Author Topic: Q's (Read 1856 times) Tweet Share

Hye · « **Reply #15 on:** April 07, 2010, 11:55:59 pm »

so y=1.3125, what does this mean? so it doesnt fit.

the.watchman · « **Reply #16 on:** April 08, 2010, 09:04:07 am »

Ok, so this means that the height of the archway over the 'corner' of the barge is 1.3125m, while the height of the barge is 1.7m, therefore the barge cannot fit. There's a diagram attached.

SO for part (c), the amount that needs to be removed is 1.7-1.3125 = 0.3875m worth of cargo, ok?

Hye · « **Reply #17 on:** June 28, 2010, 03:51:23 pm »

Differentiate:
N=100te^(-t/12)+500

the.watchman · « **Reply #18 on:** June 28, 2010, 03:54:42 pm »

$\frac{dN}{dt}=\frac{d}{dt}(100t) \cdot e^{\frac{-t}{12}} + \frac{d}{dt}(e^{\frac{-t}{12}}) \cdot 100t$

$\frac{dN}{dt}=100 \cdot e^{\frac{-t}{12}} - \frac{1}{12}e^{\frac{-t}{12}} \cdot 100t$

Factorise e^{-t/12} and numerical factors out

Hye · « **Reply #19 on:** July 06, 2010, 11:08:16 pm »

thanks

the cost of running a train at a constant speed of vkm/h is
C= 50 + v^2/100 dollars/hour

find an expression for the cost of an 800km journey

thanks

Hye · « **Reply #20 on:** July 07, 2010, 12:11:18 am »

The answer says C =800/v* (50+v^2/1000)

brightsky · « **Reply #21 on:** July 07, 2010, 12:20:09 am »

The time it takes for the train to travel 800km is $\frac{800}{v}$ hours.

Hence the cost is:

For one hour, it's $C = 50 + \frac{v^2}{100}$

Hence for $\frac{800}{v}$ hours, it's:

$C = \frac{800}{v} \cdot \left(50 + \frac{v^2}{100}\right)$

Whatlol · « **Reply #22 on:** July 07, 2010, 12:23:12 am »

Oh im sorry i misread it )=. ill delete my post.

Search the forums now!

We have moved!

Author Topic: Q's (Read 1856 times) Tweet Share

Hye

Re: Q's

the.watchman

Re: Q's

Hye

Re: Q's

the.watchman

Re: Q's

Hye

Re: Q's

Hye

Re: Q's

brightsky

Re: Q's

Whatlol

Re: Q's

Recent Posts

User:
Password: