Mathcad

Circumference of a circle is (pi)x(diameter) or 2 x (pi) x radius.  So just half that for a semicircle.

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.                    

Because you have one equation in 2 variables you can't solve this.(If you substitute and re-arrange these 3 expressions you will never be able to eliminate down to one variable.)

2 variables? Need 2 equations to solve.

slight correction,cos(30) is (sqrt3)/2.Amend accordingly,and solution drops out as 

8/(sqrt3). Which still gives you Melody's answer of course.

.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 

sin(70)=sin(40)cos(30) +cos(30)sin(40)

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)

                         =8sin(40)/(2/sqrt2sin(40)

                          =8sin(40)/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)+4sec(x)=4

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

15sec^2(x)-32sec(x)+17=0.

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.

If slope has rise of 4 in 12 then hypotenuse is square root of (16 + 144)=sqrt  160=4root10.Angle you want is arctan(4/12) or arctan(0.3333r)