CPhill

Notice that when x = -1, y = f(-1)  =  3

And when x = 1, y = f(1)  = 27

So.....average rate of change is

[ f(1) - f (-1) ]

___________  =

     1 - (-1)

[ 27 - 3 ]

______   =

      2

24

__   =

 2

12   ⇒   GOOD JOB!!!

Look at the graph....when x = - 1, then y = -1/4  = -.25

So f(x) =f (-1)  = -.25

Similarly.....when x = 1, then y = -1/2  =  -.50

So  ...f(1) =  -.50

Let the original number of grapes = x

Then the original number of oranges = 1.2x

10% increase in oranges = (1.1)(1.2)x

9% increase in number of grapes = (1/09)x

So.....when126 fruits were added

x + 1.2x + 126  =  (1.1)(1.2)x  + (1.09)x        simplify

2.2x + 126  =  1.32x + 1.09x

2.2x - 1.32x - 1.09 x  =  - 126

-.21x = -126   divide both sides by  -.21

x = 600  =   original number of grapes

1.2(600)  =  720  =  original number of oranges

f(-1)  =  -.25

f(1)  = -.50

Avg rate of change =

 [ f(1)  - f (-1) ]

___________   =

      1 - (-1) 

[ -.50  - (-.25) ]

____________   =

          2

-.25

_____   =

   2

-.125

( x + y)^2  - (x - y)^2

_________________           factor the numerator as a difference of squares

              y 

[ ( x + y) + ( x - y) ] [ (x + y)  - (x - y) ]

_______________________________

                        y

[  2x] [ 2y ]

_________  =

        y

4x =

4(6)  =

24

Avg rate of change =

[ f(-2) - f(-5) ]

___________   =

   -2 -  (-5)

[ (-2)^2 - 4 ] - [ (-5)^2 - 4 ]

_____________________    =

              3

[ 4    -   25 ] 

__________    =

       3

[ -21 ]

_____   =   

    3

-7

 2)In a sequence of ten terms, each term (starting with the third term) is equal to the sum of the two previous terms. The seventh term is equal to 6. Find the sum of all ten terms.

a1 = a1

a2 = a2

a3 = a1 + a2

a4 = a2 + a3    =  a1 + 2a2 

a5 = a3 + a4    =   2a1 + 3a2

a6 =  a4 + a5    = 3a1 + 5a2

a7 =  a5 + a6    =  5a1 + 8a2  = 6    →   a1  =  (1/5) (6 - 8a2)   (1)

a8 =  8a1 + 13a2

a9 = 13a1 + 21a2

10 = 21a1 + 34a2

So....the sum of the first ten terms =   55a + 88a2

Subbing (1) into this.....the sum is

55(1/5) (6 - 8a2) + 88a2  =

11(6 - 8a2) + 88a2  =

66 - 88a2 + 88a2 =

66

663_x  can be written as

6(x)^2 + 6x + 3   =

(3 * 2)(x)^2 + (3*2)x + 3  =

3* (2x^2) + 3 * (2x) + 3  =

3 [ 2x^2 + 2x + 1 ]

Since 3 is a factor  of 663_x.....then this is not prime for any base x  [  where x > 6 ]

This angle is in the 4th quadrant......the cos and sec will be positive...the cot will be negative

We need to find "r" .....  this   is given by :     sqrt ( 7^2  + (-2)^2 )   sqrt (49 + 4) =  sqrt(53)

cos  = x /r  =   7/sqrt (53)  =  7sqrt (53) /53

sec = reciprocal of cos  =   sqrt (53) /7

cot  = x / y  =  7/-2    = -7/ 2