CPhill

Here's a similar problem that I answered, here :

https://web2.0calc.com/questions/math_37783

We would have the following lines

y = -3x

x = -3y

These split the plane into 4 regions

See the graph, here :   https://www.desmos.com/calculator/pekveaesjs

No cylinder shown......

We  need to find two Pythagorean Triples here such that

a^2 + b^2   = 65

There are two  pairs that work    (8,1)  and ( 7, 4)

So.....we can let  AB = 8 ,BC = 1, CD = 7  and DA  = 4

We have  two right triangles  sharing the same  hypotenuse  ( AC)

The area of the quadrilateral is   (1/2) [ (8) (1)  + (7) (4) ]  =  (1/2) [ 8 + 28 ]  = (1/2) (36)  = 18

a + b   =14       (1)

Square both sides of  this

a^2 + 2ab + b^2 =  196       rearrange  as

a^2 + b^2  =196 - 2ab      (2)

a^3 + b^3  = 812

Using the formula  for  the  sum of  cubes we  have that

(a + b)  ( a^2 - ab + b^2)  = 812

sub in (1)   and (2)

(14) ( 196 - 2ab - ab)  = 812         divide both sides by   14

(196  - 3ab)  =  58     rearrange as

3ab  = 196 - 58

3ab  = 138      divide both sides by 3

ab = 46

b  =   46 / a

sub this into 1

a + 46/a  = 14     multiply through by a

a^2 + 46  = 14a      rearrange as

a^2  - 14a  + 46   =   0     complete the square

a^2 - 14a  + 49  =   -46 + 49

(a - 7)^2  = 3       take  both roots

a - 7 =    ±sqrt (3)

a =  7 + sqrt (3)        or      a  =  7 - sqrt (3)

And  b = the conjugate of either answer

So   ( a, b)  =   ( 7 + sqrt (3), 7 - sqrt (3) )  or    ( 7 - sqrt (3) , 7 + sqrt (3))

The probability is  given by :

C(7,3)  (1/4)^4 (3/4)^3  =  945 / 16384  ≈  .0577

5/24  +  l x - 1/3 l  <   5/16       subtract  3/24  from  both  sides

l x - 1/3 l   <   5/16  - 5/24       get a common  denominator (48) on the right

l x - 1/3 l  <  15/48 - 10/48

l x - 1/3 l  <  5/48

We have two equations here

x - 1/3   <   5/48                     x - 1/3   >  -5 / 48

                    add 1/3 to both sides

x <  5/48 + 1/3                     x >  -5/48 + 1/3

x <  5/48 + 16/48                 x >  -5/48 + 16/48

x <  21/48                           x  >   11/48

x <  7/16   

So the solution is       11/48 < x < 7/16

heureka answered this , here : https://web2.0calc.com/questions/a-circle-problem#r1

Here is an additional explanation for his solution :  https://web2.0calc.com/questions/graphing-circles_1

Good answer ftom  both "Guest" and Melody.....just proves there might be  more than one way to solve a problem!!!

Let  A  be the equal amount spent  by each....and we know that

[ 160  - A ]          3

_________  =   ___             cross-multiply

[ 240 - A ]           5

5 [ 160  - A ]   =  3 [ 240 - A]     simplify

800 - 5A =  720 - 3A      add 5A to both sides, subract 720 from each side

80  =  2A     divide  both side  by 2

$40  = A  = amt each spent