Asked by Riley
I am reposting this because I still don't understand it.
I need help with this problem. this is what I got so far:
P(x)=12x-(x^2/500)-680-4x-0.01x^2
P'(x)=12-0.004x-4-0.02x
P'(x)= -0.024x-8
x=8/0.024
My book says x=8/0.024 which somehow become 1000/3. They are the same when solved but how did they come up with 1000/3?
Someone answered and said to multiply numerator and denomerator by 1000/8. But WHY do I need to this? And how do I know what ot multiply by? I don't get it.
I need help with this problem. this is what I got so far:
P(x)=12x-(x^2/500)-680-4x-0.01x^2
P'(x)=12-0.004x-4-0.02x
P'(x)= -0.024x-8
x=8/0.024
My book says x=8/0.024 which somehow become 1000/3. They are the same when solved but how did they come up with 1000/3?
Someone answered and said to multiply numerator and denomerator by 1000/8. But WHY do I need to this? And how do I know what ot multiply by? I don't get it.
Answers
Answered by
Reiny
we usually don't have decimals in fractions, either we use a fraction or we use a decimal
so
8/0.024
= 8/0.024 * (1000/1000)
= 8000/24 , now reduce by a factor of 8
= 1000/3
I believe it was drwls who suggested to multiply numerator and denomerator by 1000/8.
He merely combined both of my above steps into one operation.
(I am somewhat surprised that, taking Calculus, you are confused by such a trivial operation.)
so
8/0.024
= 8/0.024 * (1000/1000)
= 8000/24 , now reduce by a factor of 8
= 1000/3
I believe it was drwls who suggested to multiply numerator and denomerator by 1000/8.
He merely combined both of my above steps into one operation.
(I am somewhat surprised that, taking Calculus, you are confused by such a trivial operation.)
Answered by
Riley
Thanks I get it now. I just couldn't figure out why they got 1000/3 and I got 8/0.024. I don't know why I didn't see that, just simple algebra.
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