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December 17, 2025, 04:54:41 am
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Author Topic: My proposed solutions for exam 2  (Read 7466 times)  Share 

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Tashi

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Re: My proposed solutions for exam 2
« Reply #15 on: November 07, 2011, 05:15:29 pm »
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Quote from: LittleMunckin on November 07, 2011, 05:09:20 pm
can you explain the last question for trig? because i got 100 more than that?

Did you get 42 from the height?
Then I found the volume of the smaller 'invisible' cone and subtracted that from the larger cone (to find the lighthouse)
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davidle_10

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Re: My proposed solutions for exam 2
« Reply #16 on: November 07, 2011, 05:19:37 pm »
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For part a) I used ratios to find h so h-18/h =2/3.5 h=42
Then to find the volume, just find the volume of the larger cone - volume of the smaller cone so
V=(pi x 3.5 x 3.5 x 42)/3 - (pi x 2 x2 x (42-18))/3
« Last Edit: November 07, 2011, 05:23:00 pm by davidle_10 »
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LittleMunckin

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Re: My proposed solutions for exam 2
« Reply #17 on: November 07, 2011, 05:19:44 pm »
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OH SHIT i didn't take it away  >:( >:( >:( >:( >:( >:( >:( >:( >:( >:( >:( >:(
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LittleMunckin

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Re: My proposed solutions for exam 2
« Reply #18 on: November 07, 2011, 05:20:57 pm »
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 :'( oh well , it's just the one mark
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MJRomeo81

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Re: My proposed solutions for exam 2
« Reply #19 on: November 07, 2011, 05:22:28 pm »
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Hahaha wow trigonometry module! Throw it out the window - what a joke!
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tom2593

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Re: My proposed solutions for exam 2
« Reply #20 on: November 07, 2011, 05:26:17 pm »
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Guys just wondering about trig question 6, proposed answer 20GB. Is there a chance that it could be 21GB? 
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davidle_10

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Re: My proposed solutions for exam 2
« Reply #21 on: November 07, 2011, 05:28:21 pm »
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q6?
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Deadshot

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Re: My proposed solutions for exam 2
« Reply #22 on: November 07, 2011, 05:30:15 pm »
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He is not paying for the 21st GB

=)
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arock

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Re: My proposed solutions for exam 2
« Reply #23 on: November 07, 2011, 05:34:01 pm »
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Quote from: tom2593 on November 07, 2011, 05:26:17 pm
Guys just wondering about trig question 6, proposed answer 20GB. Is there a chance that it could be 21GB? 
Nah, it asked for specifically the minimum amount of GB.
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natrox14

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Re: My proposed solutions for exam 2
« Reply #24 on: November 07, 2011, 05:34:21 pm »
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yea 20 GB was equal to zero dollars, the 21st GB went into minus values which become free.
So the answer is 20
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onoff312

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Re: My proposed solutions for exam 2
« Reply #25 on: November 07, 2011, 07:48:28 pm »
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By far the hardest exam out of all the exams i have practiced on 2006-2010 the Trig was far harder than any past exam it was ridiculous!
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ashleigh93

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Re: My proposed solutions for exam 2
« Reply #26 on: November 12, 2011, 06:27:29 pm »
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Does anyone knoew where I can find the actual exam? because I forgot the questions the answers, would be a great help  :D
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