anonymous
  • anonymous
ab=c Solve for a Do i need to divide both sides by b?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
yes
anonymous
  • anonymous
And that is all?
anonymous
  • anonymous
yes, so when you do that what will you get?

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anonymous
  • anonymous
a=a?
anonymous
  • anonymous
\[\frac{ab}{b}=\frac{c}{b}\] \[a=\frac{c}{b}\]
anonymous
  • anonymous
Oh! now i get it, can you set up another example so i can try?
anonymous
  • anonymous
Sure ax+y=c, solve for x
anonymous
  • anonymous
Ok, divide ax with a, then divide c with a getting ax/a+y=c/a. You then simplify that getting x+y=c/a. Then you take both sides minus y getting x=c/a-y
anonymous
  • anonymous
Is this correct?
anonymous
  • anonymous
Really close but we need to either get rid of the y first before divide or also divide the y by a
anonymous
  • anonymous
YES! Thanks for teaching me :)
anonymous
  • anonymous
aw :(
anonymous
  • anonymous
No, first start by subtracting y from both sides
anonymous
  • anonymous
Alright, so ax+y=c ax=c-y x/a=c/a-y x=c/a-y
anonymous
  • anonymous
remeber that when you divide the a you need to do it to all the terms
anonymous
  • anonymous
Where did you get z?
anonymous
  • anonymous
Oh right.. so x=c-y/a
anonymous
  • anonymous
\[x=\frac{c-y}{a}\]
anonymous
  • anonymous
zbay is correct
anonymous
  • anonymous
Inopeki got the answer i just put it in the equation maker for ease of the viewers
anonymous
  • anonymous
Yay! :D
anonymous
  • anonymous
inopeki you need to write x=(c-y)/a just to make it clearer. i see it now good work
anonymous
  • anonymous
Thanks for the help :) Is there any way you can make it a little harder to give me a challenge?
anonymous
  • anonymous
yea i will come up with one for you
anonymous
  • anonymous
Thanks.
anonymous
  • anonymous
\[\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\] Solve for T_2
anonymous
  • anonymous
umm.. ill try
anonymous
  • anonymous
Nobody mentioned division by zero ?!!
anonymous
  • anonymous
Good observation @ Foolformath
anonymous
  • anonymous
Well to begin with i need to multiply both sides with P2V2, getting P1V1/T1*P2V2, right?\[P _{1}V _{1}divT _{1}+P _{2}V _{2}=T2\]
anonymous
  • anonymous
I'll lay it out by steps for you
anonymous
  • anonymous
NOO. wait. Its
anonymous
  • anonymous
Cross multiplication
anonymous
  • anonymous
P1V1/T1*P2V2=T2
anonymous
  • anonymous
lcd = T_1T_2 \[T_2P_1V_1=V_2P_2T_1\] \[T_2=\frac{V_2P_2T_1}{P_1V_1}\]
anonymous
  • anonymous
Oh right, i see it now. You wanted me to put the things i was multiplying by(V2P2) in by the other things that were multiplied. That would mean my answer is right, but yours is simplified.
anonymous
  • anonymous
No?
anonymous
  • anonymous
No because you need to multiply both sides by the lcd of \[T_1T_2\] first then we cancel one of the varibles out because they are the same on top and bottom. then we can break T_2 apart from the rest of the left side by division, your answer would look close but in the end it would be wrong
anonymous
  • anonymous
What is an lcd? What makes T1 jump up?
anonymous
  • anonymous
lcd= lowest common denominator
anonymous
  • anonymous
What is that?
anonymous
  • anonymous
Is a denominator the 1 in T1?
anonymous
  • anonymous
The denominator is what is on the bottom of the fraction, so when we do the left side we get \[\frac{T_1T_2P_11V_1}{T_1}\] and because we have the same thing on the top and bottom we can cancel out the denominator leving us with \[T_2P_1V_1\] and we do the same operation on the right side
anonymous
  • anonymous
Why do we get T1 on both sides of the line?
anonymous
  • anonymous
because you are multiplying by t1t2 but on each side one of them will cancell
anonymous
  • anonymous
Ah
anonymous
  • anonymous
Good job that was a hard one
anonymous
  • anonymous
Thanks, when am i going to get to learn this in school?
anonymous
  • anonymous
what grade are you?
anonymous
  • anonymous
8th
anonymous
  • anonymous
Not sure when you will cover that but it's a chemistry equation compariing similar gasses to determine preasure, temp, or volume. You might touch it in highschool sometime.
anonymous
  • anonymous
Oh, cool. What should i do now?
anonymous
  • anonymous
PV=nrt solve for n
anonymous
  • anonymous
And then try this \[V=\frac{4}{3}\pi r^3\] Solve for r
anonymous
  • anonymous
Divide both sides by rt, getting PV/rt=n, am i right? By the way, why is PV upper case?
anonymous
  • anonymous
Zed, isnt that for vol├║mes of spheres or something? exept r should be ^3
anonymous
  • anonymous
good question and i don't have an answer for you but in that equation they are all upercase but the little n but i made a mistake and didn't capitilize the RT
anonymous
  • anonymous
But was i right?
anonymous
  • anonymous
Zed, im going to do yours now.
anonymous
  • anonymous
yes you were right and that was the ideal gas law by the way
anonymous
  • anonymous
Oh, that was a ^3! Divide both sides with (pi)r^2 getting V/(pi)r=4/3r, am i right sofar?
anonymous
  • anonymous
What does the law state?
anonymous
  • anonymous
okay so first step is to get r^3 by itself. what do you get?
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Ideal_gas_law
anonymous
  • anonymous
Wait. I take back my first answer. First you divide both sides by 4/3(pi) to isolate r^3
anonymous
  • anonymous
good. have you done powers before and their inverses?
anonymous
  • anonymous
I think so, what do you mean?
anonymous
  • anonymous
okay so we have \[r^3=\frac{3V}{4}\] how do I get rid of the cubed part?
anonymous
  • anonymous
Oh, i have done powers but i dont recall doing inverses, is that like squareroot and cubicroot? I guess we need to get the "cubicalroot" out of r^3.
anonymous
  • anonymous
We should be able to just divide r by r 2 times.
anonymous
  • anonymous
3* and by itself i mean the root.
anonymous
  • anonymous
not quite see let's try this example to help explain it \[x^2=4\]\[x=\sqrt{4}\]\[x=4^{\frac{1}{2}}\]\[x=2\]
anonymous
  • anonymous
What? Can you explain that?
anonymous
  • anonymous
okay so when we bring powers (eg 3) over the equation sign they turn to 1/powers (eg 1/3). if we look at squared when we bring it over it turns to 1/2 which is also known as the square root.
anonymous
  • anonymous
Why does it become 1/2?
anonymous
  • anonymous
the may be abit hard to follow but this is why. Let f(x)=y \[y=x^7\] \[\ln y=7 \ln x\] \[lnx=(lny)/7\] \[x=e^{(lny)/7}\] \[Inverse = f^{-1}(x)=e^{(lnx)/7}\] \[ f^{-1}(x)=e^{lnx^\frac{1}{7}}\]\[=x^\frac{1}{7}\]
anonymous
  • anonymous
Now usually when learning this we just accept this as fact that \[x^n=x^{\frac{1}{n}}\] where n is a positive integer
anonymous
  • anonymous
That was probably way too much information at once just use the rule in the above post and try to solve \[r^3=\frac{3V}{4}\]
anonymous
  • anonymous
So if r^2 becomes 1/2, then r^3 should become 1/3, right?
anonymous
  • anonymous
Correct
anonymous
  • anonymous
That means that i have to maake r^3 to 1/3 and convert 1/3 to 4/12(to merge it with the right side). i then convert the right sides fractions too, getting 9V/12. Now i should be able to merge them, right? so i get r=13/12?
anonymous
  • anonymous
I screwed that up right...
anonymous
  • anonymous
What?
anonymous
  • anonymous
Sorry we want to raise both sides by the power of (1/3)
anonymous
  • anonymous
r^1/3=3V^1/3/4?
anonymous
  • anonymous
so the left hand side will be \[(r^3)^{\frac{1}{3}}=r^{\frac{3}{3}}=r^1=r\]
anonymous
  • anonymous
I dont understand.
anonymous
  • anonymous
okay the right side is good. we have \[(\frac{3V}{4})^\frac{1}{3}\]
anonymous
  • anonymous
remember on the left hand side we had r^3 and when raised it to the power of 1/3 the powers need to be multiplied.
anonymous
  • anonymous
Oh! So its r^3*1/3?
anonymous
  • anonymous
yes exactly it's one of the rules for powers. here are others if you wanted to learn and practice http://www.math.com/school/subject2/lessons/S2U2L2DP.html
anonymous
  • anonymous
So the answer is r^3*1/3=3V^3/4
anonymous
  • anonymous
yes and that simplifies to r=(3V/4)^1/3
anonymous
  • anonymous
Oh!
anonymous
  • anonymous
Well I'm off to sleep. I'll be on tomorrow if you want more examples or help :)
anonymous
  • anonymous
Thanks for everything!

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