You really need to say whether you are asking about permutations or combinations.The number of permutations of six numbers from fifty (which just means arrangements) is 50x49x48x47x46x45. But within these permutations you will have chosen the same combination more than once,for example you will have picked the combination 1 2 3 4 5 6 in several different ways.(6! different ways in fact) So if you are looking for the number of combinations without repetition,,you need to divide 50x49x48x47x46x45 by 6! (6! being the total number of ways you can arrange 6 objects or numbers in this case).
(I can permute 123456 as 234561,345612 etc etc in a total of 6x5x4x3x2 ways.This is just one combination of the same group of six numbers arranged in 6! ways.) Always ask yourself when considering these problems whether you are looking at a combination (one distinct collection of objects) or a permutation (an arrangement of that particular collection of objects.)
You cannot.You have one equation with two variables,which gives you an infinite solution set.What you have written there is actually the equation of a straight line,which has an infinite set of points lying on it.For any solution,you would need another similar equation,then solving these together would give you the solution which represents where the two lines intersect.
The limit is 2,but not as demonstrated above.Since both the numerator and denominator tend to zero as x tends to zero,we need to use L'Hopital's theorem.
we have f(x)/g(x) =f'(x)/g'(x)
lim(x) tends to zero lim(x) tends to zero, and we can repeat until
we get rid of any zero's on the denominator.
Applying the theorem twice gets you there.
.Melody,I don't think simply plugging in values is what the question is really about.This is testing one's knowledge of the addition formulae,properties of triangles and manipulation of fractions. As below:
sin(a+b)=sin(a)cos(b) +cos(a)sin(b) so you can write sin70 as
Now the expression reads
8sin(40)/sin(40)cos(30)+cos(30)sin(40).One should also know (from the equilateral triangle) that cos(30)=1/sqrt2.
So now all we have to do is simplify a fraction,getting
8sin(40)/sin(70)= 8sin(40)/(1/sqrt2sin(40) +1/sqrt2sin(40)
Now the sin(40)'s cancel and all you have to do is resolve 8/sqrt2 and the answer is 4 times root 2.
tan(x)=4(1-sec(x)) Now square both sides to get
tan^2(x)=16(1-sec(x))^2 and using the identity tan^2(x)=sec^2(x)-1 on left hand side
sec^2(x)-1=16(1-sec(x))^2 Expanding brackets and collecting terms gives
Now just solve the quadratic in sec(x) to get the 2 values of the solution. Solution has real roots as b^2-4ac=1024-1020=4,but formula will need to be used as this will not factorise. I am now going for beer.