Author Topic: help (Read 7673 times) Tweet Share

khalil · « **on:** July 01, 2009, 09:31:10 am »

At a factory the time, T seconds, spent in producing a certain size metal component is related to weight, w kg, by T= k+2w^2, where k is a constant.

c) write an expression for the average time (in seconds per kg)

Thanks

TrueTears · « **Reply #1 on:** July 01, 2009, 03:43:55 pm »

We need $\frac{dT}{dw}$

$\frac{dT}{dw} = 4w s/kg$

khalil · « **Reply #2 on:** July 01, 2009, 06:50:41 pm »

I'm sorry but thats wrong.
I wish it were that easy.
The answer states: 25/w+2w

TrueTears · « **Reply #3 on:** July 01, 2009, 06:56:33 pm »

Clearly with the amount of information you provided, that is all I can come up with.

Provide the full question please.

There is no way in hell you can get that answer with the amount of information you gave me.

khalil · « **Reply #4 on:** July 02, 2009, 02:37:25 pm »

Sorry, a) also states that if a 5kg component takes 75 seconds to produce, find k.

cobby · « **Reply #5 on:** July 02, 2009, 02:51:58 pm »

Quote from: khalil on July 02, 2009, 02:37:25 pm

Sorry, a) also states that if a 5kg component takes 75 seconds to produce, find k.

whats b?

NE2000 · « **Reply #6 on:** July 02, 2009, 02:55:41 pm »

k = 25

T is given (T = 25 + 2w^2)

Divide T by w (T/w [in s/kg] = 25/w + 2w)

You get your answer (in s/kg)

lolz

khalil · « **Reply #7 on:** July 02, 2009, 04:43:31 pm »

Thanks mate

khalil · « **Reply #8 on:** July 04, 2009, 07:17:14 pm »

How would you simplify (sinx+cosx)/(sinx-cosx) to make (tanx+1)/(tanx-1)?

TrueTears · « **Reply #9 on:** July 04, 2009, 07:22:56 pm »

$tan(x) + 1 = \frac{sinx}{cosx} + 1 = \frac{cosx + sinx}{cosx}$

$tan(x) - 1 = \frac{sinx - cosx}{cosx}$

$\frac{tanx+1}{tanx-1} = \frac{\frac{cosx + sinx}{cosx}}{\frac{sinx - cosx}{cosx}}$

Can you see the pattern?

[Hint: divide your numerator and denominator by cosx and backtrack

]

khalil · « **Reply #10 on:** July 04, 2009, 08:00:29 pm »

Thanks once again !

khalil · « **Reply #11 on:** July 04, 2009, 08:18:51 pm »

Also, how would you simplify (tanx+1)/(tanx-1) to (tanx-1)/(tanx-1) + 2/(tanx-1)

TrueTears · « **Reply #12 on:** July 04, 2009, 08:22:45 pm »

Quote from: khalil on July 04, 2009, 08:18:51 pm

Also, how would you simplify (tanx+1)/(tanx-1) to (tanx-1)/(tanx-1) + 2/(tanx-1)

$1 + \frac{2}{tanx-1} = 1 + \frac{2}{\frac{sinx - cosx}{cosx}} = \frac{-cosx-sinx}{cosx-sinx} = \frac{-(cosx+sinx)}{cosx-sinx} = \frac{cosx+sinx}{sinx-cosx}$

again do you see the pattern?

[hint: first make $\frac{tanx+1}{tanx-1}$ into $\frac{sinx+cosx}{sinx-cosx}$ then backtrack.]

Mao · « **Reply #13 on:** July 04, 2009, 09:41:19 pm »

No need to do that

$\frac{\tan(x) +1 }{\tan(x) - 1} = \frac{\tan(x) - 1 + 2}{\tan(x) - 1} = \frac{\tan(x) - 1}{\tan(x) - 1} + \frac{2}{\tan(x)-1}$

khalil · « **Reply #14 on:** July 04, 2009, 10:13:20 pm »

Differentiatate (sinx+cosx)/(sinx-cosx) to 2/(sin2x+1)

Search the forums now!

We have moved!

Author Topic: help (Read 7673 times) Tweet Share

khalil

help

TrueTears

Re: help

khalil

Re: help

TrueTears

Re: help

khalil

Re: help

cobby

Re: help

NE2000

Re: help

khalil

Re: help

khalil

Re: help

TrueTears

Re: help

khalil

Re: help

khalil

Re: help

TrueTears

Re: help

Mao

Re: help

khalil

Re: help

Recent Posts

User:
Password: