CPhill

Triangle  AOB  is similar to triangle COD

Then   since  base  CD  =  38/25   of  base   AB,  then the  height  of  COD  = (38/25)  that of  triangle AOB

Then  there are  38 + 25  =    63 equal parts  to  the height of the trapezoid  ...... and 63/2  = 31.5  equal vertical parts on each side of the  midline

And  since the  height of AOB  contains 25  of  these equal parts......then  the height of triangle YOX  must = 31.5  - 25  = 6.5  of the equal parts

And triangles YOX  and COD  are similar

So

base  of  YOX  / base  of  COD     =     6.5  / 38   =    13/76

So  XY  = (13/76) CD  =    (13/76) ( 38) =  13 / 2    =  6.5  in

log ₂ 3 * log ₃ 4

By the change of base theorem we can write

log 3        log 4              log 4

____ *   ______  =     ______  =      log ₂  4  =   2

log 2         log 3             log 2

Nothing shown......

Simplify as

(2x) / (x + 2)  =  (8x  -6) / (x + 4)     cross-multiply

2x ( x + 4)  =   (8x - 6) ( x + 2)

2x^2  + 8x  = 8x^2 + 10x - 12         rearrange as

6x^2 + 2x - 12  =  0       divide through by 2

3x^2  + x - 6 =   0

         -1  ±√ [ 1^2  - 4 * -6  * 3 ]               -1  ±√73

x =   _____________________ =       __________

                   2  *  3                                        6

A + B =  90

Let A =  30 * N

30 N + 30 =  90

30N  =  90  -  30

30N  = 60°  = A

Find the height of the trapezoid

900 =  (h/2)  ( 20 + 40)

900 / 60  =  h /2

15 = h /2

h  = 30

Triangles   AXB     and  CXD   are similar

Call AB the shorter base and CD the longer

Since the base of triangle    CXD  is twice that of  triangle AXD.....their scale factor  is 2/1

So....the height of  triangle  CXD = (2/(2 + 1)) *30  =   (2/3)30 = 20

And the height of triangle AXB =  (1/3)30 = 10

Area of triangle   CXD =   (1/2)(40)(20) =  400

Area of triangle AXD = (1/2)(20)(10)  =100

And triangles  AXD  and BXC  have equal areas

So

Area of triangle BXC =    [ 900 - 400 - 100 ] /  2 =    200  units^2

9 + 27/x + 8/x^2  =   14/x + 1/x^2      rearrange as

7/x^2  + 13/x + 9  =  0     multiply through by x^2

9x^2 + 13x + 7 =   0

               -13  ±√ [ 13^2  -  4 * 9 * 7 ]           - 13 ±√-83           -13 ± i√83

x =           _____________________  =      __________  =   ____________  

                               2 * 9                                    18                         18

Draw PB

Triangle PAB is right with   AB = 5 and PA  = 1    and PB the hypotenuse

So

sqrt [AB^2  + PA^2 ]   =  PB

sqrt [  5^2 +  1^2 ] =  PB

sqrt [26] = PB

And PBT is also a right triangle with PTB a right angle  and TB = 4 and PB = sqrt (26) and PT a leg

So

sqrt  [ PB^2  -   TB^2  ]  =  PT

sqrt [ 26  - 4^2 ]   =  PT

sqrt [ 26 -16 ]  = PT

sqrt [ 10 ] (cm) = PT

Since AB = BC  then angle   A =  C

So

A + B + C  = 180

2B + B + 2B  =180

5B =  180

B =   36

And angle C =   2 * 36 =  72° 

And since CE = CD  then  angles   D and E are equal

So

D  = (180 - 72)  / 2   =   108  / 2  =    54°

OK.....that's WAY better....LOL!!!

4z^2  + cz + 9

We want to write  (  az + b)^2

Easy....

Take the square root =  4 =  2   =  a

Take the square root of 9 = 3 = b

So

(2z + 3)^2  = 

(2z + 3) (2z + 3)  =

4z^2 + 3*2z + 3*2z + 9  =

4z^2 +  12z  +  9

c  =12