CPhill

 pq^3^3 4p^2q^2 2pq^2^3 = 

8 p^4  q^17

PR  =sqrt (3^2 + 4^2)  =  sqrt (25)  = 5

PQ =  sqrt (3^2 + 5^2)  =  sqrt (34)

OR =  sqrt (4^2 + 5^2) = sqrt (41)

Semiperimeter   =  (5 + sqrt (34) + sqrt (41) )  / 2

[ PQR ]  =  sqrt  ( S ( S - 5) ( S - sqrt 34 ) ( S -sqrt 41 ) )   =   sqrt (769) /  2 

1  5  9   .....   197

3  7 11  .....   199

There will be   5 houses with one  digit  =  $5

There will be 45 houses with  two  digits =  45 (2) =  $90

There will be  50 houses with three  digits  = 50 (3)  = $150

He earns      5 + 90  + 150  =  $ 245

1 / ( 1 + 1/sqrt (3)  + 2)  =

1/ ( 3 + 1/sqrt (3) )   = 

 sqrt (3)    /  ( 3sqrt (3)  + 1 )      multiply numerator /denominator by   3sqrt (3)   - 1

sqrt (3) ( 3sqrt (3)  - 1)

__________________  =

     9*3   - 1

9 - sqrt (3)

________

      26

x^2 + y^2  = 4(x + y)  + 12

x^2 + y^2  = 4x + 4y + 12

x^2 - 4x + y^2 - 4y  = 12       complete  the square on x, y

(x^2  - 4x  + 4)  +( y^2 - 4y + 4 )   =   12 + 4 + 4

( x - 2)^2  + ( y -2)^2 =   20

This is a circle centered at (2,2)  with a radius of sqrt (20)

The greatest value of  x =    2 + sqrt (20)

On the graph......go up to the horizontal line between 350 and 400 litres.....this is  the line representing 375 liters

Now the red line crosses this  between   80  and 90  gallons

So    375 liters  ≈    85 gallons

Let M be the mid-point of AB

Then triangle PMB  is a 30-60-90 right triangle with MB  = 3   PM = sqrt 3   and   PB = sqrt (12)  = the radius of the large circle  = R

So  the radius of one of the smaller circles =  

[ R  - PM ]  / 2  =   [  sqrt (12)  - sqrt (3) ]  / 2  =   

So......the total area of the three small circles = 

3 * pi *    ( [  sqrt (12) - sqrt (3) ] / 2 )^2  =

(3/4) pi * [ 12 - 2sqrt (36) + 3]  =

(3/4)pi  * [ 12 - 12 + 3]  =

(3/4)pi (3)  =

(9/4) pi cm^2

Rearrange the given equation as

x^2 - 2x +  8  =  0

(x - 4) ( x + 2)  = 0

The roots  are   4 , -2 =   a , b

4^4 =  256 =  a^4

(-2)^4  = 16 = b^4

The sum of these roots =  -b  in the new quadratic  = -272

The product of these roots = c in the new quadratic =   4096

The new quadratic =  x^2 - 272x + 4096

No. of three digit multiples  of 5  =

(995  - 100)/5  + 1  =    180

Number of three digit multiples of  8  =

(992 - 104) / 8  + 1  =    112

Number of three  digit multiples of 5 and 8  =  those  divisible by 40

( 960 -  120)  / 40  +  1    =   22

Number of three digit multiples  of  5 and 8  =   180 + 112  -  22   =  271

Number of three digit multiples of neither  5 or 8   =    900  - 271   =  629

Let r be the original radius

We have that

New  area -  Old area =    43 cm^2

pi (r + 1)^2   -   pi r^2 =   43

pi ( r^2  + 2r + 1)  - pi r^2  = 43

r^2 + 2r + 1  - r^2  =   43 / pi

2r + 1  =  43/pi

2r  = 43/ pi  -  1

r =  (43/ pi - 1)   / 2  ≈     6 cm