CPhill

x - y  = sqrt (3)

x + y =  2

Square both sides  of these

x^2  - 2xy  + y^2   =   3

x^2  +2xy   + y^2  =   4           subtract the first equation from the second and we have that

4xy  =  1            divide both sides by  4

xy  = 1/4

Can't figure this without knowing the arc length......

OK.....sorry, Melody!!!!

Second one

Need to be  careful reading the number  of pupils.....each vertical "tick"  mark represents (1/2)  of a pupil  [ NOT  1  pupil ].......this seems to be  an odd way of presenting this...!!!!!

a)   number scoring  more than 8  =   1 + 4  =  5

b)   number of students  =   3 + 5 + 2  + 6 + 12 + 1 + 4   =  33

c)  We are assuming  that the whole class took the test  (no absentees)

First one

a)   the triangle  has two equal angles....therefore...it is isosceles

b)   acute.....it is  < 90°

999  =  

111 *  9  =

(37 * 3) * (3^2)  =

37 * ( 3)^3

The prime factors are 37  and 3

Their sum  is   40

tan (x)   =  - 2

We are  in the second quadramt.....so this implies  that   (x,y)   = ( -1, 2)

cos 2x  =   1 - 2sin^2 x

sin (x )  =  2  / sqrt [(-1)^2  + 2^2  ]  =   2/sqrt (5)

So....sin^2  (x)  =    4/5

Then  cos 2x  =     1  - 2(4/5)  =   1  -8/5  =     -3/5

The eight train cars can be arranged in  8 !  ways  = 40,320  arrangements

Sorry....I don't know anything about this topic......maybe someone else on here can help....

OK.....glad I could help a little .....