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Inopeki
Is this right? Q: Solve 13+s/p=2 for p A: p(13)=2(13)+2=28? What is s? Ill say 2. 28/13≈2.15
Goddammit this is just hopeless..
hi Inopeki do not despair! practice practice...
what made you feel like putting a 13 there? we were supposed to get a function like before
you can't start plotting points like that until we've solved it for one of the variables. The answer to the question I was looking for should have 2 variables.
13+s/p=2 what do you get multiplying both sides by p? do nothing else but that.
no, I was wrong, nevermind
13+s/p=2 for p 11+s/p=0 11=-s/p p=-s/11
See? This guy knows it. Why cant i?
Whatever. give me a new one
You can with Time Inopeki
well, there's one way. btw Rhauckus I made up this problem for Inopeki to teach him how to solve for a variable @Inopeki, he has many years doing this, you need to practice basics. It will be like that for a while.
Well that sucks.. Still, if i really need to learn this i guess i will. Give me a new example so i can practise please
a new one? solve\[A=2\pi r h\] for A, r, and h I need to get you to stop thinking in numbers.
that's three problems, by the way.
is \[\pi\] 3.14 here or is it a variable?
no, it's about 3.14159..., that's why I didn't make it something to solve for. but don't think of it as a number, leave it as pi. The number does not matter.
Ok, so the answer should not be a number?
it should be a function like r=5t+4 it has some numbers, but not all.
One question. If i, in the answer have to give you As "definition", how can that not be a number?
The formula I gave you is for the surface area of the side of a cylinder. The surface area of any particular cylinder depends on its radius and length. So the formula covers all possible cylinders, but only for some particular cylinders will you get a certain value for the surface area. It is a function, this time of both r and h.
if I tell you the radius and height of our particular cylinder, only then can you get a number.
I just studied that in school! I thought we were talking algebra here..
@Turning you train guys well I will need your help for multivariable calculus, Laplace and stuff I will let you know then :D Great job is all I can say :D
Solving for a variable is algebra. They are all connected as you will learn. Thanks Akshay!
But that still doesnt answer my question, how should i define A?
in this case, A is already defined by\[A=2\pi rh\]that is basically the definition of the surface area of the side of a circular cylinder. No more definition is needed. I wanted you to catch that the first problem, solve for A, is already done for you: we already have A=(something) so it IS solved for A. So now solve it for r, if you can. what can we divide by to get r by itself?
We can divide h with A, right?
when you say 'divide h with A' I'm not sure what you mean. Do you mean divide both sides by A or divide both sides by h ??? One is okay, one is not. You know that division is the inverse of multiplication. So ask yourself 'what is r being multiplied by?' and divide by that. Well, what is r being multiplied by?
where did you see a squared term?
I have no idea.. i realize thats wrong now cause that is for getting the volume of a cylinder..
And that concludes that i am an utter and complete retard
it's okay. so again, what, in all, is r being multiplied by?
c'mon, being hard on yourself never helped anything.
right, it's being multiplied by 2(pi)h so that is what we need to divide by no don;t take it back!
if you divide both sides by h, and both sides by pi you should isolate it so you have 2r, right?
Yeah.. pretty much what you said
But that doesnt mathematically give us anything..
Then you just divide both sides by 2 and get r
sure it does, it gives us the answer not all answers in math are numbers, in fact I hate using numbers if I can help it. about your last post, yes, so what do we get for r=?
what is on the other side of the equation when we solve for r?
what did we divide both sides by?
right. so what will A be divided by?
well we get r on both sides right?
no, just one and all alone, that's what it means to solve for r again, yes to your last post so write out r=...?
where did you get 1 ?
Nevermind, my mind is messing up my mind, if you know what i mean
If you are tired then take a break here is the answer, watch is closely: \[A=2\pi rh\]divide both sides by\[2\pi h\]giving\[r=\frac{A}{2\pi h}\]If you want to keep going then try just something simpler with NO numbers:\[ab=c\]solve for a
Maybe i should catch some sleep, its 4:50 am here..
that is a very good idea then. sleep on it and try it in the morning. I can't tell you how many problems I've solved just by sleeping on them. Don't worry, step-by-step. You will get there I know. Goodnight!
Good night, thank you for all the help!
A=2πrh divide both sides by 2πh giving r=A2πh If you want to keep going then try just something simpler with NO numbers: ab=c solve for a When you say "solve for a", do you mean i should isolate it? I guess we would divide both sides by b and then c.
I would then get a/b=c/b Simplified, a=c/b
^^^correct! now get some sleep dude! the mind needs to be fresh.
I did sleep, that answer was today.
lol I thought you stayed up all night. Told you sleeping on it would help. now solve ab=c for b
Thats easy, both sides divided by a, you get b/a=c/a Simplify: b=c/a
that's what I wanted to hear since it's so easy A=2πrh for r
Divide both sides with 2(pi)h, you get A/2(pi)h=2(pi)rh/2(pi)h Simplified: A/2(pi)h=r
awesome! I know you got it but you should write it as A/(2π)h=r parentheses make it clear what is on the bottom.
sorry, meant A/(2πh)=r
anyway, slightly different one: solve 5t=t/2+s for s, then for t
So the whole bottom part in parenthesis?
that's two problems^^^
Multiply both sides by 2+s, right?
right, because everything in parentheses is on the bottom if you wrote 1/2x I don't know whether you mean\[\frac{1}{2x}\]or\[\frac{1}{2}x\]so parentheses are nice to clear things up think you can do the other? solve \[5t=\frac{t}{2}+s\] for s a good first step is to subtract t/2 from both sides. Then s will be isolated, right?
Oh, i thought you meant t/(2+s)
I would have used parentheses if I meant that. if I had meant that then you would have been right, so you'll get one like that soon. but for now solve it as it is for s
Right, so now i have 5t-t/2=s, am i right?
totally, but can you simplify the left a little?
so I mean what is 5t-t/2 as a fraction?\[5t-\frac{t}{2}=\frac{10t}{2}-\frac{t}{2}=\frac{9t}{2}\]do you know how to simplify fractions in this way?
well did you know that you can only add or subtract the numerators when the denominators are the same?
Where did you get the 10?
so we had to get the 5 to have a 2 in the denominator. How can we do this without changing our expression? the trick is to multiply the top and bottom by the same thing, in this case 2/2. Notice that 2/2=1 so it won't change our expression, because multiplying by 1 changes nothing.\[5t-\frac{t}{2}=5t*\frac{2}{2}-\frac{t}{2}=\frac{10t}{2}-\frac{t}{2}=\frac{9t}{2}\]make sure you understand the above. here's a couple pages on fractions http://www.mathsisfun.com/fractions_addition.html http://www.mathsisfun.com/numbers/fractions-mixed-addition.html
Oh, i knew that, i still dont see where the 10 and 9 come from...
Wait, i see now! 5t *2/2 should become 10/10 though...
No, wait, since the top is the only "t" part that is the only one being multiplied. THEN we take that minus t/2 and that simplifies it to 9t/2
no it doesn't because the original denominator is 1\[5t-\frac{t}{2}=\frac{5t}{1}*\frac{2}{2}-\frac{t}{2}=\frac{10t}{2}-\frac{t}{2}=\frac{10t-t}{2}=\frac{9t}{2}\]everything is secretly over 1, times 1, and to the power of 1. we just don't write\[a=\frac{1a^1}{1}\]because it is redundant, but they are the same. You figured it out as I was typing.
I have to isolate t by getting away the 5, right?
yeah, solve it for t you can take it from\[\frac{9t}{2}=s\]there is nothing wrong with starting there if you want. It's up to you
Divide both sides by 5?
my first step if I start from 5t=t/2+s would be to multiply both sides by 2 to get rid of the fraction can you do that?
what about s? didn't it get multiplied by 2 as well?
so now you should be able to solve for t...
Now divide both sides by 10?
Then we get t=t+2s/(10)
no, because the t on the right would be divided by 10 as well let's see what happens if we do that: t=t/10+s/5 I don't think that helps... any other ideas?
hint: combine like terms, we want the t's together 10t=t+2s
subtract t from both of them?
@Inopeki yes! @arvind no I have to do 2 weeks of spanish homework in one day, I'm multitasking sorry.
TT, shouldnt you do your homework first?
I am doing it, this doesn't really distract me much, but arvind's problems can be kinda tricky, so that would distract me.
Oh, ok. Now he have 9t=2s
We have no other choice but to divide them by 9. Do we get t=-4.5s then?
how did it change to negative? and you need to leave it as a fraction for now. What did you do to both sides?
you say you divided by 9 but \[2/9\neq -4.5\]
Yeah, i have no idea what happened there..
so go again... 9t=2s
How do i divide 2 by 9?
just like I wrote it: 2/9
Oh nevermind.. so t=2/9s?
yes, good, but it is better to write t=2s/9 that way I don't think you put the s in the denominator now do the same starting from the equation when it was solved for s: s=9t/2 solve for t
that would solve it for s, right? we wanted to solve for t...
yes, which gives what as the answer??
which is what we got before, right? so mathematics is consistent, hooray! slightly different one: yz/(z+4)=7 solve for y
me too, good idea, breakfast!
Hah, i hade dinner.. You there?
yep, where were we? yz/(z+4)=7 solve for y
yz/(z+4)=7 I need to isolate y so maybe i should multiply both sides with z+4. then im left with yz=7z+4
good but you forgot to distribute the 7 yz=7(z+4) do you know what I mean by distribute?
yes, what do you get by distributing the 7? remember\[a(b+c+d)=ab+ac+ad\]you need to multiply each term in the parentheses by what is on the outside
so what should you have gotten after you multiplication? yz/(z+4)=7
Oh well, yz/(z+4)=7----->yz=7(z+4) ----->y=7(4)? Can i do that?
no you dropped the z take it from yz=7(z+4) and remember to distribute the 7 to all terms in the parentheses
good, now feel free to multiply the 7 and 4...
write thew whole expression every time
This must be like teaching physics to a mentally challenged monkey..
Without all the throwing of feces
stupid server is very slow today :( yz=7z+28 what can we do next to solve for y ?
Yeah :/ subtract 1z from both sides, getting y=6z+28
you can't subtract z because it is being multiplied by y what is the inverse of multiplication again?
Division, right. Divide both zs by 2?
will that isolate y ? why divide by 2 ?
I dont really know.. Maybe divide by z?
yes, see you do know!
yz=7z+28----->y1=7z+28?
almost, but you didn't divide the right by z... always gotta do BOTH sides
again, almost, but you need to divide each term by z you didn't divide 28 and look now there is no z in our expression! we can't be dropping variables like that. try again and make sure to divide every term yz=7z+28
what about the 7z now you forgot to divide that term by z... gotta do both
But you said i shouldnt drop that z? y=7+28/z?
you didn't you still have z in the expression, so that is right. good job, but I'm gonna make you do it again for practice please show every step, and try to write it all out in one post. so from the top: yz/(z+4)=7 solve for y please show each step clearly.
yz/(z+4)=7 yz=7(z+4) Here i took multiply both sides by (z+4). yz=7z+28 Here i simplify 7(z+4). y=7+28/z Here i divide both sides by z. y=35/z Here i simplify the right side.
that's good but you can't simplify the right side like that at the end. you also don't need to tell me what each step is, just show. I can see what you are doing. yz/(z+4)=7 yz=7(z+4) yz=7z+28 y=7+28/z ^^^ this is sufficient as far as the simplifications you tried, I will show you a way to change the expression on the right that are valid:\[y=7+\frac{28}{z}=\frac{7z}{z}+\frac{28}{z}=\frac{7z+28}{z}\]this doesn't really make it more simple though, so there's not much point in it. remember we can only add when the denominators are the same, so here we had to multiply 7 by z/z first. still, good job! another: 4s(1-1/s)=5s+t solve for s (hint: first step is to distribute)
Awesome! Ok i got this..
4s(1-1/s)=5s+t 4(1-1/)=5s+t if i take the s in 4s and put it into the same parenthesis that has 1-1/s the two s'es should "negate" eachother since one is divided by and one is multiplied by, it would be like saying this: 2*4/2. Am i right so far?
Im sorry for the delay, openstudy crashed for me
no, you lost an s we should practice distribution I think 4s(1-1/s)=4s-4s(1/s)=4s-4 one of the s's cancelled because it was multiplied by 1/s the other did not new exercise distribute: 5r(2+1/r+r)
where did the r on the outside go? you didn't multiply anything in the parentheses at all :( watch very closely every step: 5r(2+1/r+r)=5r(2)+5r(1/r)+5r(r)=10r+5+5r^2 another distribute 2x(x-1/x)
not quite, you keep forgetting to multiply EVERY term. write out the middle step: 2x(x-1/x)=2x(x)-2x(1/x)=??? continue from there
no, sorry, I think you need to practice division.. 2x(x-1/x)=2x(x)-2x(1/x)=2x^2-2 what is a(1/a)=?
a? I dont know.. This is just hopeless! Do i really need all this for physics?!
there you go anything multiplied by its reciprocal is 1 a(1/a)=1 x(1/x)=1 3(1/3)=1 what about something like a(2/a)=?
yes, good! here are a few for practice x(5/x)=? 3y(1/y)=? z(2/3z)=?
inopeki a(2/a)=2 the a's cancel sorry I couldn't correct myself, but the lag... so now that you know that try x(5/x)=? 3y(1/y)=? z(2/3z)=? again, sorry for the confusion.
5 1 2/3 I dont know.. Do i need to know much more before i cant start with trigo or calc?
The first and last are right, but 3y(1/y)=3 and I hope you don't get discouraged, but you're gonna have to master basically all of algebra, which will take at least a year. the year after you will have to master trigonometry and pre-calculus. You will not be in calculus for some time, so get prepared to settle into algebra for a while...
That sucks... Are you sure that it will take a year?
maybe not if you really delve into it, but if you want to do that you better get an algebra book and read it cover-to-cover. That's what I did over the summer when I was 15. Like I said I knew very little before that and always considered myself bad a math.
What book do you reccommend?
Is all the algebra i need included in khanacademy?
I actually just used the algebra book from my high school, I don't know what it was called. But any decent textbook on algebra should be fine. ask around. I think much algebra is in khan, but I doubt that by itself is enough. You need to practice a LOT of problems to get good. I'm not sure khan has enough.
Khans practice place has unlimited problems, i think. Or did you mean that the subjects there arent enough?
If it has unlimited practice problems than do at least 75 from each section. I don't know khan I don't use it. I still recommend getting a book though. You need good reference material.
I will, we have one universal book for all the math subjects we are supposed to cover that year. Should i just google algebra?
Maybe, but I think you should post the question on OS 'what is a good algebra book from basics to advanced algebra' I'm sure somebody has an opinion on that here.
Where were we, algebra wise?
pretty basic honestly. solving linear equations, inverse operations, distribution, etc. we haven't even done factoring yet, let alone quadratics.
So im pretty much still at square 1?
for the most part, yes. don't get discouraged though, I remind you I refused to look at math until I was 15 there is plenty of time.
But even this is so hard.. How do you expect me to get advanced level algebra when i barely understand linear equations..
You need to learn the basic rules a bit better, then you will see it makes more sense. You are trying to bite off more than you can chew at once. Learn all these rules like your very name: http://www.capitan.k12.nm.us/teachers/shearerk/basic_rules_of_algebra.htm that is the first step. it will take more than 1 day to commit all this to memory, so be patient and practice along the way with khan as you try.
Just the first part? (Algebra properties)
No, everything on the page. Dividing fractions, adding unlike denominators, everything. All the basic rules are there, so one you know them the rest will be much much easier.
I think i know all the first ones
I bet you have seen them before, but this time you really need to commit all the rules to memory, not just get the basic idea. Like I said, keep working practice problems in the meantime. When you get stuck, look at the list and try to apply a rule, that way they will have more meaning. Just memorizing the list out of context is pretty difficult and seemingly pointless.
Thanks again for all your help!
you're welcome I'll see you when this site starts working properly again. Keep it up, and good luck!