CPhill

The area is 

pi * (diameter / 2)^2  =

pi *  (3 /2)^2  =

pi *  (9/4)  =

(9 / 4) pi   m^2

We know the opposite side from the angle and the hypotenuse

Therefore....the sine inverse will let us find  x

arcsin  (32 / 58 )  ≈   33.49°  = x

(7/10)  (d - 5/6 )  = 7  multiply both sides by 10/7

d  - 5/6  = 10       add 5/6 to  both sides

d  = 10 + 5/6  =     65/6

5c / 4  - 2c / 3  =  7/10     multiply through by the LCM of 4, 3 and 10  = 60

75c  -  40c  = 42

35c = 42    divide both sides by  35

c = 42/35   =  6/5

11)The lines y = (5/12)x and y = (4/3)x are drawn in the coordinate plane. Find the slope of the line that bisects the acute angle between these lines.

Construct a circle centered at the origin with a radius of 1

The  equation  is

x^2 + y^2  = 1

We can find the x coordinate  of the intersection of the first line and this circle,thusly "

x^2  + [ (5/12)x ]^2  = 1

x^2 + (25/144)x^2   =1

[144 + 25] / 144  x^2  = 1

[169] /144 x^2  =1

x^2 = 144/169

x = 12/13

And y = (5/12)(12/13) = 5/13

So   ( 12/13, 5/13 )  is on the circle

Like wise...we can find the intersection of this circle with the  second line :

x^2  +[ (4/3)x\^2  =1

x^2  + (16/9)x^2  = 1

[9 + 16 ] / 9 * x^2  =1

25/9 x^2  =1

x^2  = 9/25

x = 3/5

And y = (4/3)(3/5)  = 4/5

So  (3/5, 4/5)  is on the circle

And a chord can be drawn on this circle with endpoints   ( 12/13, 5/13)  and (3/5, 4/5 )

And the midpoint of this chord is

  (  [12/13 + 3/5]  / 2,  [ 5/13 + 4/5 ] / 2 )  =

(99/130 , 77/130  )

And  the line drawn  from the origin to this point is the line we seek and it will have the slope

[77/130]  / [99/130]  =    77/99   =  7/9

10)The line y = (3x + 20)/4 intersects a circle centered at the origin at A and B. We know the length of chord AB is 20. Find the area of the circle.

We can write the line as

4y  = 3x + 20

3x  - 4y  + 20  = 0

The perpendicular  distance between the center of the circle and the line will bisect the chord

And this distance is

l  3(0) - 4(0)  + 20  l            20

________________  =   ______ =     4

    √[3^2 + 4^2] =                 5

A right triangle will be formed by the bisected chord, the perpendicular distance from the center to the midpoint of the chord and the radius which will form the hypotenuse.

So....the radius  is √[10^2 + 4^2 ]  =  √116  units

And the area  of the circle is

pi *  ( √116)^2  =

116 pi   units^2

f(x)  = 4x + p

g(x)  = px + 1

I think this is  (f ° g )(x)  = 12x + q

f ( g)  =   4(px + 1) + p     =  12x + q

4px + 4 + p   =  12x + q

4p  = 12      so  p  = 3

And

4 +  p  = q

4 + 3  = q  

7  = q

We have

The choice of any two toppings plus  1 with no toppings  =  106

C(n, 2)  +  1  = 106

C(n ,2)  = 105

We need to solve this

     n!

________  =   105

(n - 2)! 2!

     n!

______  =   105 * 2!

( n - 2)!

(n) (n - 1)  =  210

n^2  - n  - 210  = 0      factor

(n -15) ( n + 14)  =  0

Setting both factors to 0 and solving for n produces  n  = 15   or  n  = -14

Since n is positive....there are 15 toppings

No...I just can't add  !!....LOL!!!

We need to solve this

      n!  

________  =   365

(n -4)!* 4!

n!

_____   =   365 * 4!

(n - 4)!

n * ( n - 1) (n -2) (n - 3)  =  365 * 24

n^4 - 6 n^3 + 11 n^2 - 6n  = 8760

n^4  -6n^3 + 11n^2  - 6n - 8760   =   0

Solving this for n produces the positive solution  of  ≈ 11.239

So....we need 12 friends  to assure that the same 4 are not  invited

Proof

C(12,4)   = 495  different groups

C(11,4)  = 330    different groups  [ not enough ]

8 heads =  C(10,8)   = 45

9 heads  = C(10,9)  = 10

10 heads   = 10

So.....the number of sequences with at least  8 heads  is

45 + 10 + 1  =

56