CPhill

By the Triangle inequality, we have these possible inequalities

x/2 + 15 >  47

x/2 > 32

x > 64

47 + 15 > x/2

62 > x/2

124 > x

x < 124

x/2 + 47 > 15

x/2 > -32

x > -64

We need to take the most restictive interval that will solve all three inequalities....this is

64 < x < 124

And the number of integers in this interval is   123 - 65 + 1  =  59

It cannot be negative because the right side of the original equation is positive......thus...x is positive

OK.....!!!

Let the number of dimes = N

Then the number of nickels  = N + 6

So

number of dimes * value of each + number of nickels * value of each = total money

So  we have

N * .10  +   (N + 6) * .05   =  1.35    simplify

.10N  + 05N + .30  = 1.35

.15N + .30  = 1.35      subtract  .30 from both sides

.15N = 1.05      divide both sides by .15

N = 7    (dimes)

And the number of nickels  =  N + 6  = 7 + 6  =  13

Correct......2 hours......!!!

The probabilty is given by :

Prob 2 hits  + Prob 3 hits + Prob 4 hits

C (4,2) (.37)^2 (1 -.37)^2  + C(4,3) (.37)^3 (1 -.37) + (.37)^4 ≈ .4724  = 47.24%

Here's the image :

Let D = (0,0)

Let C =   ( 24sin 60, 24 sin 60) =  (12, 12sqrt3 )

And by symmetry, angle BDC  = 60/2  = 30°

We can find sin DBC as follows

sin BDC /13  =  sin BCD/24

sin30/13 = sinDBC/24

1/26 =sinDBC/24

24/26 = sin DBC

12/13 = sin DBC

So

5/13 = cos BCD

And we can find BD using the Law of Cosines

DC^2  = BD^2 + BC^2 -  2(BD*BC) cosBCD

24^2  = BD^2 +13^2  - 2(BD *13)(5/13)

576 -169  =  BD^2 - 10BD

407 = BD^2 - 10BD

BD^2 - 10BD - 407 = 0

The solution to this is

BD =  5 + 12sqrt3

So  B =   (0 , 5 + 12sqrt 3 )

So X is the midpoint of   B and D  =   

[ 12/2,  [ 5 +12sqrt(3) + 12sqrt(3)] /2  ]  =

[ 6 , 2.5 + 12sqrt(3) ]

So....by symmetry... A is  (-12, 12sqrt3) 

So....Y  is the midpoint of D and A  =  (-6, 6sqrt(3) )

So....the distance^2  X to Y  =

[ 6 - (-6) ]^2  +  [ 2.5 + 12sqrt (3) - 6sqrt(3) ]^2  =

12^2  +  [ 2.5 + 6sqrt(3) ]^2  =

144 + 6.25 + 30sqrt(3)  +108  ≈  

310.21 units

angle addition postulate

alternate exterior angles

definition of a bisector

substitution property of equality

To   have integer solutions.....the discriminant must  either possibly  = 0   or possibly be a  perfect square.....so

a^2 - 4(5a) = 0

a^2 - 20a  =  0

a ( a - 20) = 0

Setting each factor to 0 and solving for a   produces  a =  0   or  a = 20

Or

a^2 - 20a   = perfect square

a (a - 20)  = perfect square

This happens when   a = 20  , a = 0 , a = -16  or a  =  36

So....4 values for a  produce integer solutions