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Monday, January 16, 2006

Scribe Post!!!


At the start of the class, we talked a bit about pi day and stuff. Then we did four problems on the board. Homework
is exercise 63 skip numbers 8b and 19. That's it.

Just kidding!

Ok, at the start of the class,we talked a bit about pi day. It's going to be held on March 14, and the school is going to have some sort of a contest where we have to find a coin somewhere around the school grounds. To find it, you have to solve three or more puzzles and the clues will lead to where the coin can be found. It seems like an interesting activity, and I'd join, but I'm on the committee. Otherwise, Mr. K encourages us to join in the activity.

Then we solved four problems on the board:

a) For a), we were asked to find out what is the mass of ethanol when the volume of ethanol is 250 mL.
Answer: Well, as given in the table, the ma
ss of the beaker without ethanol is 90 (grams?). For every 50 mL of ethanol, the mass of the beaker increases by 39 (grams?). So when the volume of ethanol is 250 mL, the mass of beaker with liquid is 285. Since the dry mass of the beaker is 90, we subtract 90 from 285, leaving us with 195 g of ethanol.

b)I can't remember what was the question for this one... I don't usually write the questions, which I know is irresponsible for
my part, but I'll try to remember.. correct me if I'm wrong. We were asked to find out what is the mass of the ethanol when there is 200 mL of ethanol.
Answer: When there is 150 mL of ethanol, the mass of the beaker with ethanol is 207. Well we have to add another 50 mL of ethanol, and when we add 50 mL of ethanol, the mass of the beaker with liquid will increase by 39, so we add 39 to 207, which gives us 246. Then we take the dry masss of the beaker
, which is 90, and subtract it from 246. It gives us 156 g of ethanol.


For this part, it involves a distance/time graph (surprise, surprise for all who are taking Science! We HAVEN'T done this before.. *sarcasm* haven't we had enough of these yet?)
The questions are:
a) Find the eq'n* of the line.
b) What's the slope?
c) What's the domain of the equation?
d) If t = 25, what's the distance?
e) If d = 850, what's the time?
f) If d = 1500, what's the time?

a) d(t) = 10t
b) m = 10
c) x is greater than or equal to zero
(x 0), or [0, )
d) d = 250
e) t = 85
f) t = 150

3.) How many multiples of 5 are there starting from 15 to 450?
450 = 5(n-1)+15
450 = 5n-5+15
450 = 5n+10
450-10 = 5n
440 = 5n
n = 88

So.. there are 88 multiples of 5 starting from 15 to 450.

4.) Once again, I don't quite remember the question. But I remember the general details. The 18th term is 262, common difference is 15. We need to find the first term and the 100th term in this sequence or series.
So, solving for the first term:
t18 = 262
d = 15
tn = t1 + (n-1)d
262 = t1 + (18-1)15
262 = t1 + (17)15
262 = t1 + 255
262-255 = t1
7 = t1

Solving for the 100th term:
tn = t1 + (n-1)d
t100 = 7 + (100-1)15
t100 = 7 + (99)15
t100 = 7 + 1485
t100 = 1492

There you have it! That's pretty much everything about what we did in class.

And now, the scribe for tomorrow is.. *drumroll* Sith Lord Darth SAMUS! Haha I know how much you like Star Wars and Darth Vader.. see you later people!


At 1/17/2006 4:20 PM, Anonymous Anonymous said...

Samus is not from "Star Wars." Samus is from Metroid.

At 1/17/2006 8:17 PM, Anonymous Anonymous said...

I know but the real SAMUS in our class likes Star Wars.


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