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Topic: **Homework area rules****Question:**
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i really really really need help with this math homework problem...the new term has just started and i'd hate to leave a bad impression on my geometry teacher--i heard she's one of the strictest and harshest in my school.
i have literally asked everyone i know and no one is sure about how to do this. please help me...TEN POINTS WILL BE GIVEN TO THE FIRST CLEAR ANSWER THAT I GET. thank you so much in advance, even if you're not picked as the best answerer, i will except any and all help. thank you.
http://farm5.static.flickr.com/4034/4350037682_19ac33cd72_b.jpg
(hopefully the link works. zoom in to see full size.)
THANK YOU TO EVERYONE!
i dont need anyone to write this whole thing out for me--just a quick explanation of what i need to do.
thank you again.
to the first answerer: i'm still going over your problem to understand it a little better but-- what are you referring to as "a" and "b" (both lowercase)? the only a and b shown in the diagram are points.
also: i've learned trig in middle school but we haven't gone over it yet so they don't "allow" us to use it.
thanks again.
JUST TO CLEAR THINGS UP: when ah, bk, cl, etc are referred to in the problem, they mean the lengths of the sides multiplied by the lengths of the altitudes...not actual lines. (in other words, a, h, b, k, c, and l are not points but represent actual lines.

May 23, 2019 / By Dandrenor

First lets get nomenclature right The triangle vertices are A,B and C. A, B C are also the (interior) angles at the respective vertices. a is the line opposite A a is also the length of the line opposite a. Similarly with b and c h is the altitude from A to a and is perpendicular to a j is the altitude from B to b and k is he altitude from C to c j, k and h are also the lengths of the altitudes. First we must prove the sin rule. h = c * sin B also h= b * sin C Therefore c*sin B= b* sin C Therefore c/sinC=b/sinB Area1 = a *h Area1 = a * b* sinC k=a * sin B Area2= k * c Area 2 = c*a*sinB = a*c*sin B But we know that c*sinB = b*sinC Therefore Area 2 = a * b* sinC Which is the same as Area1 Proof completed.

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Good luck you need math in almost everything and I don't think you ll pass if you don't get a passing grade i suggest you get your books open and don't study everything just the basics of the chapters but its really important Wishing you luck.:)

Failing geometry is not the end of the world. Throwing away the opportunities you are being offered with both hands, is. If you go to somewhere like India of Africa where so many children compete for any school space you will realise how reckless (or ridiculous) you are being. If your school has a counsellor go and see them and sort out how you are going to get yourself back on track.

I'll borrow a diagram from: http://polymathematics.typepad.com/photo... The proof doesn't require that you know what an orthocenter is, or prove that it exists. I just don't feel like creating my own diagram, and this one was the closest to your setup that I could find. Let e = CK and f = CH. Then we have e/b = f/a. (They are both equal to cos C, if you know trigonometry. If not, use the fact that triangles ACK and BCH are similar: they both contain angle C and a right angle.) Rearranging, we find that ae = bf, so a^2 e^2 = b^2 f^2. Now use the Pythagorean Theorem: a^2 = f^2 + k^2, so (f^2 + k^2) e^2 = b^2 f^2. Do a little algebra: b^2 f^2 - e^2 f^2 - e^2 k^2 = 0 b^2 f^2 - e^2 f^2 - e^2 k^2 + b^2 k^2 = b^2 k^2 (k^2 + f^2)(b^2 - e^2) = b^2 k^2 a^2 * (b^2 - e^2) = b^2 k^2 Use the Pythagorean Theorem again: b^2 - e^2 = h^2, so a^2 h^2 = b^2 k^2. Taking square roots, we conclude that ah = bk. The hint says that you will need two diagrams. This is stupid. Since we didn't make any particular assumptions about b and k, the proof to show that ah=cl is exactly the same. No mathematician would ever write a separate proof that ah=cl. I would say something like, "By an identical argument, ah=cl." Though if your teacher is really that harsh, you might not want to point out how dumb her hint is, so maybe you should write a second proof.

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Assume the next 2 axioms: A1: a million+X = a million. A2: two*X = X. Given A1 and A2, reply the next: a million.) a million+6 = ? two.) two*eight = ? Just considering I recognise the axioms to be unsuitable does now not imply that I can not preserve them as actual lengthy sufficient to accurately finish the solutions are, below the ones axioms: a million.) a million two.) eight

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The answer to 1 is B. 2 c 3b If you want homework help, i have an online homwork help service that i offer. http://myonlinetutor.wordpress.com/ Its an easy way to get your homework done when you dont understand it.

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