CPhill

Here's one  more way

Sum of arith series =  ( first term + last term) (number of  terms)  / 2

( 24  + 60  )  ( 6)  / 2 =   252

So

252  - 24 - 60  =  252  - 84 =   168

Surface area =  4pi *radius^2

3364 pi  = 4 pi (5x^2 - 7x + 1)^2

3364  / 4  =  (5x^2 -7x + 1)^2

841 = (5x^2 -7x + 1)^2 

Take the positive sqrt  of  both sides   ....the negative sqrt leads to a discrminant < 0

29 = 5x^2 - 7x + 1

5x^2 - 7x - 28   = 0

Q Formula   leads to

x =   ( 7 + sqrt [ 49 + 20*28 ] ) / 10  =   [ 7 + sqrt (609) ] /  10   ≈  3.168

ab - 3a + 4b  = 13

ab + 4b  = 13 + 3a

b ( a + 4)  = 13 + 3a

b  =  (13 + 3a)  / (a + 4)

a must be odd for  b to  be  an integer

No integers avaiable

The function aproaches  3   when a is  large

When a = 1   b =  16/ 5  =  3.2  = b

And any odd "a"  > 1  will  result in   3 < b < 4     (not an integer)

Very nice, DS2011  !!!!

d = 8

First term = 2

To find the nth term  =   first term + d (n -1)

2009th term  =    2 + 8 ( 2009 - 1)    = 16066

"a"  is  correct

Any vertical line will intersect the graph only once {Vertical Line Test }

y  =   x^2 -18x + 27 + c

We want to find the vertex  of

x^2 -18x + 27

x coordinate of the  vertex   =  -(-18) /(2 * 1)   = 9

Put this  back into  the  function  to  find the y coordinate ofthe  vertex

9^2 - 18(9) + 27 =  -54

So  ....to find c

-54  + c = 1

c  = 55

d is correct

The the quantity under the root must be positive

Therefore

x  -1  > 0

x > 1

****

AD = sqrt [ AB^2 -BD^2 ] =   sqrt [ 5.6^2  - 3.2^2 ] 

DC =   6 - 3.2 =  2.8

AC =sqrt [ AD^2 + DC^2 ]   =  sqrt  [ 5.6^2 - 3.2^2  + 2.8^2 ]  =  5.38