CPhill

\( f(x) + b \cdot g(x) \)

If b = 0 .....the smallest possible degree is  just  5

-7pi/4.....convert to  + degrees thusly

-7/pi / 4 +  2pi   =   -7pi/4  + 8pi/4  = pi/4

So

tan (-7pi/4)  = tan (pi/4)  = 1

sec (pi/6)  =  1 / cos (pi/6)  =  1 / (sqrt (3) / 2 )  =  2 / sqrt (3)

cos (23pi/6)

23pi/6 =  3pi + 5/6 pi  .....3pi puts us back at 180°.....(5/6) pi  adds  150° to this  = 330°

This is a  30° reference angle  in the 4th quadrant

cos (30°)  =  sqrt (3)  / 2

Not sure if this is the intended question or not......triangles ABM and AMC  are under the same  heights and have equal  bases....so their areas are equal

So  [ AMC ]  = (1/2) [ ABC ] =  (1/2) (180)  = 90

(x^3 + x^2 + x + 1)(x^4 + 3x^3 + 8x + 7x + 1)

"x^2" terms occur with the multiplications of

x^2 * 1    + x *7x  +  x * 8x   =

x^2  + 7x^2  + 8x^2  = 

16x^2

 (x^3 + x^2 + x + 1)(x^4 + 3x^3 + 4x^2 + 8x + 7)

Note that  "x" terms  will only occur with multiplications of

x * 7   +  1 * 8x   =

7x  +  8x  = 

15x

-5pi / 6  =  Q3

-35pi / 4  =  -8pi - (3/4)pi =   Q3

57 pi / 8  =  7pi + (1/8)pi  =  Q3

3pi/4  = Q2

13pi/6 = 2pi + (1/6)pi = Q1

-5pi / 11  =  Q4

                      x^2

x^2 - 2x  - 5  [ x^4 - 2x^3 + kx^2  ]

                       x^2 - 2x^3 -  5x^2

                     _________________________

                                        (k +5)x^2

k + 5 = 0

k = -5

5n /( n + 2)  =  6

5n = 6 (n + 2)

5n = 6n + 12

-n = 12

n  = -12

2. When the set of natural numbers is listed in ascending order, what is the smallest prime number that occurs after a sequence of five consecutive positive integers all of which are non prime?

24,25,26,27,28

Smallest prime  = 29

BE will  equal CE

To find CE

angle CED = 90  -15 = 75

CD / sin 75 =  CE / sin 90

10 / sin 75  = CE  =  10 (sqrt 6 - sqrt 2)  ≈  10.35 = BE