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April 17, 2026, 03:39:43 pm

Author Topic: VCAA 08 q9c  (Read 1085 times)  Share 

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lacoste

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VCAA 08 q9c
« on: November 05, 2009, 09:52:48 pm »
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Do we actually have to find what x is?

nala

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Re: VCAA 08 q9c
« Reply #1 on: November 05, 2009, 09:57:58 pm »
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We're required to find the value of x which gives the minimum, however we're not required to find this minimum. That's how I understand it. :)

lacoste

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Re: VCAA 08 q9c
« Reply #2 on: November 05, 2009, 09:59:48 pm »
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We're required to find the value of x which gives the minimum, however we're not required to find this minimum. That's how I understand it. :)

what the hell does that mean, finding the minimum but not the minimum ? :S

when you did it did you get 4000^1/3 by hand ?

cheers

nala

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Re: VCAA 08 q9c
« Reply #3 on: November 05, 2009, 10:03:48 pm »
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The x-value that you find (and I did get the same answer as you), when substituted back into the SA equation from part b, will give the minimum SA, but we don't need to do this. The x-value is the length on the triangle edges.

is that making any sense? lol

lacoste

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Re: VCAA 08 q9c
« Reply #4 on: November 05, 2009, 10:05:48 pm »
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so what is the actual answer to c?

nala

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Re: VCAA 08 q9c
« Reply #5 on: November 05, 2009, 10:07:59 pm »
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The value of x is 4000^(1/3).

lacoste

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Re: VCAA 08 q9c
« Reply #6 on: November 05, 2009, 10:09:42 pm »
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The value of x is 4000^(1/3).

yeh thanks, can you show me the working out on how to get this?

im stuck on that so i used the calc lol

lacoste

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Re: VCAA 08 q9c
« Reply #7 on: November 05, 2009, 10:14:39 pm »
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gooooooot it nvm cheers ~!!!!!!!!!!111