CPhill

\(f(x) = \frac{3x - 7}{x + 1}\)

Write

y  =  [ 3x - 7]  /   [ x + 1]       get x by itself

y [ x + 1]  = 3x - 7

yx + y  =  3x - 7

y + 7  =  3x - yx

y+ 7  =  x (3 - y)

[y + 7 ] / [ 3 - y ]  =  x     "swap" x and y

[ x + 7 ] / [ 3 - x ]   = y   =  f^-1(x)

I'm using  "e" for  "x"

Let   a < b < c < d < e

So we have......

a + b + c + d + e  = 320   (1)   

Where "a" is the smallest integer and "e" is the largest

And

c + d + e  =  283    (2)  

a + e  = 119     (3)

Sub (2) into (1)

a + b + 283  =  320

a + b  =  37

Let  a  =  1

Then.....as large as "e" can be is

1 + e  = 119

e  = 118

Let  a  = 36

Then......  a possibility for as small as "e" can be  is

36 + e  = 119

e  = 83

However     c < d < e

And  c + d + e  = 283

So....e will be minimized when  the sum of c and d are as large as possible

If      c  = d = e = 95

95 * 3  =  285....but  c and d  are < e.....so....c and d < 95  and  c < d

So....the largest that the sum of c and d can be  is  93 + 94  =   187

But.....

93 + 94 + 95   =  282   ......too small

Then e cannot be < 96  because  93 + 94 + 96  = 283

So.....as  small as e  can be  is  96   and as large as e can be  is  118

So.....the positive difference   between  the greatest positive value of  e   and the least positive value of e  is   118 - 96   =  22

Thaks Omi......here's another approach....

By the Law of Cosines, we have

9^2  = 7^2 + 8^2  -  (2 * 8 * 7)cos(DCB)

-32   / -112  =  cos(DCB)

2/7  = cos(DCB)      (1)

And since angle DAB is supplemental to DCB, cos DAB  = -cos(DCB)

So....by the Law of Cosines again, we have

9^2 =  7^2  + AB^2  - (2 * 7 * AB) (-cosDCB)

[32- AB^2] = (14AB) cos(DCB)

[32 - AB^2] / [14AB ]   = cos(DCB)   (2)

Set  (1)  = (2)

2/7  =  [32 - AB^2] / [ 14AB ]

2 = [32 - AB^2] / [ 2AB]

4AB = 32 - AB^2

AB^2 + 4AB - 32  =  0      factor

(AB + 8) (AB - 4)  = 0

Setting both factors to 0 and solving for AB produces  AB  = -8  or AB = 4

Reject the first answer, accept the second

Dropping a perpendicular line from B to intersect DC at E.....then, by symmetry,  EC  =  2 

The height of the trapezoid  is given by √[BC^2 - EC^2 ] =  √[7^2 - 2^2]  = √45 = 3√5

So....the area is

height / 2   *  (sum of the bases )  =

[ 3√5  ( 8 + 4) ]  / 2  =    18√5  units^2  ≈  40.25 units^2

We have  5 odd integers....

P (both odd)   =  5/9  * 4/8    =    5/9 * 1/2   =   5/18

The guest has assumed replacement......my answer  assumes without replacement

Wow  !!!....  that's  nice guest   !!!!

Note that   A  is 2   units to the left  of  B

So.....D  is 2 units to  the left of C  =  ( 4 - 2, 0 )  =  (2, 0)

So.....the sum of its coordinates   =  2

a)  Lower limit   = 1st term =   1     

Upper limit  =  18th term  =  18

b)  We need to find the 18th  term.....this  is  3 + 7(18 - 1)  =

3 + 7 * 17   = 

  122

The sum  is  given by :

[ 1st term + 18th term ]  *  number of terms / 2  =

[ 3 + 122 ]  *  18  / 2  =

[ 125 ]  * 9  =

1125  beads  !!!

a)  A decline of 10%  per year  means that the common ratio between terms  is ⇒  r  = .9

So.....geometric

b) aⁿ  =  1250 (.9)^n-1     where n  is the value at the beginning  of the nth year

c)  6th year   =   1250 (.9) ^{6 - 1}   =  1250 (.9)⁵  = $738.11 

A travels at 81 mph

B travels at    324/ 4   =  81 mph

So...

The distance Train A traveled in 1 h is equal to the distance Train B traveled in 1 h.

18.

Since  each successive term  is greater than the previous term.....this series will diverge and will not have a definite sum.......another way to see this is that the common ratio  is 3......for a series to converge, r  must have an absolute value < 1

19.

a)     a₁  = 9,    a_n  = a_n-1 + 2

b)   a_n  =  9  + 2 (n - 1)

c)  a₁₂  =  9 + 2(12 - 1)   =   9 + 2*11  =  9 + 22   =   31 seats