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 Posted: Tue Jan 02, 2007 5:42 pm
Just kidding! I'm in 8th grade and I'm taking honors math, which means I'm a year ahead in math. The first marking period was all fine and dandy and I got a 97% average! Now it's the second half of the second marking period and I'm only doing okay. I have 85%.
I have a test on Thursday! (I know it seems like it's right after the break, but no worries, we have been working on this for a while.)
The test is on "transforming formulas". The way our text book describes it, it's really pretty easy. The way our teacher gives it to us, it's a lot harder. I was wondering if any of you know what I'm talking about, and if you do, could you tell me? A sample problem (I already know the answer to this) is A=1/2(b*h)
That's the equation for the area of a triangle, in case you were wondering... =]
But what we would have to do is find the value of let's say h. Then you distribute and use the properties of equality and receive the answer h=2A/b.
If that makes any sense, could you try to help me? I don't know if I was specific enough... If I wasn't, tell me and I'll find more sample problems.
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Posted: Tue Jan 02, 2007 5:49 pm
I'm pretty good at it. I'll be a bit rusty, as I haven't really focused on that for...4 years now. But I ran though a couple just now, so I'm all for it again.
What is the issue?
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Posted: Tue Jan 02, 2007 6:49 pm
If you had, for example, d=r(t), and you wanted to find t, you would try to make t and only t on one side. To do that, you would inverse what was done- since r and t are multiplied, you would divide by r to find t. You always do this to BOTH sides, so then it would look like:
d=r(t)
d/r=t
The A=(1/2)Bh would be like this:
A=(1/2)Bh (find h)
A/B=(1/2)h
2A/B=h (2A since divded by 1/2 is multiply by 2/1, which is 2)
All you do is the inverse. Just like doing normal equations, but with variables instead of real numbers.
Just wait until you take number theory smile
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Posted: Tue Jan 02, 2007 7:07 pm
MyOwnBestCritic If you had, for example, d=r(t), and you wanted to find t, you would try to make t and only t on one side. To do that, you would inverse what was done- since r and t are multiplied, you would divide by r to find t. You always do this to BOTH sides, so then it would look like: d=r(t) d/r=t The A=(1/2)Bh would be like this: A=(1/2)Bh (find h) A/B=(1/2)h 2A/B=h (2A since divded by 1/2 is multiply by 2/1, which is 2) All you do is the inverse. Just like doing normal equations, but with variables instead of real numbers. Just wait until you take number theory smile
I wish I could help, but, well... numbers scare me. Critic's post scares me...
So, good luck. Remember, the teacher is there to help. Ask for a study session during lunch, or something. It really helps.
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Posted: Thu Jan 04, 2007 7:12 am
Yeah... Maths... sweatdrop
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Posted: Thu Jan 04, 2007 3:21 pm
Yup. That is the basics of it.
Now try doing that while you have to take in the rotation of the Earth and gravity while solving for the final velocity of a rock, thrown up by someone with a certain amount of force.
The beauty of AP physics.
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Zabuza Demon of Mist Crew
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Posted: Fri Jan 05, 2007 4:17 am
You got it.
Zabuza: Oh, boy.
It looks confusing, but now that I took a second look at it.. Not so much. Then again, sometimes I don't trust my own judgement..
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Posted: Sun May 27, 2007 4:38 pm
You forgot pre-calc! (that's what I'm taking currently, even though I'm not doing too well)
I remember my 8th grade algebra...best I can tell you is that algebra proofs, despite being occasionally difficult, are a really good way of learning the properties. If you can manipulate a series of variables, you shouldn't have a problem manipulating numbers, because they're only easier and more straightforward.
But then again, I enjoy proofs. It's pretty much the only part of math I'm good at because it's about logic rather than actual numbers (which scare me)
what critic said will help you quite a bit. I know that memorizing all the properties is difficult, but once you "get" them they're very easy to use. If you have a problem understanding the theory behind something, feel free to ask, because I think I have a pretty decent understanding of the "why" of lower-level math. No, ask me why a discriminant is important, and I'll tell you that I still don't know what the hell "h" stands for blaugh
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Intellectual Elocutionist
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Posted: Tue Aug 21, 2007 2:21 am
Is it not just a question of rearranging the formula and doing the same thing to each side?
So:
A = 1/2 (b*h)
2A = b*h
h = 2A/b
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