Assignment 17

course mth164

ÔñŸÂŠWŒÁ鎅öòï°ªºÆ˜˜assignment #017

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017.

Precalculus II

07-24-2007

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13:12:13

Query 11.3.26 a = -2, r = 4.

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RESPONSE -->

ok

confidence assessment:

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13:13:09

Give the 5th term and the nth term of the geometric sequence.

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RESPONSE -->

-512

confidence assessment: 3

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13:14:27

Query 11.3.42 sum (4 * 3^(k-1), k, 1, 4)

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RESPONSE -->

ok

confidence assessment:

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13:14:57

Give the sum and explain how you obtained it.

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RESPONSE -->

160

I used the formula 4((1-3^4)/(1-3))

confidence assessment: 3

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13:16:32

Query 11.3.66 (5th edition 70) ball dropped from 30 feet, .8 of height on each bounce.

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RESPONSE -->

ok

confidence assessment:

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13:17:36

What height will the ball bounce to after the third time it hits the ground and how did you obtain your answer?

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RESPONSE -->

15.36 feet

confidence assessment: 3

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13:18:11

What height will the ball bounce to after the nth time it hits the ground and how did you obtain your answer?

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RESPONSE -->

30(.8^n)

confidence assessment: 3

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13:19:59

After how many bounces will ball's height first be less than six inches?

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RESPONSE -->

19 bounces

confidence assessment: 3

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13:20:34

What total distance to the ball travel before it stops bouncing?

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RESPONSE -->

150 feet

confidence assessment: 3

You don't show how you obtained your results so I can't tell what you did correctly and what you might have overlooked. I suspect that you counted only half of the distance on each full bounce.

After the 1st bounce it will bounce back up to .8 * 30 ft; after the 2d bounce the height will be .8^2 * 30 ft; after the 3d bounce height is .8^3 * 30 ft.

After the nth bounce the height is .8^n * 30 ft.

The ball falls 30 feet, then it bounces to height 24 feet and back, then to height 19.2 feet, etc.; the distance it travels is 30 + 48 + 38.4 + etc..

In symbols the distance is 30 + 2 * ( .8 * 30 + .8^2 * 30 + .8^3 * 30 + ... + .8^19 * 30), which upon factoring out .8 * 30 from the second part becomes

30 + 2 * (.8 * 30) ( 1 + .8 + .8^2 + ... + .8^18), or

30 + 48 * ( 1 + .8 + .8^2 + ... + .8^18).

The sum ( 1 + .8 + .8^2 + ... + .8^18) is, by the formula, equal to

a * (1 - r^18) / (1 - r) = 1 * (1 - .8^18) / (1 - .8) = 4.91, approximately.

So the total distance through the 18th bounce is

30 + 48 * 4.91 = 265.7, approximately.

Adapting the sum 30 + 48 * ( 1 + .8 + .8^2 + ... + .8^18) to the total of all bounces we get 30 + 48 * ( 1 + .8 + .8^2 . . . + .8^n + . . . ).

The sum 1 + .8 + .8^2 + . . . + .8^n + . . . is 1 / (1-r) = 1 / (1-.8) = 5. So the total distance on all bounces is

total distance = 30 + 48 * 5 = 270.

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13:22:33

Query 11.5.6 evaluate the expression C(100,98).

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RESPONSE -->

ok

confidence assessment:

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13:22:52

What did you obtain when you evaluate the expression, and how did you obtain it?

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RESPONSE -->

4950

I used my calculator 100 nCr 98

confidence assessment: 3

Calculator solutions to this problem are unacceptable, except that the calculator can be used for the very last step where the integers that don't divide our are multiplied. The expressions must be written out and evaluated in a manner similar to the following:

100! / (98! * 2!) = 100 * 99 * 98 * 97 * .... * 1 / [( 98 * 97 * ... * 1) * (2 * 1)].

The 98 * 97 ... in the denominator divides out the same mess in the numerator leaving

100 * 99 / ( 2 * 1) = 50 * 99 = 4950.

Combinations and permutations never result in fractional answers. The denominators ALWAYS divide out completely.

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13:23:05

Query 11.5.18 expand (2x+3)^5.

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RESPONSE -->

ok

confidence assessment:

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13:26:21

What do you get when you expand the expression, and how did you used Binomial Theorem to obtain your result?

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RESPONSE -->

32x^5+240x^4+720x^3+1080x^2+810x+243

I used the binomial theorem to find what to multiply the terms by in each exponent.

confidence assessment: 2

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13:26:36

Query 11.5.32 third term in expansion of (x-3)^7.

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RESPONSE -->

ok

confidence assessment:

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13:27:47

What is the third term in your expansion, and how did you use the Binomial Theorem to obtain the result?

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RESPONSE -->

189x^5

I found C(7,2) and multiplied it by (-3)^2 to get the coefficient of x^5

confidence assessment: 3

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13:28:15

Comm on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

It was tedious to expand the next to the last problem but easier than if you had to do it by hand.

confidence assessment:

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Good responses. See my notes and let me know if you have questions. &#