jordan

Three with Windows XP around here. Wow YT, time for an upgrade? I've never even heard of Amiga, just know it's "girl friend" in spanish...which raises a few questions.

Hey thanks for the help. coquina: I do try to alternate the endurance with the sprints but I guess it's still all the same muscle groups in the end. I'll try throwing in a few different types of workouts in between. Would peanut butter be a good food to eat?

Hey, I need some help here. Recently I've gotten into riding my bike. Don't know why, just found it and started riding it and got hooked. But now I have this problem: it's getting more and more difficult each day. I try to do an intense, hilly ride one day, a few days of a more flat, easy workout and then back to the intense one and so on. But what I've found is, especially after the intense rides, it's a lot harder to ride the next day. I'm going to assume it's because I'm not eating right after a workout. I usually finish and drink a lot of water and then don't usually eat much (because the water is sort of filling). Is there anything I should eat special before and after these sorts of workouts that would help?

Use the fact that after 13 minutes the mass is now .0425. Put that into the y=Cert equation with C as the initial amount and y as the final amount (half the origional). Solve for r. Then go back and use your new value for r to solve.

For physics class we had to make these journal entries about different physics principles. My best friend and I decided it would be great to do them as a movie. We got four of them done and needed one more so we took the camera to a huge, snow-covered hill and built a three foor ramp at the bottom. We decided to demostrate momentum so he went to the top of the hill while I stood at the base of the ramp. I wasn't aware at the time but he took a running start from the top and was going pretty fast by the time he got to the bottom. The goal was for me to try and stop him (to show transfer of momentum) but he just leveled me. Completely drove me into the ground. We got that one on video. Later that day we built the ramp up higher to see what cool stuff we could do. It got to be so we could get a good four or five feet off air of the ramp. He went down first and landed but never got up for some reason. I was going down behind and didn't stop (figured he'd move). Well, he didn't and I ended up landing on him from several feet in the air while still going with good forward speed. He couldn't walk the rest of the day so we went home. Next day we went back and I was repairing and building up the ramp. He decided to come down without telling me. I heard something so I turned around, saw him coming and tried to jump over him. He caught me feet, took them out from under me and in a flash my face slammed into the ground and I was being dragged down the hill. There were some classic moments sledding.

Some parts are calculus based, but most aren't. I don't really know what IB Physics cover though.

http://www.prenhall.com/giancoli/ We use the top left picture.

Speaking of which, where are Voyager 1 and 2 now? Can they still send pictures or are they just defunct peices of metal floating around out there?

Hey insane, what type of car is that on your desktop?

Tangent isn't a continuous function. Is there a circumfrence or diameter for the other two wheels? And which one is the "right" pedal and which is the "left"? Your basic method is going to be to see how many revolutions (or what fraction of a revolution) the big wheel is going to have to make to move 1m. Then translate that into revolutions for the small wheel with the pedals. Find where that puts the desired pedal and then use trig ratios to find how far off the ground it is. I'm not sure there is enough information given in the problem to solve that right now.

Oh yeah, stupid mistake. Don't know why I put 0 instead of -1. Sorry about that.

Ok, the picture helps. According to the picture, the height of the pedal will vary between 100mm and 440mm depending on where it is. It can never excede these values and will hit every value in between. Therefore, sine and cosine are helpful because they range from 0 to 1 with every value in between but none outside. The sine curve, for example, starts at 0 for 0 degrees and then increases to 1 at 90 degrees and then beck to 0 for 180 degrees and so on for whatever values of x degrees you can want. The pedal starts at 100mm and goes up to 440mm and then back to 100mm and so on. Therefore, the sine curve (or cosine) can be manipulated to give the height of the pedal so that it remains within the bounds that the picture shows the pedal needs to remain in.

You might need to add a little more of the previous problems first. It looks like this one picks up at the end of it and I don't exactly know what is going on with the pedals.

Yeah, I'm taking that tomorrow too. We sat around for a while in a little study session after school. The goal was to learn solids of revolutions (because I had missed those few classes) and we got through the basics but we got sidetracked and somehow ended up deriving the equation for the volume of a sphere from what we had just been talking about. From there it drifted away from math entirely. We didn't really get much done. Whatever though, I've never gotten very nervous about these tests. Good luck to you though.

I assume you meant 9^2 and not 9^1? That's quite a step from the last sentence. But basicly it says that when you're adding up the columns after multiplying, the ones column will always be a one because 9^2=81. Now you've lost me. I don't understand how those two are equal. And what does mod 2 mean versus mod 20 or mod 1000?

Impressive, but would you be able to explain that to someone like me with only a background of around calc II?

Ok, this is the first thing that came to mind when I looked at that one from Digital Fortress, so I thought I'd just make a quick sentence out of it. If it's too hard I can do a few things to make it easier, but if I put all the right spaces in it would probably be too easy. Oh well, just for fun then: 1200192243108315876751323432481275343267512007531083243759 72243108120019275972751587759727510837683277510831921323192 On second thought, that shouldn't be too hard at all. By the way, that's one continuous line of numbers split up only so it fits into the post.

In quick response to Dak and for everyone in the thread, the tags are [ hide ] and [ /hide ], minus spaces of course.

Since we're on the topic of crytpo so much now, here's one that I could never figure out (but then again I never really tried). It's from Dan Brown's "Digital Fortress" so I'm sure it could easily be googled, but that's not that point. So, on to the code: 128-10-93-85-10-128-98-112-6-6-25-126-39-1-68-78 This one's a little different than the others so far, but good luck.

It was a squirtgun?

I looked up quaternions on mathworld (since this thread isn't going anywhere else), but I don't follow what that means. Could anyone explain it in simpler terms?

But on the interval of 6-14 the funtion ranges between 80 and 90. Nope. So I only need to set the first derivitive equal to zero?

Also, can anyone tell me what [math]\oint[/math] is called and/or means?

Yes, this is homework, but it's under the calc section because I'm looking for a discussion more than just quick answers. 1) a) Find the everage temperature, to the nearest degree, between t=6 and t=14 for all [math]F(t)=80-10\cos(\frac{\pi t}{12})[/math] [math]\frac{1}{8}\int_{6}^{14}80-10\cos(\frac{\pi t}{12})dt[/math] [math]\frac{1}{8}[80t+\frac{10\pi}{12}\sin(\frac{\pi t}{12})]_{6}^{14}[/math] [math]\frac{5}{4}[8t+\frac{\pi}{12}\sin(\frac{\pi t}{12})]_{6}^{14}[/math] [math]\frac{5}{4}[8(14)+\frac{\pi}{12}\sin(\frac{7\pi}{6})]-\frac{5}{4}[8(6)+\frac{\pi}{12}\sin(\frac{\pi}{2})][/math] This yeilds 79.5 or so. The only problem [math]80 \leq F(t) \leq 90[/math], so the average can't be in the 70s. Can anyone find where the mistake is? b)An air conditioner cooled the house whenever the outside temperature was at or above 78 degrees. For what values of t was the air conditioner on? [math]78=80-10\sin(\frac{\pi t}{12})[/math] [math].2=\sin(\frac{\pi t}{12})[/math] [math]\frac{12\arcsin(.2)}{\pi}=t[/math] That gives me .7 something, which again is wrong acording to the graph. Plus, I should have 2 answers. Any clues on where this one went wrong? c)Cost of cooling this house is $0.05 per hour for each degree over 78. What was the total cost for cooling the house for one 24 hour period? I plan to get the answers from part b and use them as the bounds when I integrate the function. That would give me the area under the curve for the tiime the air conditioner was on. Then multiply that by .05 and that should give the cost, right? 2) a) A curve is defined by 2y3+6x2y-12x2+6y=1. Show that [math]\frac{dy}{dx}=\frac{4x-2xy}{x^2+y^2+1}[/math] I got that part. b) Write an equation for each horizontal tangent line to the curve. A horizontal tangent line would be were the second derivitive would be 0, but taking the second derivitive of that seems insane. Is there an easier way to do this one? Also, if there are any easier/quicker meathods to any previous problems could you also point those out. Thanks.