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09-06-2009, 01:15 PM #1
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robjdixon
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Maths Help Plox

Hey guys, I know many of you are geniuses. Can you help with my maths please?

I have like 4 questions that I can't do.

1.
Make f the subject of

1/f = 1/u + 1/v

Basically I have no idea with this one. If I multiply by fuv, then it goes completely wrong. I remember something about just inverting it somehow so it is f/1 = u/1 + v/1, but don't know if this is possible

2. Simultaneous equations. I need to solve these.

2x - y+3=0 and y² - 5x²=20

Thanks

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09-06-2009, 01:38 PM #2
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laserpyro
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Re: Maths Help Plox

i think this may be right
1/f=1/u+1/v
1/f=1/uv
1=f 1/uv (times both sides by f/1)
UV=F (Times both sides by uv/1)
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09-06-2009, 04:01 PM #3
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daguin
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Re: Maths Help Plox

EVERY "C" on my transcripts is from a math class.

Peace,
dave

09-06-2009, 10:30 PM #4
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Re: Maths Help Plox

2x - y + 3=0 and y² - 5x² = 20
Separate variables of one equation: y = 2x + 3
Substitute to eliminate variable: y² - 5x² = 20 = (2x + 3)² - 5x²
Simplify: (2x + 3)² - 5x² = 4x² + 12x + 9 - 5x² = -x² +12x + 9 = 20
Solve: -x² +12x = 11
By inspection: x = 1.
Plug back in: y = 2x + 3 = 2(1) + 3 = 5
x = 1, y = 5

The other problem:
1/f = 1/u + 1/v
Multiply each part of the right side by 1: 1/f = (1/u)*(v/v) + (1/v)*(u/u)
Simplify: 1/f = v/(uv) + u/(uv) = (u + v)/(uv)
Invert both sides: f = (uv)/(u + v).
That seems simple enough to me, do they want it any more simplified than that?

09-06-2009, 10:38 PM #5
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Re: Maths Help Plox

Quote:
 Originally Posted by laserpyro i think this may be right 1/f=1/u+1/v 1/f=1/uv <-----You can't do this 1=f 1/uv (times both sides by f/1) UV=F (Times both sides by uv/1)
1/u + 1/v doesn't equal 1/(uv)

For example, 1/4 + 1/8 doesn't equal 1/32, it equals 3/8.

09-07-2009, 02:19 AM #6
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Cyparagon
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Re: Maths Help Plox

My results mirror yours for the first problem. But for the second, you didn't use the quadratic equation. You're right up untill -x² +12x = 11. From there we have

-x² +12x - 11 = 0

-b ± √(b² -4ac)
---- -2a

-12 ± √(12² -4(-1)(-11))
--------2(-1)

12 ± √(144 - 4x11)
--------2

12 ± √(100)
-----2

12 ± 10
---2

Which solves as the points (1,5) AND (11,25).

This becomes more evident when we graph both of the equations. The left graph is as it appears with xmin/max of 40 and ymin/max of 40. The right is zoomed in to better show where the lines intersect.
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09-07-2009, 04:07 AM #7
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nikokapo
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Re: Maths Help Plox

1. 1/f = 1/u + 1/v

You can just invert it (everything ^-1): f = u+v
But you can always just:

1 = (1/u + 1/v)*f --> 1/(1/u + 1/v) = f --> u + v = f

I don't know if that's what you were looking for here, I can do math in Spanish...now... in English...

2. PBD's approach is correct.

09-07-2009, 05:12 AM #8
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Cyparagon
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Re: Maths Help Plox

Quote:
 Originally Posted by nikokapo 1. 1/f = 1/u + 1/v You can just invert it (everything ^-1): f = u+v
It doesn't work like that. Suppose the numbers for f, u, and v were 2, 4, and 4 respectively. 1/2 = 1/4 + 1/4, but 2 ≠ 4 + 4

You don't invert "everything," per se, you invert both sides, but that's just a shortcut for:

w/x=y/z

(*z):
wz/x=y

(*x):
wz=xy

(÷w):
z=xy/w

(÷y)
z/y=x/w

Now let's plug my number set (2,4,4) which works in the original equation into PBG's answer:

f = (uv)/(u + v)

2 = (4*4)/(4 + 4)
2 = 16/8
Bingo

Last edited by Cyparagon; 09-07-2009 at 05:29 AM. Reason: grammar

09-07-2009, 06:02 AM #9
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Re: Maths Help Plox

Quote:
 Originally Posted by Cyparagon My results mirror yours for the first problem. But for the second, you didn't use the quadratic equation. You're right up untill -x² +12x = 11. From there we have -x² +12x - 11 = 0 -b ± √(b² -4ac) ---- -2a -12 ± √(12² -4(-1)(-11)) --------2(-1) 12 ± √(144 - 4x11) --------2 12 ± √(100) -----2 12 ± 10 ---2 Which solves as the points (1,5) AND (11,25). This becomes more evident when we graph both of the equations. The left graph is as it appears with xmin/max of 40 and ymin/max of 40. The right is zoomed in to better show where the lines intersect.
Doh, there are 2 valid solutions. The first one (x=1,y=5) is so obvious though, but you're correct, either is perfectly valid.

Of course in our wonderful technological world, a calculator will solve all that for you, and save you time for more important things. But you should definitely learn to do it all by hand.

09-07-2009, 08:03 AM #10
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Re: Maths Help Plox

Yeah, my math courses were a while back. Glad to see you were put on the right track!
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09-07-2009, 11:41 AM #11
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Re: Maths Help Plox

Quote:
 Originally Posted by daguin EVERY "C" on my transcripts is from a math class. Peace, dave
Hey Dave...

does that mean all "other" subjects were better or worse....

Jerry
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09-07-2009, 03:15 PM #12
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Elektrotechniker
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Re: Maths Help Plox

1/f=1/u+1/v i dont think u can invert the part 1/u + 1/v without first doing v+u/v*u.
So u do first v+u/v*u and then invert so it would be v*u/v+u. Like
1/f = 1/u + 1/v =>
1/f = v+u/v*u =>
f = v*u/v+u

It's the same process as trying to find a resistance to replace 2 others which are in parallel connection.
1/R = 1/R1 + 1/R2 =>
1/R = R1+R2/R1*R2 =>
R = R1*R2/R1+R2

Last edited by Elektrotechniker; 09-07-2009 at 03:18 PM.

09-09-2009, 05:09 PM #13
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nikokapo
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Re: Maths Help Plox

Oh Chuck I just realized I made a BIG mitake, LOL I can't believe I passed Math Analysis II and couldn't invert that simple equation

1/f = 1/u + 1/v

f = 1/(1/u + 1/v)

Sorry for the previous mistake, man

09-09-2009, 06:54 PM #14
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Re: Maths Help Plox

BIG doesn't begin to describe it...
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