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 Posted: Wed Mar 22, 2006 7:17 pm
Soleq Cal Poly > Davis. So ha. Plus, I only went to Sac St. because their teaching program is superior to Davis'. I really don't care for either school. Lol, Cal Poly SLO was actually my first choice school, before I changed what I was studying. I originally wanted to be a computer engineer....but I decided to keep computer stuff as a hobby instead of a career. But I guess the point is to go to a school that has a good program for what you want to do. And I do hear good stuff about the teaching program at Sac State. My school has good environmental biology and entomology stuff, so I like it here 3nodding
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Posted: Thu Mar 23, 2006 11:49 am
I can't decide if I like him or not.
I used to. And I love talking to him.
But then we I'm not talking to him all the other things come into play like other's opinions. Ugh.
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Posted: Thu Mar 23, 2006 12:06 pm
Wendy_Chan This is an Algebra II, applications of quadratic equations problem: "Two ferryboats leave opposite shores of a lake at the same time. They pass each other when they are 800 meters from the nearest shore. When it reaches the opposite side, each boat spends 30 minutes at the dock and starts back. This time the boats pass each other when they are 400 meters from the nearest shore. Assuming that each of the boats travels at the same speed in both directions, how wide is the lake between the two ferry docks?" My friend read half of that and immediately said, "Mr. Hunter has no life." gonk I got that the lake was 1200 meters from the distance formula. If the lake is "x" meters wide, one boat would be going "x - 800" meters while the other went 800 meters in the same direction, and in going back, the one that went 800 meters went "x - 400" meters and the one going "x - 800" meters went 400 meters. Because they started out at the same time, I counted the thirty minutes as irrelevant (because I'm dumb and don't know what else to do >D<) and so if distance = rate x time for all of them, you can set x - 800 to 400 and get 1200. D: But if the first boat is x-800 meters away from the closest shore the first time, isn't it only 400 meters away? And then doesn't this become the closest shore? Also, if this is a problem using quadratics, you think they would be in there somewhere. I think I'm missing something. I don't understand how two boats could make identical passes at the same speed and still be different distances away from shore.
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Posted: Thu Mar 23, 2006 1:15 pm
Thomas Neo Anderson Wendy_Chan This is an Algebra II, applications of quadratic equations problem: "Two ferryboats leave opposite shores of a lake at the same time. They pass each other when they are 800 meters from the nearest shore. When it reaches the opposite side, each boat spends 30 minutes at the dock and starts back. This time the boats pass each other when they are 400 meters from the nearest shore. Assuming that each of the boats travels at the same speed in both directions, how wide is the lake between the two ferry docks?" My friend read half of that and immediately said, "Mr. Hunter has no life." gonk I got that the lake was 1200 meters from the distance formula. If the lake is "x" meters wide, one boat would be going "x - 800" meters while the other went 800 meters in the same direction, and in going back, the one that went 800 meters went "x - 400" meters and the one going "x - 800" meters went 400 meters. Because they started out at the same time, I counted the thirty minutes as irrelevant (because I'm dumb and don't know what else to do >D<) and so if distance = rate x time for all of them, you can set x - 800 to 400 and get 1200. D: But if the first boat is x-800 meters away from the closest shore the first time, isn't it only 400 meters away? And then doesn't this become the closest shore? Also, if this is a problem using quadratics, you think they would be in there somewhere. I think I'm missing something. I don't understand how two boats could make identical passes at the same speed and still be different distances away from shore. Yeah, I don't get if the boats are both going at the same speed, or if each boat is going at its own speed which remains constant throughout the problem.
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Posted: Thu Mar 23, 2006 2:57 pm
AraTeran Thomas Neo Anderson Wendy_Chan This is an Algebra II, applications of quadratic equations problem: "Two ferryboats leave opposite shores of a lake at the same time. They pass each other when they are 800 meters from the nearest shore. When it reaches the opposite side, each boat spends 30 minutes at the dock and starts back. This time the boats pass each other when they are 400 meters from the nearest shore. Assuming that each of the boats travels at the same speed in both directions, how wide is the lake between the two ferry docks?" My friend read half of that and immediately said, "Mr. Hunter has no life." gonk I got that the lake was 1200 meters from the distance formula. If the lake is "x" meters wide, one boat would be going "x - 800" meters while the other went 800 meters in the same direction, and in going back, the one that went 800 meters went "x - 400" meters and the one going "x - 800" meters went 400 meters. Because they started out at the same time, I counted the thirty minutes as irrelevant (because I'm dumb and don't know what else to do >D<) and so if distance = rate x time for all of them, you can set x - 800 to 400 and get 1200. D: But if the first boat is x-800 meters away from the closest shore the first time, isn't it only 400 meters away? And then doesn't this become the closest shore? Also, if this is a problem using quadratics, you think they would be in there somewhere. I think I'm missing something. I don't understand how two boats could make identical passes at the same speed and still be different distances away from shore. Yeah, I don't get if the boats are both going at the same speed, or if each boat is going at its own speed which remains constant throughout the problem. If they were going their own speed that remained constant, it still wouldn't be possible for them to meet at different distances from the shore because they left at the same time. Like if boat A was going 20 knots and boat B was going 10, they would still both meet like X miles from shore, no matter which one either left from. I don't think there is enough information to solve, else I'm overlooking something.
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Posted: Thu Mar 23, 2006 3:15 pm
*Is glad he's a math teacher and not a math student*
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Posted: Thu Mar 23, 2006 5:02 pm
Soleq *Is glad he's a math teacher and not a math student* Gotta love the "Teacher's Edition" of the math books, eh?
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Posted: Thu Mar 23, 2006 5:29 pm
-waves- o.o well then a small rant
OMFG SYSTEM OF A DOWN HAS GONE DOWN BADLY WITH THERE NEW ALBUM!!!
that will be all Xd
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Posted: Thu Mar 23, 2006 7:16 pm
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Posted: Thu Mar 23, 2006 9:14 pm
I'm ruining my life because I cant make Mike happy and be happy myself at the same time.
Why cant I just have him as a friend.
Why.
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Posted: Fri Mar 24, 2006 8:51 am
Would you like a hug, Leviticus? xp
As for the math problem, forget it, guys. I had a nervous breakdown in the guidance counselor's office (because I'm stupid like that) and they got Mr. Hunter in there. He said not to worry about it. He said don't even THINK about it anymore, because we'd go over it in class.
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Posted: Fri Mar 24, 2006 1:13 pm
The joys of being a teacher.
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Posted: Fri Mar 24, 2006 5:53 pm
>: ( I hope you get date-raped.
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Posted: Fri Mar 24, 2006 6:31 pm
Thomas Neo Anderson >: ( I hope you get date-raped.
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Posted: Fri Mar 24, 2006 8:19 pm
I certainly hope you're not referring to me.
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