Double check what the text book actually wrote if it wrote (d/dx)*y to mean dy/dx then throw it away.
More likely you have misinterpreted the text. Is it possible to scan the page and include it in your next post?
3. acos(h/r) is theta, ie half the angle of the sector at the centre. Need to double your result.
2 is submerged area + wet area above water
5 is submerged area
for 6 you are working out wet area above water
thomas49th is taking what is called General Secondary Certficate Examination (GCSE) in Maths in England. It includes number, algebra, statistics and geometry and is taken at the age of 16. The examination is graded A,B,C,D,E,F, U (unclassified). Students achieving grades A or B will be allowed...
NO ABSOLUTELY NOT. Just as Hallsof Ivy said.
The webpage above is for vector displacement not distance between points.
Lots of people have been pointing you in the right direction including me. Sometimes to follow advice you have to let go of the idea of what you were attempting was right...
The question show in from an examination paper for GCSE. Thomas49th will have been given past examination papers for practice. The examination will take place in May.
Thomas49th you will not have been shown the distance formula for GCSE but you will have covered pythagoras and should be able...
Depends on the wording of the question
verify that x=180 is a solution of ..... use the first method
show that x=180 is a solution of...... use the second method.
To verify x=180 is a solution
LHS is sin(360)=0
RHS is 2 -2cos(360)=2-2=0
LHS=RHS so x=180 is a solution
to obtain x=180 as a solution
(1-cos2x)=(tanx)(sin2x)
sin2x=2(1-cos2x)
sin2x=2(tanx)(sin2x)
sin2x-2tan(x)(sin2x)=0
sin2x(1-2tanx)=0
sin2x=0 or 1-2tanx=0
This gives that a player on average will lose $24.90 on a $100 dollar bet.
Your questions asks for the expected net winnings for the winner so this is not the same thing. I was assuming this question was really about expected values and as you had had difficulty in finding anything about...
Techno - I agree with Matt Grime Gonzo never changed his mind about what he wanted proof of. However there was a lack of clarity (for me) about the statement.
I took this to mean
For any positive integer k, you can find k points on any circle such that each point is a rational distance...
Below is a link to an image showing what I mean by transforming a complex polygon to a simple one. I belive that sufficient repeations of this process on even a very overlapping complex polygon would result in a simple polygon.
http://img101.imageshack.us/img101/5027/comptosimpps4.png [Broken]
I'll return to my idea in post #23
Let n_1, n_2, .....n_k be a set of k "angles" such that their sum is 180(k-2)
Let m_1, m_2, .....m_k, be defined by
m_i = 180 - n_i for 1<=i<=k ( some m_i may be negative)
the sum of all m_i is 360
rotations when m_i>0 will be clockwise...
Giving it more thought this arranging causes problems with the induction
still restricting to the case when all angles <180
for n>3
from the set of n angles choose 3, a_{1}, a_{2}, a_{3} such that
a_1 + a_2 + a_3 = 180 + d_1 + d_2
so that
a_1-d_1 + a_2 -d_2+ a_3 = 180
take the set of...
Ok Doodle Bob, I see the flaw in my method, halving the angles does not work as you have to take exactly 180 from the n angles so that ther n-1 form a polygon. Mea Culpa
When all the angles are <180 you do not need them in a certain order.
For n>3
take the angles a_1, a_2, a_n
form the n-1 polygon with angles
a_{2}/2, a_{n}/2 and the angles a_3,...... a_{n-1}
form the triangle with angles a_1, a_{2}/2, a_{n}/2 and add to the n-1 polygon placing the...
I was thinking convex polygons and so thought that n_1, n_2, .....n_k were all <180.
Reading your posts again I now realise you are thinking concave or convex polygon and so some n_i >180.
and in this case m_i = 180 - n_i is not true.
But when n_1, n_2, .....n_k are all <180 then it...
Another approach
Instead of the interior angle n_1, n_2, .....n_k
Consider the exterior angles m_1, m_2, .....m_k,instead where
m_i = 180 - n_i for 1<=i<=k
and sum of all m_i is 360
Start with a line segment A_0 to A_1
at A_1 rotate clockwise through m_1 and draw a line...
Gonzo - At first I was not entrirely sure what it was you wanted to prove (or find a counter example for). Following our discussions I am now clear on that. Like you however I am not clear how Techno's statements help.
Techno - Thank you for answering my question re: unit circles and...
k=4 is possible
draw a rectangle with sides 3 and 4, hence diagonals 5. All vertices lie on circle radius 2.5. Do not see how this helps in general though.