CPhill

y= x

y =2x

We can  find  m thusly 

  m  =  tan [ (arctan(slope of one  line) + arctan(slope of second line) / 2  ]  

  =      tan ([ arctan (2)   + arctan(1)] /  2)   ≈ 1.387

Since AB  = AC and BAC = 30

Then angle  ACB = (180 -30) / 2  = 75

And since BAC = 30.....then  minor arc BC = 60...so  angle BCE = (1/2)(60) = 30

So angle ACD  =180  - BCE - ACB   = 180  - 75 - 30 =  75°

Semiperimeter of top triangle =  19/2 = 9.5

Semiperimeter of bottom triangle  = 20/2 = 10

Area =  sqrt [ 9.5 ( 9.5 - 9) (9.5 - 7) (9.5 -3) ] + sqrt [ 10 (10 - 9) (10 -6) (10 - 5) ] =

sqrt [ 9.5 * .5 * 2.5 * 6.5 ]  + sqrt [ 10 * 4 * 5  ] =

sqrt [ 77.1875 ] +  sqrt [ 200]  ≈  22.93

WY  (by graph) ≈  5.34

Draw IZ  and YI

Angle AZY = 44°

So   minor arc YZ = 2 * angle AZY = 2 (44)  =   88°

And central angle YIZ intercepts this arc....so  its measure = 88°

Let 12 o'clock represent  0°....let the hands advance in a  positive angular  direction

At  3 o'clock   the hour  hand is  at 90°   and the  minute  hand is  at 0°

The hour hand  moves   15°   more   in  30 minutes.....so it is at 105°   at 3;30

And the minute hand is  at180°    at 3:30

So....the acute angle is   180 - 105  =   75°

(90 + d) / (60 + d)  = r , (165 + d) / (90 + d)  = r

Setting equal  for r  and simplifying

(90 + d)^2  = (60 + d) ( 165 + d)

d^2 + 180d + 8100 =  9900 + 225d + d^2

-45d = 1800

d = 1800 / -45  =  -40

r = [ 90 - 40 ] / [ 60 - 40 ] =  50 / 20   =  5/2

(1000) (.9)^(n - 1)  =  1

.9^(n -1)n  = 1/1000      take the log of  both sides

log .9^n  = log (1/1000)

(n-1) log (.9)  =  log (1/10000

n -1 = log (1/1000) / log (.9) ≈  65.56

n = 65.56 + 1 =   66.56

66 terms will be > 1

If Alice receives $n$ dollars from Bob, then she will have $7$ times as much money as Bob. 

A + n  = 7 ( B - n)        (1)

If, on the other hand, she gives $n$ dollars to Bob, then she will have $2$ times as much money as Bob

A - n  =  2 ( B + n)      (2)

Subtract  (2)  from (1)

2n = 5B - 9n

11n = 5B

n = (5/11)B

Using (1)

A + (5/11)B  = 7 (B - (5/11)B)

A + (5/11)B  = 7B - (35/11)B

A = 7B - (40/11)B

A =  (37/11)B

A / B  =  37  / 11

pi =  180°

On the unit circle,  the terminal point of this angle  can  be represented  by (x , y)  =  (-1. 0)

tan (pi)  =   y /  x  =   0   / -1  =   0

At first   J / S  =  3 /2  ⇒     J = (3/2)S        (1)

After the changes we have

[J + 6 ] / [ S - 1 ]  = 18/7     (2)

Sub (1) into (2)

[(3/2)S + 6 ] / [ S - 1 ] =  18/7

(3/2)S + 6 =  (18/7) [S-1]

(3/2)S + 6 =   (18/7)S - 18/7

6 + 18/7 =   (18/7)S - (3/2)S           multiply though by common denominator of 2,7  = 14

84 + 36  =  36S - 21S

120 =  15S

S =  120 / 15  =  8

J = (3/2)(8)  =  12

After changes   J + S  =  [ 12 + 6 ]  + [ 8 -1] =  25