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March 24th, 2013, 08:33 AM   #1
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Transposition problem

D = (2(s-an)) / (n(n-l)) Solve for a

Next I got

Dn(n-l) = 2(s-an)

I have no idea is this correct or what I should do from here.
Appreciate any help.
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March 24th, 2013, 08:58 AM   #2
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Re: Transposition problem

Quote:
Originally Posted by adamfoley
D = (2(s-an)) / (n(n-l)) Solve for a
Next I got
Dn(n-l) = 2(s-an)
I have no idea is this correct or what I should do from here.
It is correct...WHY are you not sure?

You need to isolate the "a"; complete the right side:
Dn(n - l) = 2s - 2an
Isolate the "a" term:
2an = 2s - Dn(n - l)
Divide by 2n:
a = (2s - Dn(n - l)) / (2n)

When you say "I have no idea.....", what do you mean?
Don't understand your math teacher? Skip classes?
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March 24th, 2013, 10:26 AM   #3
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Re: Transposition problem

Thanks Denis,

I have a page of transposition questions with no answers, I did the question you answered and got the exact same answer but what's confusing me is
wolfram alpha's answer of a = 1/2 d (l-n) + s/d
I assumed I was going wrong somewhere. Btw I don't have a teacher and I am not in school
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March 24th, 2013, 01:09 PM   #4
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Re: Transposition problem

Quote:
Originally Posted by adamfoley
I have a page of transposition questions with no answers,
I did the question you answered and got the exact same answer but what's confusing me is
wolfram alphas answer of a = 1/2 d (l-n) + s/d
That's same answer, arranged differently. Last term should be s/n (NOT s/d)
Quote:
Btw I don't have a teacher and I am not in school
OK. That changes things some
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March 25th, 2013, 03:50 AM   #5
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I suspect the original equation was d = (2(s-an)) / (n(n-1)).
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March 25th, 2013, 06:12 AM   #6
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Re:

Quote:
Originally Posted by skipjack
I suspect the original equation was d = (2(s-an)) / (n(n-1)).
Agree. However, since solving for a, then easier/faster if:
let k = n(n - 1)
d = 2(s - an) / k
.....

On a "timed test", do you think this would be ok? I'd sure do it!!
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February 12th, 2015, 03:16 AM   #7
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Hi Guys,

Can someone help me get L from this formula please?

i= v_s/R (1-e^((-Rt)⁄L) )
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February 12th, 2015, 03:55 AM   #8
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You should be thinking about what you need to do. To "solve for L" means to get L on one side of the equation, by itself, with all other numbers and letters on the other side. And you do that by "reversing" what has been done to L. If you were given a value of L (as well as the other values in the formula) what you do to calculate i?

Using the usual laws of arithmetic and algebra, to find i, you would first find "-Rt" then divide L into that. Next, take the exponential, subtract it from 1, and finally multiply by v_s/R.

To "reverse" that, do the opposite of each operation, in the opposite order.
-
That is, since the last thing you would do in calculating this was "multiply by v_s/R", the first thing you should do to solve for L is "divide by v_s/R". And, of course, to keep the equation balanced, you must always do the same thing to both sides of the equation.

So to solve for L, the first thing you should do is divide both sides by v_s/R which is te same as "multiply by R/v_s. That gives iR/v_s= 1- e^{-Rt/L}.

Now, since this says "subtract e^{-Rt/L} from 1, instead we add it to both sides. That leaves us with iR/v_s+ e^{-Rt/L}= 1.

The left side of that now has iR/v_s add to the term with L in it so subtract iR/v_s from both sides to get e^{-Rt/L}= 1- iR/v_s.

Here, on the left, we have and exponential. You should have learned that the "inverse function" to the exponential function is the natural logarithm. To "reverse" that exponential, take the logarithm of both sides:
-Rt/L= ln(1- iR/v_s).

We are almost done. Since L is in the denominator of a fraction, undo that by taking the reciprocal: L/(-Rt)= 1/ln(1- iR/v_s)

Finally, L has been divided by -Rt so multiply both sides by -Rt:
L= -Rt/ln(1- iR/v_s)
i= = v_s/R (1-e^((-Rt)⁄L) )
Thanks from brezzy86
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February 12th, 2015, 04:40 AM   #9
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Thanks a million mate.
That was giving me a headache!

I couldn't get the L by itself as I didn't have the Rt before the ln.
I thought ln has to go before everything once it is taken over from the rhs of the equation.
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