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Help me with two math problems?
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IllusoryDeath Away
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Post: #1
Help me with two math problems?

Long story short: I hate math. Math hates me. It all evens out....except for the fact that it doesn't end up failing at the end of the year.

They are probably pretty easy to figure out, but my teacher skipped the review notes that day on how to do them.
Lucky me. Rolleyes

So, who wants to help me pass math studies for this triterm? Biggrin

[Image: MXLI.jpg?t=1253399582]

[Image: MXLII.jpg?t=1253399565]

Thanks in advance to anyone who can help me out.

EDIT: Images deleted so that no one can trace me to my PB account or any of that nonsense. Yes, I am that paranoid. Suspicious
09-21-2009 12:32 PM
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Eidolon Away
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Post: #2
Re: Help me with two math problems?

Quote:[Image: MXLI.jpg?t=1253399582]
This one can be solved using the Pythagorean theorem. (a^2 + b^2 = c^2)

(5.6/2)^2 + (1.2)^2 = (PQ)^2

7.84 + 1.44 = 9.28

Now if we take the square root of 9.28, we will have the length of PQ.

And since those ticks on some segments of the truss mean they are the same length as the others, it's a simple matter of adding the lengths together.

(PQ/2) = length of one of those ticked segments.

(PQ/2)*6 = length of all ticked segments.

Then add the 5.6 m and 1.2 m lengths to that, and you're left with the answer to the second question in the image.

Answers:
Hidden stuff:
The length of PQ is 3.04630924234556331426.

The length of total metal in one truss is 15.93892772703668994278.

(best round those, methinks.)
Quote:[Image: MXLII.jpg?t=1253399565]
The Pythagorean theorem will also come into play here, to find the length of the unmarked side of the triangle (let's call it 'x').

First we have to find the length of the base of the triangle by subtracting the lengths of protruding metal (both are 3 cm):
20 - (3*2) = 14

Now for the theorem:
14^2 + 8^2 = x^2

196 + 64 = 260

sqrt(260) = x

After that, you can add all of the lengths together, excluding the two protruding 3 cm parts, since those are part of the 20 cm length.

Oh, and multiply that by 2500.

Answer:
Hidden stuff:
To the nearest tenth of a centimeter, the amount of metal required to make 2500 of these right angle brackets is 110311.3

Remember that by no means am I a mathematician, so you may want to look these over to see if they're right.
09-21-2009 01:27 PM
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CrayolaColours Offline
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Post: #3
Re: Help me with two math problems?

Hmmm... I think I'm now more confused about the Pythagoream theorem that I was before. But I'm sure that's because I'm just stupid with math.

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Quote:Riddle me this, riddle me that. Give me a straight answer, you pain in the ass cat.
09-21-2009 01:36 PM
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IllusoryDeath Away
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Post: #4
Re: Help me with two math problems?

Thank you, Eidolon. JSYK, the answers came out to be:

The length of PQ is 3.05
The length of total metal in one truss is 15.9
&
The amount of metal required to make 2500 of these right angle brackets is 110311.3.

Your explaining helped greatly.
09-21-2009 11:11 PM
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