### Scribe

The first thing we did in this morning's class was hover around Mr. K's computer and look at our sexy wiki. Remember guys, our constructive modification + significant contribution are due this

Then we got onto sequences. Mr. K wrote two examples on the board for us to solve.

We got to use our calculators today, I am not sure if they were used the other day. With that said, make sure the calculator is in sequence mode!

-All you have to do is press mode and arrow over to seq. Then yeah, hit enter.

__Question the first:__

My cat is sick. His name is Little John and he has a cold. The vet has given him some medicine. Each day he gets a pill with 35 mg of medicine. His body eliminates 25% of the medicine each day and then he gets another pill.

a. How much medicine will be in his body in 5 days?

b. Will the amount of medicine in his body stabilize? How many days will it take and how much medicine will be in his body?

Answers:

a. After 5 days, there will be 106.77 mg of medicine.

-In the second column, the reason you multiply by .75 is because Little John's body eliminates only 25% of the medicine, leaving the other 75% in his body. Then, when he has another pill (35 mg), it's added with the medicine still in his body.

-.75 represents what you multiply U by, U being the term. N is the column. I believe so, anyway. Also, to get to that screen all you have to do is press Y=.

-I really like this, because you can go up as many terms as you'd like and then store them into one of the lists. 1 represents the first term, and 50 is the term I chose to go up to. I would assume you guys know all about this, as we did it quite a bit with periodic functions. The only difference is that we have it in sequence mode and we put the u. Ohh, to get the U it's 2nd mode 7.

b. The medicine does stabilize after 36 days, at 140 mg.

-This represents the first week of medication.

-This is where it stabilizes. Hurray! :D

- This is the graphed data, as you can see, it stabilizes

__Question the second:__

A small forest pf 4000 trees is under a new forestry plan. Each year 20% of the trees will be harvested and 1000 new trees are planted.

a. Will the forest ever disappear?

b. Will the forest size ever stabilize? If so, how many years and with how many trees?

Answers:

-.80 is what you multiply by the first term, as 20% is being cut down, leaving the other 80% in tact. 1000 is what we'd add on to what we're left with after the first year. 4000 is what we started with, as it's stated in the question.

-See the difference? The "ipart" means that we won't have answers with decimals. We can't have part of a tree.

a. As long as they're replanting, the forest won't disappear. In fact, the amount of trees inceases before it stabilizes.

b. The forest size does stabilize at 27 years with 4996

__This afternoon:__

We worked in our textbooks this afternoon, starting on page 264. I'll go over the first three questions, as I think that's where the majority of the class was by the end of class. I could be wrong, but too bad. :P

__Une:__

1. Use a calculator to generate the following sequences and find the indicated term.

a) 5, 7, 9, 11 ...; the 14th term

*n*Min=0

u(

*n*)=u(first term)(

*n*-1)+2 (The sequence goes up by 2)

u(

*n*Min)={5} (Our sequence starts at 5)

-To get the 14th term you go to the home screen, u

__2nd function, 7__. Then (14) ... You could do (1, 14) ... but the first way is faster. Oh yes, and the answer is 31.

b) 5, 15, 45, 135, ...; the 12th term

*n*Min=0

u(

*n*)=u(first term)(

*n*-1)*3(multiplying by 3)

u(

*n*Min)={5}(Again, our sequence begins at 5)

- The answer is 885735. To get the answer you do the same thing you'd do for the first problem, except instead of 14 it'd be 12. I'll spare you the play by play from here on. ;)

c)3.7, 2.3, 0.9, -.5, ...; the 15th term

*n*Min=0

u(

*n*)=u(

*n*-1)-1.4

u(

*n*Min)={3.7}

- The answer is -15.9

d) 2.4, 3.6, 5.4, 8.1, ...; the 9th term

*n*Min=0

u(

*n*)=u(

*n*-1)*1.5

u(

*n*Min)={2.4}

- The answer is 61.51

e) 12, 3, .75, .1875, ...; the 10th term

*n*Min=0

u(

*n*)=u(

*n*-1)/4

u(

*n*Min)={12}

- The answer is .000046

__Deux:__

A salesperson at a car dealership receives a base salary of $275 per week, plus $50 for every car sold. Each week, the salesperson sells one more car then the previous week. In his first week, he sold one car.

a. Create a graph to show the salesperson's earnings each week.

b. Use your graph to determine the week in which earnings will exeed $1000, if the pattern continues.

-This table is simple enough, all you do is add the base salary to the money made for each car sold. Week 15 is when more than $1000 is made.

c. solve the problem using another method.*n*Min=0

u(*n*)=u(*n*-1)+50

u(*n*Min)={275}

-Quit to the home screen.

u(1,15)

{325 375 425 475 525 575 625 675 725 775 825 875 925 975 1025}__Trois:__

A colony contains 100 ants at the start of May. Suppose the population doubles every month and no insects die.

a. How many insects are there at the beginning of July?

I did this in my head. May-July is a span of 2 months. There will be 200 at the beginning of June, and therefore 400 at the beginning of July.

b. How many insects are there at the beginning of September?

Again, did this in my head. July-September is again, a span of two months. We had 400 insects at the beginning of July, therefore by August we would have 800. This means that by the beginning of September we'd have 1600 insects.

If you wanted to, you could do 100x2^{4}. That would give us 1600 as well.

Two terms to keep in mind are:

Arithmetic Sequence:

There is a common difference between terms. For example; 3.7, 2.3, .9, -.5. The common difference is -1.4 as you're taking -1.4 from the one term to get the next, or adding 1.4 to get one term from the previous.

Geometric Sequence:

There is a common ratio between terms.

For example; 2, 4, 8, 16. The common ratio is 2, as you're multiplying by 2 to get the next term, or dividing by 2 to get the previous term. __And finally:__

Our homework is finishing up Checking Your Skills on pages 264 through 265.

The scribe for tomorrow is: Mr. M. Sherif~~sucker~~!

^{}

Congratulations on making it into the Scribe Post Hall of Fame.

Mr. Harbeck

Sargent Park School

Posted by Mr. H | 6/05/2006 7:49 AM

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