Assignment 14

course mth 163

assignment #014014.

Precalculus I

10-29-2008

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12:41:26

`questionNumber 140000

`q001. Note that this assignment has 5 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).

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RESPONSE -->

Just plug in and solve,

a(1)=5

a(2)=9

a(3)=17

a(4)=33

confidence assessment: 2

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12:41:38

`questionNumber 140000

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 we are given 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 we are given 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 we are given a(3) = 16 we now have

a(4) = 17 + 16 = 33.

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RESPONSE -->

I understand

self critique assessment: 2

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12:43:10

`questionNumber 140000

`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)?

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RESPONSE -->

Solve again getting

a(1)=7

a(2)=16

a(3)=35

a(4)=74

confidence assessment: 2

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12:43:18

`questionNumber 140000

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.

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RESPONSE -->

I see

self critique assessment: 2

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12:45:11

`questionNumber 140000

`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?

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RESPONSE -->

The slope is rise over run

The rise varies 10,18,26, and 34

The run is 2

Slope is 1-3=6

3-5 is 9

5-7 is 13

7-9 is 17

confidence assessment: 2

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12:45:18

`questionNumber 140000

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 = 6.

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.

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RESPONSE -->

I see

self critique assessment: 2

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12:46:46

`questionNumber 140000

`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?

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RESPONSE -->

Weight is defined by volume so we know that w=3/64*24^3 Then find that w=748

confidence assessment: 2

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12:47:01

`questionNumber 140000

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 = 748.

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RESPONSE -->

I see

self critique assessment: 2

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12:48:50

`questionNumber 140000

`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.

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RESPONSE -->

The 2nd is 3 times the 1st.

So we multiply 22 *3 to get 66.

confidence assessment: 2

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12:49:10

`questionNumber 140000

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.

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RESPONSE -->

I totally did that wrong but I understand how the answer was achieved.

self critique assessment: 2

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&#Very good responses. Let me know if you have questions. &#