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

- anonymous

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

- katieb

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- UnkleRhaukus

nope

- anonymous

Which part isnt right?

- anonymous

Goddammit this is just hopeless..

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## More answers

- TuringTest

hi Inopeki
do not despair!
practice practice...

- TuringTest

what made you feel like putting a 13 there?
we were supposed to get a function like before

- anonymous

And didnt i?

- TuringTest

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.

- TuringTest

13+s/p=2
what do you get multiplying both sides by p?
do nothing else but that.

- anonymous

I can go check

- TuringTest

no, I was wrong, nevermind

- UnkleRhaukus

13+s/p=2 for p
11+s/p=0
11=-s/p
p=-s/11

- anonymous

See? This guy knows it. Why cant i?

- anonymous

Whatever. give me a new one

- UnkleRhaukus

You can with Time Inopeki

- TuringTest

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.

- anonymous

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

- TuringTest

a new one?
solve\[A=2\pi r h\] for A, r, and h
I need to get you to stop thinking in numbers.

- TuringTest

that's three problems, by the way.

- anonymous

is \[\pi\] 3.14 here or is it a variable?

- TuringTest

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.

- TuringTest

A=2(pi)rh

- anonymous

Ok, so the answer should not be a number?

- TuringTest

not at all

- anonymous

Interesting..

- TuringTest

it should be a function like
r=5t+4
it has some numbers, but not all.

- anonymous

One question. If i, in the answer have to give you As "definition", how can that not be a number?

- TuringTest

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.

- TuringTest

if I tell you the radius and height of our particular cylinder, only then can you get a number.

- anonymous

I just studied that in school! I thought we were talking algebra here..

- Akshay_Budhkar

@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

- TuringTest

Solving for a variable is algebra. They are all connected as you will learn.
Thanks Akshay!

- anonymous

But that still doesnt answer my question, how should i define A?

- TuringTest

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?

- anonymous

We can divide h with A, right?

- anonymous

No.. that cant be it.

- TuringTest

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?

- anonymous

The height.

- TuringTest

is that all?

- anonymous

Actually no

- TuringTest

what else?

- anonymous

pi^2

- anonymous

right?

- TuringTest

where did you see a squared term?

- anonymous

I have no idea.. i realize thats wrong now cause that is for getting the volume of a cylinder..

- anonymous

And that concludes that i am an utter and complete retard

- TuringTest

it's okay.
so again, what, in all, is r being multiplied by?

- TuringTest

c'mon, being hard on yourself never helped anything.

- anonymous

2 and then pi

- TuringTest

and...?

- anonymous

height?

- anonymous

wait!

- TuringTest

right, it's being multiplied by
2(pi)h
so that is what we need to divide by
no don;t take it back!

- anonymous

if you divide both sides by h, and both sides by pi you should isolate it so you have 2r, right?

- TuringTest

true!

- anonymous

Yeah.. pretty much what you said

- anonymous

But that doesnt mathematically give us anything..

- anonymous

Then you just divide both sides by 2 and get r

- TuringTest

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=?

- anonymous

What do you mean?

- TuringTest

what is on the other side of the equation when we solve for r?

- TuringTest

what did we divide both sides by?

- anonymous

2(åi)h?

- anonymous

(pi)*

- TuringTest

right. so what will A be divided by?

- anonymous

well we get r on both sides right?

- anonymous

2(pi)h

- TuringTest

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=...?

- anonymous

1

- TuringTest

where did you get 1 ?

- anonymous

Nevermind, my mind is messing up my mind, if you know what i mean

- TuringTest

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

- anonymous

Maybe i should catch some sleep, its 4:50 am here..

- TuringTest

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!

- anonymous

Good night, thank you for all the help!

- TuringTest

anytime :)

- anonymous

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.

- anonymous

Actually, just b.

- anonymous

I would then get a/b=c/b
Simplified, a=c/b

- TuringTest

^^^correct!
now get some sleep dude!
the mind needs to be fresh.

- anonymous

I did sleep, that answer was today.

- TuringTest

lol I thought you stayed up all night.
Told you sleeping on it would help.
now solve
ab=c
for b

- anonymous

Thats easy, both sides divided by a, you get b/a=c/a
Simplify: b=c/a

- anonymous

no!

- TuringTest

yes^^

- anonymous

actually, yes

- TuringTest

that's what I wanted to hear
since it's so easy
A=2πrh
for r

- anonymous

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

- TuringTest

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.

- anonymous

Yes!
Oh ok.

- TuringTest

sorry, meant
A/(2πh)=r

- TuringTest

anyway, slightly different one:
solve
5t=t/2+s
for s, then for t

- anonymous

So the whole bottom part in parenthesis?

- TuringTest

that's two problems^^^

- anonymous

Multiply both sides by 2+s, right?

- TuringTest

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?

- anonymous

Oh, i thought you meant t/(2+s)

- TuringTest

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

- anonymous

Right, so now i have 5t-t/2=s, am i right?

- TuringTest

totally, but can you simplify the left a little?

- anonymous

5t-t/(2)?

- anonymous

It cant be 4t/2

- TuringTest

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?

- anonymous

Nope

- TuringTest

well did you know that you can only add or subtract the numerators when the denominators are the same?

- anonymous

Yeah

- anonymous

Where did you get the 10?

- TuringTest

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

- anonymous

Oh, i knew that, i still dont see where the 10 and 9 come from...

- anonymous

Wait, i see now! 5t *2/2 should become 10/10 though...

- anonymous

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

- TuringTest

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.

- TuringTest

now solve it for t

- anonymous

5t=t/2+s?

- anonymous

I have to isolate t by getting away the 5, right?

- TuringTest

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

- anonymous

Divide both sides by 5?

- TuringTest

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?

- anonymous

10t=t+s, i guess?

- TuringTest

what about s? didn't it get multiplied by 2 as well?

- anonymous

oh right, 10t=t+2s

- TuringTest

so now you should be able to solve for t...

- anonymous

Now divide both sides by 10?

- anonymous

Then we get t=t+2s/(10)

- TuringTest

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?

- anonymous

I really dont know...

- TuringTest

hint: combine like terms, we want the t's together
10t=t+2s

- anonymous

subtract t from both of them?

- AravindG

turing can u hlp me?

- TuringTest

@Inopeki yes!
@arvind no
I have to do 2 weeks of spanish homework in one day, I'm multitasking sorry.

- anonymous

TT, shouldnt you do your homework first?

- TuringTest

I am doing it, this doesn't really distract me much, but arvind's problems can be kinda tricky, so that would distract me.

- anonymous

Oh, ok. Now he have 9t=2s

- TuringTest

good, now what?

- anonymous

We have no other choice but to divide them by 9. Do we get t=-4.5s then?

- TuringTest

how did it change to negative?
and you need to leave it as a fraction for now.
What did you do to both sides?

- TuringTest

you say you divided by 9 but \[2/9\neq -4.5\]

- anonymous

Yeah, i have no idea what happened there..

- TuringTest

so go again...
9t=2s

- TuringTest

leave as a fraction

- anonymous

How do i divide 2 by 9?

- TuringTest

just like I wrote it: 2/9

- anonymous

wait!

- anonymous

Oh nevermind.. so t=2/9s?

- TuringTest

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

- anonymous

2s=9t?

- TuringTest

good, what next?

- anonymous

Divide by 2?

- TuringTest

that would solve it for s, right?
we wanted to solve for t...

- anonymous

Oh right

- anonymous

Divide by 9

- TuringTest

yes, which gives what as the answer??

- anonymous

2s/9=t

- TuringTest

which is what we got before, right?
so mathematics is consistent, hooray!
slightly different one:
yz/(z+4)=7
solve for y

- anonymous

I see!

- anonymous

I have to eat, brb

- TuringTest

me too, good idea, breakfast!

- anonymous

Hah, i hade dinner.. You there?

- TuringTest

yep, where were we?
yz/(z+4)=7
solve for y

- anonymous

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

- TuringTest

good but you forgot to distribute the 7
yz=7(z+4)
do you know what I mean by distribute?

- anonymous

the parenthesis?

- TuringTest

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

- anonymous

I knew that :D

- TuringTest

so what should you have gotten after you multiplication?
yz/(z+4)=7

- anonymous

Oh well, yz/(z+4)=7----->yz=7(z+4) ----->y=7(4)? Can i do that?

- TuringTest

no you dropped the z
take it from
yz=7(z+4)
and remember to distribute the 7 to all terms in the parentheses

- anonymous

i have 7z and 7*4=yz

- anonymous

7z+7*4=yz

- TuringTest

good, now feel free to multiply the 7 and 4...

- anonymous

28

- TuringTest

write thew whole expression every time

- TuringTest

the*

- anonymous

yz=7z+28?

- anonymous

This must be like teaching physics to a mentally challenged monkey..

- anonymous

Without all the throwing of feces

- TuringTest

stupid server is very slow today :(
yz=7z+28
what can we do next to solve for y ?

- anonymous

Yeah :/ subtract 1z from both sides, getting y=6z+28

- TuringTest

you can't subtract z because it is being multiplied by y
what is the inverse of multiplication again?

- anonymous

Division, right. Divide both zs by 2?

- TuringTest

will that isolate y ?
why divide by 2 ?

- anonymous

I dont really know.. Maybe divide by z?

- TuringTest

yes, see you do know!

- anonymous

yz=7z+28----->y1=7z+28?

- anonymous

y*

- TuringTest

almost, but you didn't divide the right by z...
always gotta do BOTH sides

- anonymous

Ah,, y=7+28=35?

- TuringTest

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

- anonymous

y=7z+28/z?

- TuringTest

what about the 7z
now you forgot to divide that term by z...
gotta do both

- anonymous

But you said i shouldnt drop that z? y=7+28/z?

- TuringTest

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.

- anonymous

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.

- TuringTest

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)

- anonymous

Awesome!
Ok i got this..

- anonymous

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?

- anonymous

Im sorry for the delay, openstudy crashed for me

- TuringTest

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)

- anonymous

5(2+1/r+r)

- TuringTest

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)

- anonymous

2x(x)-1/x?

- TuringTest

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

- anonymous

2x(x)(-2x)(1/x)?

- TuringTest

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)=?

- anonymous

a? I dont know.. This is just hopeless! Do i really need all this for physics?!

- anonymous

1

- TuringTest

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)=?

- anonymous

2a?

- TuringTest

yes, good!
here are a few for practice
x(5/x)=?
3y(1/y)=?
z(2/3z)=?

- anonymous

5x
3y
2/3z

- TuringTest

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.

- anonymous

5
1
2/3
I dont know.. Do i need to know much more before i cant start with trigo or calc?

- TuringTest

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...

- anonymous

That sucks... Are you sure that it will take a year?

- TuringTest

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.

- anonymous

What book do you reccommend?

- anonymous

Is all the algebra i need included in khanacademy?

- TuringTest

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.

- anonymous

Khans practice place has unlimited problems, i think. Or did you mean that the subjects there arent enough?

- TuringTest

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.

- anonymous

I will, we have one universal book for all the math subjects we are supposed to cover that year. Should i just google algebra?

- anonymous

Algebra book*

- TuringTest

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.

- anonymous

Yeah, ill do that.

- anonymous

Where were we, algebra wise?

- TuringTest

pretty basic honestly.
solving linear equations, inverse operations, distribution, etc.
we haven't even done factoring yet, let alone quadratics.

- anonymous

So im pretty much still at square 1?

- TuringTest

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.

- anonymous

But even this is so hard.. How do you expect me to get advanced level algebra when i barely understand linear equations..

- TuringTest

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.

- anonymous

Just the first part? (Algebra properties)

- TuringTest

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.

- TuringTest

once you know them*

- anonymous

Alright.

- anonymous

I think i know all the first ones

- TuringTest

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.

- anonymous

Thanks again for all your help!

- TuringTest

you're welcome
I'll see you when this site starts working properly again.
Keep it up, and good luck!

- anonymous

Thanks :)

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