Tuesday, September 25, 2007

Circular Functions Pre Test

Good Evening, this is Miller reporting today on our blog. Unfortunately our beloved teacher Mr. K was not in class today. Instead we had the charming Miss Nicholson. During our first period class we had a quiz yet again. I can't remember what was on the quiz since we handed it in afterwards. Once we were done the quiz we were allowed to study for the afternoon pretest. The test looked like this:

Here are the answers we received in class today: (n) = pie ( I couldn't find a symbol for it)
(€)= theta
1) b) 2n/3, because sin€ is only equal to √3/2 in the first and second quads.

2) b) 2.41, y=2

3) It asks for cos(k)= -√3/2 between the intervals of 450 degrees and 630. 5n/6 is what cos is when it is at -√3/2. The measure of that angle is 150 degrees. This means the have to add 360 to it in order for it to be in the correct intervals. The answer is 510 degrees.

4) Here we are looking for the B of the equation. AsinB(x-C)+D. We find out the period of the graph and then use the equation B=2n/period.
= n/8
5) A) Here we are asked to find the period of the function p(t)=100-20cos5nt/3. I arranged it so it would look like p(t)= -20cos5n/3+100. We take the B of this equation and use it to divide 2n.
2n/1 * 3/5n
= 6/5
B) Our sinusoidal axes is now at 100. Since the A of the function is -20 it means that it will go 20 below 100 and also 20 above 100.
Max= 120
Min= 80

c) This is what the graph of the cos should look like:

Tommorow I believe we are going to do something new if Mr. K is there. Reminder that our test is on Thursday. The next lucky blogger is Lina.


Miss Nicholson said...

"Charming"... I'll take it. :)

Good work with correcting the mistake I made on the board in the last question (when I put 3/4 instead of 3/5... funny how basic algebra gets complicated when you're thinking so hard!)

duckie said...

i need help in english