Well, we started the class by solving a geometric sequence, given the 2nd and 3rd term. We had to find the 8th term given this. Since it was a geometric sequence, we know that there's a common ratio. To find this common ratio, we found the product of the 2 terms. Which was 12/24 = 1/2. We then found where y = 0, if we were to graph this. We found the 1st term by going backwards, which means you would have to multiply the 2nd term (24) by 2. 24 x 2 = 48. Then to find the term of 0, we would double that, which was 96. We then multiplied 96 by (1/2)

^{8}because we need to find the 8th term, and it had a common ratio of 1/2. The answer was 0.375

Last class we found out where the infinite geometric sequence came from, but today we learned that when the common ratio is greater than 1, then it is a diverging series, because it will keep going on and on, thus meaning it is infinity. So when |r| > 1 then r

^{∞}= ∞

When the common ratio is less than 1, it's called a converging series, because it hones in on one value, which is 0. So when |r|<>∞ = 0.

This was pretty much all we learned, so we did some questions after. There was a question about a super ball. At this point Paul pulled out a super ball for Mr. K to use, and it gave the class a little interaction. The ball would start at 200 cm and drop to the ground, it would rebound 3/4 of the distance it fell. We had to then find the total distance travelled by the 4th bounce. This question was easily accomplished using a diagram.

The initial distance being 200 cm, we can multiply by 3/4 to find the distance it bounces back up, then multiply that distance by 2, because it travels that distance twice, bouncing up and falling down, then we multiply the distance from the second bounce by 3/4 (150 x (3/4) = 112.5) to get the distance going up, then multiply by 2 for it going up and down,(150 x (3/4) = 112.5 x 2 = 225 cm) then multiply the distance from the 3rd bounce by 3/4 again (112.5 x (3/4) = 84.375) then multiply that by 2 because it goes up that distance and down hat distance, so it would be (84.375 x 2 = 168.75) add all the values together to get 893.75

You can also use the geometric sequence equation to find this also.

After this we had to find the distance travelled until the ball stopped, so we used the infinite geometric series which was 1200, given all the variables, and you would add the initial value which was 200, and that would be 1400 cm. This was the last question.

That was all we did for our last class. I know, we need something special, because I'm the last scribe and this whole course has been oh so special, because of this reason, I wrote a poem. It's a Kyrielle poem. With 8 syllables each line, and rhyming scheme being aabB ccbB ddbB. Enjoy.

One More Step

Unluckily today's the last,

Of this most enjoyable class.

Though it's not the end for us yet,

As we move on just one more step.

Mr. K, we give you homage,

For expanding on our knowledge.

Never dull, you were fun instead.

As we move on just one more step.

Leaving with a fantastic smile,

As every thing's been so worthwhile.

Now to continue, and to prep.

As we move on just one more step.

-Francis Bowers

## 5 comments:

You honour me sir. It has been my very great pleasure to be your teacher.

Thank you.

Its been very great to be your student.

Xd

Very heartwarming? Lol.

Francis,

That was an amazing poem you made, you deserve some major props for that. Today was not really the last class , tomorrow will be the last class. I kinda feel sad because it is going to be my last class with Mr. K, while you guys still have the chance to become his student next year IF you take AP Calculus. I will definitely miss his skits that make the class very interesting.

-m@rk

He's got the skills to pay the bills.

Post a Comment