CPhill

B.  the product of 7  and a  number x   [ "product"  means "multiply" ]

"A"  is correct....

If ABC  is equilateral and AB  = 2....then the perimeter    = 6

"D'   i s the correct  choice....

Let  B  =(0,0)   A  = (0,6)   and C  = (8,0)   M  =(0,3)   N  = (4,0)

Let  the slope of MC  = [ 3 -0 ]  / [ 0 - 8]  =  -3/8

And the equation of MC is   y  = -(3/8) x + 3     (1)

Let the slpoe  of AN  = [ 6 - 0 ] / [ 0 - 4 ] = -6/4  = -3/2

And the  equation of An  is   y  = (-3/2)x + 6    (2)

Set (1)  = (2)  to find the x coordinate of P

(-3/8)x + 3  = (-3/2)x + 6

(-3/8 + 3/2)x  = 3

(9/8)x  = 3

24/9 = x  = 8/3

So...the y  coordinate of P  is  (-3/2)(8/3)  + 6 =  -24/6 + 6=  -4 + 6  = 2  = altitude of triangle PNC

The area of triangle APC  = area of triangle ANC  - area of triangle  PNC

Area of ANC  =(1/2) (AB) (NC)  = (1/2)(6)(4)   = 12  units^2

Area  of PNC  = (1/2) (2)(NC) = (1/2) (2) (4)  =4  units^2

So...area of triangle APC =  12 -  4  = 8 units^2

If AK  is an altitude  drawn to  base  BC.....then angles AKB  and AKC  form  right angles

So......using the Pythagorean  Theorem......

√[AK^2 + BK^2] =  AB  = √ [ 6^2 + 8^2  ] =  √100  = 10  units

Ans

√] AK^2 + CK^2 ] =  AC  = √ [ 6^2 + 6^2 ] = √72  = 6√2  units

And  CK + BK  = BC  =  6 + 8  = 14

So...the perimeter is  [ 10 + 6√2 + 14  ] =   24 + 6√2  units

What  is  " P ??? "

33 + 24  + 8  =  65

Thus....this would be 15 students too many

So...15  play both

33 - 15  = 18  play only hockey

24 - 15  = 9   play only  baseball

18 + 15   = 33

9 + 15  = 24

:

r =      5 / [ 1 + cos θ]

Note :  r = √ [ x^2 + y^2]       and    cos θ  =  x / r   

So we have

r = 5 / [ 1 + x/r ]

r [ 1 + x / r]  =   5

r  + x  = 5

 √ [ x^2 + y^2]  +  x  = 5      subtract  x from both sides

 √ [ x^2 + y^2]   =  5  - x      square both sides

x^2 + y^2 =  25  - 10x  + x^2       subtract x^2 from  both sides

y^2  = 25 - 10x

I think this can be solved by a graph as well, Coldplay...here :  https://www.desmos.com/calculator/myszb3kghi

The total possible area for (x, y) is a 3 by 3 square = 9 units^2

The feasible region of  (x, y)  values occur inside the boundary points (0,1), (1,0), (2,3), (3,2) and (3,3)

We have an  upper rigtt triangle with an area of (1/2) (product of the leg lengths)  = (1/2) (1) (1)  = 1/2

And we have a rectangle of length  √ [ (3 - 1)^2 + (2 - 0)^2 ] = √ [2^2 + 2^2] = √8

And a height of  √[ (2-3)^2 + (3 - 2)^2] = √ [1^2 + 1^2 ] = √2

So...the area of this rectangle is  √8 * √2 = √16  = 4

So....the toal area of the feasible region is  [ 4 + 1/2]   =  4.5 units^2

So...the probability that a triangle exists  is   4.5 / 9   =  1/2