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yes, thank you!

Your question does not ask for the inverse of any of the functions....sooooo....

f(x) = 3x-4      then  for f (g(x) )  put   g(x) in place of 'x'

   f(g(x)) = 3 (5x+c) - 4    = 15x + 3c - 4 

and 

g(f(x)) = 5 (3x-4) +c  = 15x - 20 + c         

Now your question is missing something.....I will assume you want the two to be equal for all x 

   15x -20 + c = 15x + 3c - 4 

  -20 + c = 3c -4 

    -16 = 2c 

         c = - 8           

Nah

If we know  an  included angle between two known sides,  the area = 

(1/2) ( product of the sides)* sine (of included angle)

So

(1/2) (6 * 6)  sin 45°

Can you take it  from  here  ???

                     A

                         14

          14               Y

                                8

B                                  C

Since BY is an angle bisector

AB / AY  = BC /  CY

14 / 14  = BC/ 8

BC = 8

       A

   6

B           8                C   30

AB^2  = BC^2 + AC^2  - 2 (BC * AC) cos (30°)

36  = 64  + AC^2  - 2( 8 * AC) *(sqrt (3)/2)

-28 = AC^2  - 8sqrt (3)AC

AC^2 -  8sqrt (3) AC + 28  = 

AC^2 - 8sqrt (3) AC  =  -28

AC^2  - 8sqrt (3) AC  + 48  = -28 + 48

(AC - 4sqrt (3) )^2  =   20                take the positive root

AC - 4sqrt (3)  = sqrt (20)            

AC  =   sqrt (30) + 4sqrt 3  =   sqrt (20) + sqrt (48)

[ABC]  = (1/2) (BC)(AC) sin (ACB)  =

(1/2) (8) ( sqrt (20) + sqrt (48) (1/2)

2 [ sqrt (20) + sqrt (48) ]  =     2 [  2sqrt (5) + 4sqrt (3) ]  = 4sqrt (5) + 8sqrt (3)   ≈  22.8

30 m is the hypotenuse of a right triangle

sin = opposite leg / hypotenuse 

sin 25 = height / 30 

30 sin 25 = 12. 67 m    but you have to add the one meter hand height for a total  = ~ 14 m 

minor arc FB  = 360 - 160 - 140   =  60

So angle FAC  =  30

Angle FCA =  (1/2) ( minor arc  GH - minor arc FB)  = (1/2) (160 -60)  =  50 

FC / sin FAC    =  AF / sin FCA

10 / sin 30°   = AF /sin 50°

AF  =  10sin 50° / sin 30° =  10 sin 50° / (1/2)  =  20sin 50°  ≈   15.3

Height of sign  = Height from ground to top of sign  -  Height from ground to roof

                                             (a)                                -            (b)

                             tan (42°) =  a / 8                             tan (31°) = b / 8

    a  -  b   =  

8   tan ( 42°)  -  8 tan (31°)  =

8 ( tan (42° ) - tan (31°)  ≈  2.4 m

       P

    5             8

Q         M              R

            11

Law of Cosines

PR^2  =  QR^2  + PQ^2  - 2(QR * PQ) cos (PQR)

 8^2 = 11^2  + 5^2  - 2 ( 11 * 5) cos (PQR)

cos (PQR)  =  [ 8^2 - 11^2  - 5^2 ] / [ -2 * 11 * 5]   =  41/55

QM  =  11/2 =  5.5

Law of Cosines again 

PM^2  = QM^2 + PQ^2  - 2 (QM * PO) cos (PQR)

PM^2  = 5.5^2 + 5^2  - 2 ( 5.5 * 5) ( 41/55)

PM = sqrt [ 5.5^2 + 5^2  - 2 ( 5.5 * 5) ( 41/55) ]  ≈  3.77

A(2) = 3 - 40 x  4 x 4 = 3 - 640 = -637.

A(1) = 3 - 5 x 4 x 2 = 3 - 40 = -37

-637 - ( - 37 ) = -637 + 37 = -600

Answer: -600

The only  restriction  to the domain is an x value that makes the denominator = 0

But  notice  that  

x^2 + 7   = 0

x^2  = -7

And no (real) x makes this true....so....the domain is all real  numbers  → (-inf, inf)

A

                N

        1         

B            M              C

In a right triangle, the length of  a median drawn to  the hypotenuse =  (1/2) the hypotenuse length

So   BN  = (1/2) AC

        1 = (1/2) AC

        2 = AC  = the hypotenuse

We have a lower power polynomial  divided by  a higher power polynomial

When  we have this situation  we  will have a horizontal asymptote at  y  = 0

So  the range is   (-inf, 0) U (0, inf)

x^2 - 2  >  0         (1)

(x - sqrt 2)  ( x + sqrt 2)  >  0     make into an equality

(x -sqrt 2) ( x + sqrt 2)  = 0

x -sqrt 2 = 0                  x + sqrt 2  = 0

x =sqrt 2                        x = -sqrt 2

We have three possible solution intervals

(-inf, -sqrt 2)        x  = -2  is in this inerval  and it   makes (1)  true

(-sqrt 2 , sqrt 2)   x = 0   is in the  interval and it  makes (1) false

(sqrt 2, inf)          x = 2  is in this  interval and it makes (1) true

Answer

(-inf, sqrt 2)  U ( sqrt 2, inf)   =   "d"

Answer= D

Answer: B

AC  is a diameter

arc ABC  =180

angle ABC measures (1/2)  of this arc  = 90

minor arc BC  =  2 * minor arc AB

So  let minor arc AB  = x

So   minor arc BC  = 2x

So minor  arc AB + minor arc BC  =180

            x              +  2x   =  180

3x  = 180

x  = 180/3   = 60

If minor arc AB  = 60   then  angle ACB  is  (1/2)  of this   =  30

Hmm... I see that we already have all we need to find the area, WX and XY. we just multiply 2 by 3 and get 6.

Answer: 6

Connect AC ,  AO

Angle CAB  = 90  = angle ACB

So arc CDB =  180 = arc DCA

Then BC and AD are  diameters

Angle ABC = (1/2) 100  =  50 = Angle ADC

ABODC   is a  pentagon....the sum of  its interior angles  = 540

So  angle 

BOD =  540  - 2(90) - 2(50) =   540  - 180 - 100  =   260

So angle BOD = 360  - 260  =  100   ( just  as DS2011 found !!!

I am trying to draw quadrilateral BCED but am failing to get the given lengths of the side to make sense. I don't know how to draw the quadrilateral. If you could upload an image of it, and where points B, C, E, D, and A are that would help a lot.

C (n , 14) =  100

   n!

____________      =  100  different sets

(n -14)! * 14!

n! / (n -14)!  =  100 * 14!   

Solving this by hand is difficult

WolframAlpha  shows  n  to be ≈ 15.9

C (15 , 14)  = 15  different  sets

C (16, 14)  = 120  different sets

So  n  = 15 days

I haven't really learned this stuff at school, but I can assume that if AB and CD are parallel and chord AC is 100, then chord BD must also be 100, so that makes the measure of angle BOD = 100°

Answer: 100°