CPhill

sin x  =  9/10  = y/r

So....cos x / r

To find x

  x  = ±√[r^2- y^2 ]  =  ±√[ 10^2  - 9^2]  =  ±√[100 - 81 ]  = ±√19

So....depending upon where x  lies

cos x  = √19 /100  ≈ .04    or    cos x  ≈  -.04  

Converting rads to degrees, we have

tan ( 60° -  A)  =  cot (60°)

And since they are co-functions....cot (60°)  =  tan(90° - 60°)  =  tan (30°)

So

tan (60 - A)  = cot(60)  = tan (30)

So

tan (60 - A)  - tan(30)

So   A  =  30°  =  pi / 6

Here's the first one

Let    a > c > b

And  since  a > b, then  l a - b l  =  a - b

And since  c > b, the l b - c l  =  c - b

And since a > c, then l c - a l  = a - c

Then  we have that

(a - b) + ( c - b) + ( a - c)  = 20  simplify

2a  - 2b  = 20       divide through by 2

a - b  =  10

And this is the max value for  l a - b l

 B.It shifts the graph down 5 units.

We have 5 photos and we want to choose only 3

So  C(5,3)  =  10

Call the pics  1, 2, 3, 4 and 5

Here are the possible sets :

(1,2,3)    (1,4,5)

(1,2,4)    (2,3,4)

(1,2,5)    (2,3,5)

(1,3,4)    (2,4,5)

(1,3,5)    (3,4,5)

For the first one....KJL  aand KML  are inscribed angles that intercept the same arc....inscribed angles that intercept the same arc are equal

So  Rarinstraw's  working is correct   !!!!

Second one...in this situation....V  will equal (1/2)  the difference  of arcs ZY and WX....so we have

V  = (1/2) (ZY - WX)

V  = (1/2) ( 130 - 50 )

V  = (1/2) (80)

V  = 40°

                    O        A       B     AB     Total

Male (M)       210    174   74     42     500

Female (F)   315     261   111   63     750     

Total             525     435   185  105   1250

 P(F l A )  =  261/435  =  3/5

If independent  P(F and A)  = P(F) * PA) 

P(F and A)  = 261 / 1250

P(F) = 750/1250      P(A)  = 435/ 1250

So....  P(F) * P( A)  =   (750/1250) * (435/1250)  = 261/1250

So  independent

P( F and A)  = P(F) * P(A)

So...  "C"  is correct 

We have

Area  = L * W    

20 = (2x - 1) (x - 2)   simplify

20  = 2x^2 - x - 4x + 2     subtract 20 from both sides and we have

2x^2 - 5x - 18  = 0      factor

(2x - 9) ( x + 2)  =   0

Only  the first factor set to 0  will yield a positive result for x

2x  - 9  =  0

2x  = 9

x = 9/2 

So....the length  is  2(9/2) - 1   =    9 -  1   =   8

Nothing shows up by the probabilities......

First one

This is an equilateral triangle....all sides are equal....so....this implies that

2x + 5  =  4x  - 1       add 1 to both sides, subtract 2x from each side

6  = 2x       divide both sides by 2

3  = x

Second one....again...we have an equilateral triangle...so...

5x  - 4  = 3x      add 4 to both sides....subtract 3x from both sides   

2x  = 4   divide both sides by 2

x  = 2

So  AB  =  5(2) - 4   =  6

Third one.....this is  a 30-60-90  right triangle....the hypotenuse is opposite the 90 degree angle

And the side oppposite the 30 degree angle is 1/2 of the hypotenuse

So  x  =  (1/2) *8   =  4

Last one.....this is a 45-45-90 right triangle

The side opposite the right angle  - x -  is  √2 times as long as either of the other two sides

So

x  = √2 * 6   =  6√2