CPhill

HAHAHA!!!!....that's a lot of midpoints   !!!!

Midpoint of  PQ  =  (8  + 48) / 2  =  56/2  = 28  =  B

So  midpoint of   BQ   =  ( 28   + 48)/ 2  =  76/ 2  =  38  =  C

And midpoint  of PC  =  (8 + 38) / 2  =   46/ 2   =  23   =  D 

3x + 2y = 5

5x + 3y = 7

We can choose to eliminate y thusly  :

Multiply the first equation  by  3   and the second by -2

9x     +   6y   =   15

-10x  -   6y    =  -14      add these

-1x   =  1       divide both sides by -1

x  =  -1

And using   3x +  2y  =  5   to find y  we have

3(-1)  + 2y   = 5

-3 + 2y   =  5       add 3 to both sides

2y  =  8     divide both sides by 2

y  =  4

So  {x, y}   =  {-1, 4 }

4^x  - 2^x  =  56

(2^2)^x  - 2^x  = 56

2^(2x) -  2^x  -  56  = 0

(2^x)^2 - 2^x  - 56   = 0

Let  u    = 2^x         so      u^2  =  (2^x)^2

u^2  -  u  -  56   = 0       factor

(u - 8) ( u + 7)  = 0

So    either

u  -  8   =   0               or           u +  7    =  0

u =  8                                        u  =  - 7             

Substitute back

2^x  =  8                                  2^x   =  - 7       this isn't possible

x  =  3

Check

4^3  -  2^3   =  56  ??

64   -  8     =  56        correct  !!!

One thing about these identities is that there may be more than one way of proving them

So...let's try this....

 2cos^2 (x) = cos(2x) +1 

2( 1 - sin^2 (x))  =  cos (x + x)  + 1

2 - 2sin^2 (x)    =   cos(x)cos(x) - sin(x)sin(x)  + 1      subtract 1 from both sides

1 - 2sin^2(x) = cos^2(x) - sin^2(x)          add 2 sin^2x to both sides

1  =  cos^2(x)  + sin^2(x)        which is true

5 + 3/5   =   28/5

So  we have

(28 / 5)  / (4 / 5)     invert the second fraction  and multiply

(28 / 5) * (5 / 4)   = 

( 28 / 4 ) *  ( 5 / 5)  =

7   *    1   =

7

Since DC  can be considered as one base of parallelogram   ABCD  and  EG is the height drawn to this base.....the area  is  DC * EG =  27 * 16  = 432 units^2

But BC  can be considered as another base  and  FH  is a height drawn to this base....and since the area is constant......we have that

432  = BC * FH     but   AD  = BC  = 18  ....so....

432  = 18 * FH      divide both sides by 18

24  = FH

Yeah, Julius....I forgot about  the second  "≤" ....2pi is also a solution   

Thanks for noticing this  !!!

Good job, Ghosty  !!!

tan^2 (x) * sin (x) - sin (x) / 3  = 0     get a common denominator

[ 3tan^2 (x) * sin (x) - sinx] / 3  = 0     mutiply both sides by 3

3tan^2 (x) * sin (x) - sin (x)  = 0        factor out sin(x)

sin (x)  [  3tan^2 ( x)  - 1]   =  0

We have two equations here.....either.......

sin (x)  = 0       and this happens at   0  and   pi 

Or

3tan^2 (x)  -  1   =  0       add 1 to both sides

3tan^2 (x)   =   1           divide both sides by 3

tan^(2) x  =  1/3            take both roots

tan (x)  =  1/√3      and this happens at  pi/6   and 7pi/6

And

tan (x)  =  -  1/√3   and this happens at  5pi/6  and 11pi/6

So.....the solutions are    0, pi/6, 5pi/6, pi, 7pi/6 and 11pi/6

 (x^2y-3)/[2-y^(-3) ] = 

(x^2y-3)/( 2 -  1 / y^3 )  =

( x^2y - 3) / [  (2y^3 - 1) /  y^3) ]

[ y^3 (x^2y - 3) ] / {2y^3 - 1)

[ x^2y^4 - 3y^3 ] / [ 2y^3 -1 ] 

We can't solve the second one for anything because we have to have an equation for that