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#1 Help Me ! » tetrahedron... how many right angles can the 4 surface-areas have? » 2006-11-20 10:32:51
- hmm_marie
- Replies: 4
there are 4 triangular surfaces of a tetrahedron. how many right angles can they have?
there can obviously be none, right? in a regular tetrahedron, for example...
there can be 4 at the very most, because every surface can only have one right angle as the sum of the angles has to be 180° and therefore no 2 right angles can exist in a triangle.
but how do i show that there can be 1, 2, 3 and 4? is that possible?
that only 1 triangular surface can have a right angle seems quite plausible to me... but i can't prove it.
i think 3 right angles in the same corner of the tetrahedron are possible too. but i can't prove it, as I don't know how to show that the other triangle can't possibly have a right angle.
could anyone please help me on this one?
#2 Re: Help Me ! » prime numbers » 2006-10-11 01:23:17
yaaaaayy!!! cool, thank you:):) sooooo much
#3 Help Me ! » prime numbers » 2006-10-10 04:35:15
- hmm_marie
- Replies: 2
hey... i have another tough one here... i hope someone can help me... 
so one has to find all possible positive full (=whole?) numbers for n, for which 6n²+5n-4 is a prime number.
so far i factorised the term to make 6 (n - (1/2)) * (n + (4/3)) has to be eual to the prime number.
now i thought that as soon as one of the factors was equal to "1" then the other factor would have to be equal to the actual rime number itsself. and that would obvieously make the solution a prime number. as 1 * p = p
when (n + (4/3)) = 1 then n =-(1/3)
when (n - (1/2)) =1 then n = 1/2
but if one has to take the "6" into consideration? then 6 (n - (1/2)) = 1 then n = 2/3 etc......
but these are NOT positive full/numbers?!?!? as they are supposed to be in the question.
but if one takes the quation from the question and tries it out and puts in lets say n=1 then 6*1² + 5*1 - 4=7 which is a rime number.
oh i dont get it. can anyone help me? whats n and how do i prove it?
#4 Re: Help Me ! » tangent of two parabolas, how do i solve this problem? » 2006-10-10 04:24:06
hey thank you very much... i got nearly that far aswell. my problem there was that i didn't know how to get a tangent for BOTH parabolas facing downwards... so when i took the derrivative it was -2x + b1 and for the other one -2x + b2. do they have to be set as egual to get the COMMON tangent? if so, b1 and b2 have to be the same... and in that case they would have to be above each other?? oh my gosh i am sorry, but i dont get it...
#5 Help Me ! » tangent of two parabolas, how do i solve this problem? » 2006-10-09 04:17:47
- hmm_marie
- Replies: 5
A parabola with the equation y= ax² + bx + c, a>0 touches the two parabolas p1 and p2 with the equations y= -x²+(b1)x + (c1) and y= -x² + (b2)x + (c2) at the points A and B. One has to show that the common tangent of the parabolas p1 and p2 is parallel to the straight line AB.
I really don't get this one... could anybody help me please?
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