CPhill

Kinda' like the last one....we want to find

[ f(6) - f(3) ] / [ 6 - 3 ]  =

[  (6)^2 + 3(6) + 2  -  ( (3)^2 + 3(3) + 2 )  ]  /  3    = 

[ 36  + 18 +  2  -  9 -  9  - 2 ]  /  3    =

[ 36] / 3  =   12

We want  to find this :

[ -log (4*8)  + 5   -  ( - log(4*3)  + 5) ]  /  [ 8 - 3 ]  =

[ - log(4*8)  + log (4*3) ]  /  [ 8 - 3  ] =

[ log (12)  - log (32)  ] /  5 =

[ using  log a  - log b   =  log (a/b)  ]

log ( 12 / 32)  / 5   ≈   -0.085  ≈  -0.09

AIDS / A  =  SEER

This means the same as

S E E R

          A

______

A I D  S

Here's the logic of the solution

The last multiplication   S x A  =  A   implies that S  = 1

The first multiplication  R x A   results  in an ending value of 1

And....only the product 7 * 3   results in an ending  "1"

Thus  R = 3  and A  = 7  or vice-versa

This means that  EA + 2  =  D 

And  EA  + ??  =   I

But there is no "carry"  because  SA  could not = A  if this were true 

So...   ??  =  0 

Thus...  EA  <  10

But  if A  = 7  and EA  < 10... guess that  E  could = 1

But this is impossible because S  =1

And E  cannot = 0  because the second multiplication (EA) would have to equal E if this was true

So.....A must equal  3

And E could only   = 2

And the "carry"  from R x A  = 2

Which means that

2  + EA  =

2 + 2(3)  =  8  =  D

And since there is no "carry"  on the  next multiplication ... I  must  = EA  =  6

And R  = 7

So we have that

1 2  2  7

           3

_______

3  6  8  1

Or

AIDS / A   = SEER

3681 / 3   =   1227

y  =  0.155x^2 - 4.074x +  38.910

The average rate of change is given by :

[  217.2   - 238. 4 ]  /  [ 4 - 2 ]  =

[ -21.2 ]  /  2  =

-10.6    ⇒  Profits decreased by an average of $10,600 per year from Year 2 to Year 4

The polynomial will have zeroes of

3, 4 + 2i, 4- 2i, 1 + √7 and 1 - √7

So we have

(x - 3) (x - (4 + 2i)  )  ( x  - (4 - 2i) )  ( x - (1 - √7)  ) ( x - (1 + √7) )  =

Breaking this down, we have

( x - (4 + 2i ) (x - ( 4 - 2i)  =  (x^2 -x(4 + 2i) - x (4 - 2i) +  (4 + 2i) (4 - 2i)  =

(x^2  -4x - 2ix - 4x + 2ix  +  16 + 4)  =  ( x^2 - 8x + 20)

And

( x - (1 - √7)) ( x - ( 1 + √7)  =  

x^2 - x(1 - √7) - x ( 1 + √7) +  (1 - √7) (1 + √7) =

x^2  - x + √7x - x - √7x + 1 - 7  =

(x^2  - 2x  - 6)

So  we have

(x - 3) ( x^2 - 8x + 20)  (x^2 - 2x - 6)  = 

This expands to :

x^5 - 13 x^4 + 60 x^3 - 82 x^2 - 144 x + 360

Correct, Melody...!!!

f(x)   =  18  /  [ 6 + 5e^-0.1x ]

So

f(6)  =    18 / [ 6 + 5e^{-0.1 * 6} ]  =   18 /  [  6 + 5e^-0.6 ]   ≈  2.1

3.

y  =  (5/6) (x + 4) + 3   ⇒  (5/6)x + 10/3 + 3  ⇒  (5/6)x + 19/3 

y  =  3

y = (2/7) (x - 5) + 6  ⇒  (2/7)x - 10/7 + 42/7  ⇒  (2/7)x + 32/7 

So....a possible piece-wise function is

               (5/6)x + 19/3         if   -10 ≤ x ≤ -4

y  =            3                         if     -4 < x  ≤ 5

                (2/7)x  + 32/7        if   5 < x  ≤ 12

f^-1  does exist

 f^{-1}(f^{-1}(f^{-1}(1)))  =

(  f^-1  °    f ^-1   °   f^-1 ) (1)  =

( f^-1 °  f ^-1) (2)  =

f^-1 ( 5)  =

3