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I can see where 2/3 comes from.  I read of the question as the arclength. (not just the distance)

My error.   

I turned a trivial question into something much more complicated.

And please do not post AoPS questions here. 

It is against their code of conduct. 

Plus they claim their questions are copyright so you may be breaking the law. 

3x + 4y = -14.       re-arrange  as      y = -3/4 x - 14/4      slope , m = -3/4      perpindicular slope = - 1/m = 4/3

new line becomes

y = 4/3 x + b      sub in the given point to find 'b'

7 = 4/3 (-5) + b       so b = 13 2/3

y = 4/3 x  + 13 2/3        you can finish....I hope !    

Drop the vertical on the right  (see image below)

  find '30'   via pythagorean theorem

     then find  '12'   from     94 - 30 - 52 = 12

        then find   20   from pythagorean theorem (again)

Now you have the length of all four sides       52  + 34 + 94 + 20 = __________ units

thats wut i got just checking!!

Sum  =         first term  /  (1 -  r)

2296  =   328 / ( 1 -  r )

2296  ( 1 - r)  =  328

2296  - 2296r  =  328

2296 -  328  =   2296 r

1968 =  2296 t

r  =   1968 / 2296   =    6  / 7

Linear function   y = mx +b   , 

y =  (3/4)  x -14   .....L₁

Let  m = slope of L₁  = (3/4)  

m₂ =  slope of  L₂

if the lines are perpendicular :     slope   ( m₁ * m₂ ) =  ( - 1 )   

   (3/4) m₂   = -(1) ,     m₂   =  (-1)* (4/3)  =   -(4/3)

Then L_2:   y =  -(4/3)x + b.......L₂

_{Find b:   Plug in (x = -5, y  = 7 ) into}    y = -(4/3)x +b.....L₂

Thanks, catmg!!!!!.....I wanted to try  this one, as well

M -1  =   men's hands   that  Mr. Li  shook

M -1  =   men's hands   that  Mr. Li  shook

W -1  =  women's  hands that Mrs. Li shook 

M  =  men's  hands that Mrs. Li shook 

So

(M - 1)  / W   =    3    ⇒    M - 1  =  3W   ⇒   M = 3W  + 1     (1)

M / (W -1)  =  4     (2)

Sub  (1)   into (2)

(3W + 1)  / (W - 1) =  4

3W + 1  =  4 (W - 1)

3W + 1  =  4W  - 4

5  = W

3(5)   + 1  =  M  =  16

M + W  =    16 + 5   = 21   (just as catmg found  !!!)

w = # of women

m = # of men

Mr. Li is a man. 

He shook hands with m - 1 men. 

He shook hands with w women. 

3w = m - 1

Mrs. Li is a women. 

She shook hands with m men. 

She shook hands with w - 1 women. 

4(w-1) = m

3w = m - 1

4(w-1) = m

4w - 4 = m

3w = (4w - 4) - 1

w = 5

3(5) = m - 1

m = 16

16 + 5 = 21

=^._.^=

Let $a$ denote the first term of the AP and let $d$ be the constant so that the $n$th term is $a+(n-1)d$.  The sum of the first five terms is $5\cdot(a + a+4d)/2 = 5\cdot(a+2d) = 20 +10d$.  The sum of the next five terms is $5\cdot((a+5d) + (a+9d))/2 = 5\cdot(a+7d) = 20 +35d$.  The former is one-fourth of the latter, so $20+10d=(20+35d)/4$, which solves to $d=-12$.  Therefore, the 15th term is $a+14d = 4-14\cdot 12=-164$.

Rewrite  as

16t^2   - 26t    -  105   =  0           factor  as

(8t  +15)  (2t  -  7)    

The second factor  set to  0  and solved for t  gives  us what we need......t   =  7/2 sec  =  3.5 sec

agreed!

Very  nice logic, xC   !!!!

Let   the   middle  term  =  bx

Completing the  square  on this term  would  give us

(b  / 2)^2    = 4

b^2  / 4   =  4

b^2   =   16       take both roots

b = 4      or     b  =   -4

Since EF  ll  GH.......then  EH  =  GF

Then 

x^2 - 2x  =  48  - 4x

x^2 + 2x  - 48  =  0

(x + 8)  ( x - 6)  =  0

The second  factor  set to  0 ad solved  for  x gives  x =   6

So

EF  = (6)^2  - 2(6)  =   24°  =   GH

Then  arc  EPF  =  360  - 2*24  -  70   =  242° 

Try expanding (x+2)^2 and (x-2)^2 and see what you get. :))

=^._.^=

Thank y really appreciate it 

I'm really sorry, it's fixed. 

=^._.^=

It said it was wrong but thank you.

It's 2.

pi x - 17 < 20

pi x < 34

x < 10.8225361302

Largest integer solution is 10. 

=^._.^=

Edit:

20 + 17 is not 34

pi x < 37

x  < 11.7774657

Largest integer solution is 11. 

I'm so sorry I messed up the first one. 

This one is right.

=^._.^=

Nice job mps.

However, I think you forgot the 2xy part. 

=^._.^=

When the ball hits the ground, the height, y, will be zero

0 = -16t^2 + 26t + 105       Use the quadratic formula to solve for t ....throw out the negative value....

Let the function $f$ be $f(x)=ax^2+bx+c$.   The zeroes of $f$ is the roots of the quadratic equation $x^2+\frac{b}{a}x+\frac{c}{a}=0$.  We are given that the zeroes of $f$ are at $x=2$ and $x=4$.  So the zeroes are the roots of the equation $(x-2)(x-4) = x^2-6x+8 = 0$.  We are also given that $f(3)=6$.  From these facts we get a system of three equations in three unknowns $a,b,c$:
\begin{eqnarray*}
    \frac{b}{a} &=& -6 \\
    \frac{c}{a} &=& 8 \\
    9a+3b+c &=& 6.
\end{eqnarray*}
This solves to $a=-6$, $b=36$, and $c=-48$.  Thus, $f(x)=-6x^2+36x-48$.

Yes! thank you guys for helping me understand this problem. Aops gave me a hint and i managed to solve it without using pi!

The answer is 2/3 because like you picture melody, say the first point is exactly on a vertex of the triangle, then the other point could not land in the bottom third of the circle, so the probability is 2/3 that the triangle is shorter than a side of the triangle.

$x^2 + y^2 + 6 =25$

$x^2+y^2= \boxed{19}$

-mps