Showing posts with label SharmaineD.. Show all posts
Showing posts with label SharmaineD.. Show all posts
Tuesday, January 15, 2008
BOB
Probability was a difficult unit. I dont know what exactly my muddiest point is but all I know is that I always get confuse what to use for every question. Each question has its own technique to answer them. Good luck everyone!
Monday, January 14, 2008
Thursday, December 6, 2007
BOB
Okay, for the combinatorics unit, my muddiest point are the circular permutation and the card problems. I freaked out when I see these kind of problems and don't know what to do. I am really confuse when to use the pick and choose formula though I know the rules. Even though it seems like an easy question, I totally messed up. I simply don't like this unit. =p
Good luck everyone!
Good luck everyone!
Tuesday, November 27, 2007
Permutations
Hey guys! It's Sharmaine and I'm the scribe for today. It's my first time using slideshare for my scribe post, so I hope it'll help.
Next scribe is... Precious.
Next scribe is... Precious.
Tuesday, November 20, 2007
BOB
Okay, I thought this unit was easy not until Mr. K introduce the natural log. I think my muddiest point are the word problems especially the half-life ones. I get confuse which formula to use. Sometimes I make mistakes with using log or ln in solving for x. The change of base law is pretty easy and the one when you need to solve for x that was an exponent and they have to be in the same base. Good luck everyone!
Thursday, November 1, 2007
BOB
In this unit, I found the double angle identities more difficult than just proving and solving for x. I get confuse of what identities to use. Sometimes I still made small mistakes with proving and solving for x or by doing the factoring that made me screwed up the whole thing. I just hope I'll do well in the exam tomorrow.
Tuesday, October 30, 2007
Thursday, October 18, 2007
Blogging on Blogging
Okay..In this unit, I'm a little confuse of graphing f^-1 (x) and 1/f(x). I always do it the other way. Sometimes, i reflect my graph using y=x instead of finding the reciprocal graph. I'm also having a hard time with the problem solving questions but as Mr. K gave us time and let us practice with these questions, I'm getting used to solving it. Well, I think that's it. I'm okay with stretches and compressions of the graph, even and odd functions. I guess I'll do well in the exam if I keep practicing solving the inverses and reciprocal questions.
Monday, October 15, 2007
Group 6: Sea Port Function
At a sea port, the depth of the water, h meter, at time, t hours, during a certain day is given by this formula.
h(t)=1.8sin[2π(t-4.00)/12.4]+ 3.1
(a)State the: (i)period=12.4 (ii)amplitude=1.8 (iii)phase shift=4.00
(b)What is the maximum depth of the water?
3.1 + 1.8 = 4.9
The maximum depth of the water is 4.9m.
When does it occur?
h(t)=1.8sin[2π(t-4.00)/12.4]+ 3.1
4.9=1.8sin[2π(t-4.00)/12.4]+ 3.1
Let θ = 2π(t-4.00)/12.4
4.9=1.8sin θ + 3.1
1.8=1.8sin θ
1 = sin θ
1.5708 = θ
(12.4/2π)2π(t-4.00)/12.4 = 1.5708 (12.4/2π)
t-4.00 = 3.1
t=7.1
The maximum depth of the water occurs at 7:06am.
(c)Determine the depth of the water at 5:00am and at 12:00noon.
h(t)=1.8sin[2π(t-4.00)/12.4]+ 3.1
For 5:00am
h(5)=1.8sin[2π(5-4.00)/12.4]+ 3.1
= 3.9735
The depth of the water at 5:00am is 3.9735m.
For 12:00noon.
h(12)=1.8sin[2π(12-4.00)/12.4]+ 3.1
=1.6766
The depth of the water at 12:00noon is 1.6766m.
(d)Determine one time where the water is 2.25 meters deep.
h(t)=1.8sin[2π(t-4.00)/12.4]+ 3.1
2.25=1.8sin[2π(t-4.00)/12.4]+ 3.1
Let θ = 2π(t-4.00)/12.4
2.25=1.8sin θ + 3.1
-.85=1.8sin θ
-.4722 = sin θ
-.4918 = θ
(12.4/2π)2π(t-4.00)/12.4 = -.4918 (12.4/2π)
t-4.00=-.9706
t=3.0294
The water is 2.25 meters deep at around 3:00am.
Group 6 Members: Luis, Joe, Sharmaine, Precious, Ivanna, Roslyn
Correct me guys if I'm wrong.
h(t)=1.8sin[2π(t-4.00)/12.4]+ 3.1
(a)State the: (i)period=12.4 (ii)amplitude=1.8 (iii)phase shift=4.00
(b)What is the maximum depth of the water?
3.1 + 1.8 = 4.9
The maximum depth of the water is 4.9m.
When does it occur?
h(t)=1.8sin[2π(t-4.00)/12.4]+ 3.1
4.9=1.8sin[2π(t-4.00)/12.4]+ 3.1
Let θ = 2π(t-4.00)/12.4
4.9=1.8sin θ + 3.1
1.8=1.8sin θ
1 = sin θ
1.5708 = θ
(12.4/2π)2π(t-4.00)/12.4 = 1.5708 (12.4/2π)
t-4.00 = 3.1
t=7.1
The maximum depth of the water occurs at 7:06am.
(c)Determine the depth of the water at 5:00am and at 12:00noon.
h(t)=1.8sin[2π(t-4.00)/12.4]+ 3.1
For 5:00am
h(5)=1.8sin[2π(5-4.00)/12.4]+ 3.1
= 3.9735
The depth of the water at 5:00am is 3.9735m.
For 12:00noon.
h(12)=1.8sin[2π(12-4.00)/12.4]+ 3.1
=1.6766
The depth of the water at 12:00noon is 1.6766m.
(d)Determine one time where the water is 2.25 meters deep.
h(t)=1.8sin[2π(t-4.00)/12.4]+ 3.1
2.25=1.8sin[2π(t-4.00)/12.4]+ 3.1
Let θ = 2π(t-4.00)/12.4
2.25=1.8sin θ + 3.1
-.85=1.8sin θ
-.4722 = sin θ
-.4918 = θ
(12.4/2π)2π(t-4.00)/12.4 = -.4918 (12.4/2π)
t-4.00=-.9706
t=3.0294
The water is 2.25 meters deep at around 3:00am.
Group 6 Members: Luis, Joe, Sharmaine, Precious, Ivanna, Roslyn
Correct me guys if I'm wrong.
Thursday, September 13, 2007
Unit Circle
Hey guys!(: To start off my post, Mr. K was away today but as usual there's a substitute teacher. We had our first quiz on Circular Functions. After we finished the quiz, the sub handed out a worksheet called "Working with the Unit Circle". Together with the worksheet, we were also asked to do Exercise 5, questions 11-20.
Don't forget to finish the worksheet and hand it in tomorrow morning! (:
Next scribe is Jessica!
Don't forget to finish the worksheet and hand it in tomorrow morning! (:
Next scribe is Jessica!
Labels:
Circular Functions,
Scribe Post,
SharmaineD.
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