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geekhack Community => Off Topic => Topic started by: Computer-Lab in Basement on Wed, 22 July 2015, 12:26:39
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so is there any way to solve for p in this equation?
p=(g*p)+c
p is on both sides, so i'm guessing no?
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i think this is how you do it, I'm out of college now so everything is rusty :)
p=gp+c
0=gp+c-p
0=(gp-p)+c
0=(g-1)p+c
-c=(g-1)p
-c/(g-1)=p
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Divide per p:
1= g+c/p
1-g=c/p
not sure if that's helping.
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i think this is how you do it, I'm out of college now so everything is rusty :)
p=gp+c
0=gp+c-p
0=(gp-p)+c
0=(g-1)p+c
-c=(g-1)p
-c/(g-1)=p
This one is correct, I believe. -c/(g-1) where g ≠ 1.
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i think this is how you do it, I'm out of college now so everything is rusty :)
p=gp+c
0=gp+c-p
0=(gp-p)+c
0=(g-1)p+c
-c=(g-1)p
-c/(g-1)=p
how did you get from the second line to the third line? i'm confused...
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i think this is how you do it, I'm out of college now so everything is rusty :)
p=gp+c
0=gp+c-p
0=(gp-p)+c
0=(g-1)p+c
-c=(g-1)p
-c/(g-1)=p
how did you get from the second line to the third line? i'm confused...
just think of the minus p like a negative p that you can move around. He's just moving it so he can pull out the "p" from both "p" and "gp" so there's only 1 "p"
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i think this is how you do it, I'm out of college now so everything is rusty :)
p=gp+c
0=gp+c-p
0=(gp-p)+c
0=(g-1)p+c
-c=(g-1)p
-c/(g-1)=p
how did you get from the second line to the third line? i'm confused...
i just grouped together the p terms in parentheses. so from the second to third step it is basically this.
0=gp+c-1*p
0=gp-1*p+c Rearranged
0=(gp-1*p)+c Parentheses
0=(g-1)*p+c
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i think this is how you do it, I'm out of college now so everything is rusty :)
p=gp+c
0=gp+c-p
0=(gp-p)+c
0=(g-1)p+c
-c=(g-1)p
-c/(g-1)=p
how did you get from the second line to the third line? i'm confused...
just think of the minus p like a negative p that you can move around. He's just moving it so he can pull out the "p" from both "p" and "gp" so there's only 1 "p"
ok that makes sense, thanks
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thanks for the help guys, my boss is happy...
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thanks for the help guys, my boss is happy...
:thumb:
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is this what math is like
****in hell meth sounds like more fun than this
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Computer lab.. how old are ya.. please tell me..
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are you trolling us ... I thought you were programmer by trade.. (http://www.cute-factor.com/images/smilies/onion/efb50fe2.gif)
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Divide per p:
1= g+c/p
1-g=c/p
not sure if that's helping.
This is exactly the same as the answer above you by (aznairjordon) if you pull a negative 1 from (1-g)
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thanks for the help guys, my boss is happy...
You got paid to figure that out?
I'd happily be your company's contract remedial algebra consultant.
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thanks for the help guys, my boss is happy...
You got paid to figure that out?
I'd happily be your company's contract remedial algebra consultant.
Don't listen to this mean-ee Computer Labz...
Math is good for u and fun 2 learn..
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thanks for the help guys, my boss is happy...
You got paid to figure that out?
I'd happily be your company's contract remedial algebra consultant.
Don't listen to this mean-ee Computer Labz...
Math is good for u and fun 2 learn..
That's why I offered to consult, perhaps tutoring would be better.