have

a(4) = 2 * 35 + 4 = 74.

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question:  `q003.  What are the average slopes of the graph of y = x^2 + x - 2 between the x = 1 and x= 3 points, between the x = 3 and x = 5 points, between the x = 5 and x = 7 points, and between the x = 7 and x = 9 points?  What is the pattern of this sequence of slopes?

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution: 

At x = 1, 3, 5 , 7 and 9 we find by substituting that y = 0, 10, 28, 54 and 88.

 

The x = 1, 3, 5, 7 and 9 points are therefore (1,0), (3,10), (5,28), (7,54) and (9,88).

 

The run from one point to the next is always 2.  The rises are respectively 10, 18, 26 and 34.  The slopes are therefore

 

slope between x = 1 and x = 3:  slope = rise / run = 10 / 2 = 5.

slope between x = 3 and x = 5:  slope = rise / run = 18 / 2 = 9.

slope between x = 5 and x = 7:  slope = rise / run = 26 / 2 = 13.

slope between x = 7 and x = 9:  slope = rise / run = 34 / 2 = 17.

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question:  `q004.  If a solid stone sphere 4 inches in diameter weighs 3 pounds, then what would be the weight of a solid stone sphere 2 feet in diameter?

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution: 

The volume of a sphere is proportional to the cube of its diameters, and weight is directly proportional to volume so we have the proportionality

 

w = k d^3, where w and d stand for weight and diameter and k is the proportionality constant.

 

Substituting the known weight and diameter we get

 

3 = k * 4^3, where we understand that the weight is in pounds and the diameter in inches.  This gives us

 

3 = 64 k so that

k = 3 / 64.

 

Our proportionality equation is now

 

w = 3/64 * d^3. 

 

So when the diameter is 2 feet, we first recall that diameter must be in inches and say that d = 24, which we then substitute to obtain

 

w = 3/64 * 24^3.  A simple calculation gives us the final weight

 

w = 648.

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question:  `q005.  Two boxes are each constructed of a single layer of cardboard.  The first box is 12 inches by 18 inches by 24 inches and weighs 22 ounces; the second is 36 inches by 54 inches by 72 inches.  Using proportionality determine the weight of the second box.

 

 

Your solution:  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution: 

The ratios 36/12, 54/18 and 72/24 of the corresponding sides are all the same and all equal to 3, so the dimensions of the sides of the second box are 3 times those of the first.

 

Since the thickness of the cardboard is the same on both boxes, only the dimensions of the rectangular sides change.  The only thing that matters, therefore, is the surface area of the box.  The proportionality is therefore of the form w = k x^2, where w is the weight of the box and x stands any linear dimension.

 

It follows that w2 / w1 = (x2 / x1)^2.  Since as we just saw x2 / x1 = 3, we see that

 

w2 / w1 = 3^2 = 9.  Since w1 = 22 oz, we write this as

 

w2 / 22 oz = 9.  Multiplying both sides by 22 oz we see that

 

w2 = 22 oz * 9 = 198  oz.

 

STUDENT QUESTION

 

I reduced the sides of each box by dividing each side measurement by 6. Then since we were looking for weight, I used
the y = kx^3 formula to calculate the weight of the second box. I used y as the weight and x as the volume (l*w*h).
Obviously my calculation was way off. Why wouldn’t this work?

INSTRUCTOR COMMENT

 

Presumably the cardboard is of the same thickness for both boxes. So the amount of cardboard is determined by the surface area of the the box, not the volume.

If the box was filled with cardboard as well as being constructed of cardboard, then the proportionality with volume would be appropriate and your solution would be correct.

 

Question:  `q006.  Between t = 1 and t = 3 the function y = .5 t^2 - 5 t + 9 has average slope -3; between t = 3 and t = 5 the average slope is -1; and between t = 5 and t = 7 the average slope is 1. 

 

Your solution: 

 

Confidence Assessment:

 

If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily.  If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

 

Question:  `q007.  A sphere of radius .5 meters has mass 1100 kg and requires 11 ounces of paint to cover its surface.  What would be the mass of a sphere of the same material whose radius is 1.3 meters, and how much paint would it take to cover its surface with a coat whose thickness was the same as that used to cover the first?

 

Your solution: 

 

Confidence Assessment:

 

Self-critique rating:

 

 

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