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Here's the Calculus approach.....this may have been answered elsewhere

Let the hypotenuse form the base

And since it's a right triangle, we have that the area = L1 *L2 / 2   where L1 and L2 are the legs

But, by the Pythagorean Theorem L2 =  sqrt( 1 - L1^2)

And let L1  = x

So....the area is given by

A(x) =  [x * sqrt( 1 - x^2 ]  / 2      ....so ....taking the derivative, we have

A' (x) =   (1/2) [ sqrt(1 - x^2)  +  x *(1/2) (1 - x^2)^(-1/2) (-2x)] 

A' (x) = (1/2) [  (1 - x^2)^(1/2) - x^2(1 -x^2)^(-1/2)]  set the derivative to 0

(1/2) [  (1 - x^2)^(1/2) - x^2(1 -x^2)^(-1/2)]  = 0      multiply both sides by 2

 [  (1 - x^2)^(1/2) - x^2(1 -x^2)^(-1/2)]  = 0

[ 1 - x^2]^(-1/2)  [ ( 1 - x^2 - x^2] = 0     multiply through by [ 1 - x^2]^(-1/2) 

[1 - 2x^2]  =  0

2x^2   = 1

x^2  = 1/2

x = 1/√2  = L1

And L2  =   sqrt (1 - L1^2)  = sqrt(1 - x^2)  = sqrt(1 - 1/2)  = sqrt(1/2) = 1/√2

So  ....L1  = L2     and the max area   = (1/2)L1* L2  = (1/2)(1/√2)(1/√2)  = 1/4 m^2

We could take the second derivative to prove this is a max., but it's messy.....much better to graph the area function and note where the max occurs.....here it is :

https://www.desmos.com/calculator/hmmyq9o0io

Note that when x = L1 = L2 = 1/√2     the area is maximized at 1/4 m^2

BTW.....the geometric approach which I presented above seems way more straightforward...!!!

Compute the definite integral:
 integral_0^1 sqrt(1-x^2) dx
For the integrand sqrt(1-x^2), substitute x = sin(u) and  dx = cos(u)  du.
Then sqrt(1-x^2)  =  sqrt(1-sin^2(u))  =  sqrt(cos^2(u)).
This substitution is invertible over 0<u<pi/2 with inverse u = sin^(-1)(x).
This gives a new lower bound u = sin^(-1)(0) = 0 and upper bound u = sin^(-1)(1) = pi/2:
  =   integral_0^(pi/2) cos(u) sqrt(cos^2(u)) du
Simplify cos(u) sqrt(cos^2(u)) assuming 0<u<pi/2:
  =   integral_0^(pi/2) cos^2(u) du
Write cos^2(u) as 1/2 cos(2 u)+1/2:
  =   integral_0^(pi/2) (1/2 cos(2 u)+1/2) du
Integrate the sum term by term and factor out constants:
  =  1/2 integral_0^(pi/2) cos(2 u) du+1/2 integral_0^(pi/2) 1 du
For the integrand cos(2 u), substitute s = 2 u and  ds = 2  du.
This gives a new lower bound s = 2 0 = 0 and upper bound s = (2 pi)/2 = pi:
  =  1/4 integral_0^pi cos(s) ds+1/2 integral_0^(pi/2) 1 du
Apply the fundamental theorem of calculus.
The antiderivative of cos(s) is sin(s):
  =  (sin(s))/4|_0^pi+1/2 integral_0^(pi/2) 1 du
Evaluate the antiderivative at the limits and subtract.
 (sin(s))/4|_0^pi = (sin(pi))/4-(sin(0))/4 = 0:
  =  1/2 integral_0^(pi/2) 1 du
Apply the fundamental theorem of calculus.
The antiderivative of 1 is u:
  =  u/2|_0^(pi/2)
Evaluate the antiderivative at the limits and subtract.
 u/2|_0^(pi/2) = pi/(2 2)-0/2 = pi/4:
Answer: |  =  pi/4

55 X 25 = 1375

Assuming that the center of the island is "dead ahead,"  we have a triangle with a height of 6 mi, and assuming that the mermaids were an equal distance from the center of this island, this triangle is isosceles with an apex angle of 120 degrees.  And the base of this triangle is their distance from each other.....and the height of the triangle bisects this 120 degrees....so we actually have two congruent 30-60-90 triangles formed by this bi-section........the distance from one mermaid to the center of this island is opposite the 60 degree angle in one of these triangles and the distance to the island - 6 miles - is opposite the 30 degree angle......so...the distance that one mermaid is from the center of the island = the base of one of these 30-60-90 triangles =  6√3 miles

So....the total distance between them = 2*6√3 miles = 12√3 miles = about 20.8 mi.

2z - 4 +4 = 12 +4

add 4 to both sides

2z /2 = 16 /2

divide both sides by 2

ANSWER: z = 8

Huzzah! haha 

4       7      13         25        49 

    3        6        12        24

 The next number is 49

http://www.cymath.com/answer.php?q=2t^2-25t-250

Good, haha, i'm not as concerned about it now :P

Ok good that's what I got haha 

\(75^2=70\times80+5^2=5625\)

STEPS:

-2(x+3) < 8                1^st: Multiply both sides by -1 (reverse the inequality)

2x + 6 - 6 > -8 - 6      2^nd: subtract 6 from both sides

2x /2 > -14 /2            3^rd: divide both sides by 2

X > -7                         SOLUTION

We can actually just plug-and-chug

Start putting in numbers for X and see that it goes to infinity

1: 1/2-4 = -7/2

2: 1-1 = 0

3: 3/2-4/9 = 19/18

4: 2-1/4 = 3/2

5: 5/2-4/25 = 117/25

Etc.

You can see it is progrssively getting bigger.

10,000: 5,000-1/25,000,000 = 4,000 and 24,999,999/25,000,000

Do you mean like this?

\(\sqrt{\frac{X}{Y}*\frac{A}{B}}\)

Or like this?

\(\sqrt{\frac{\frac{X}{Y}}{Z}}\)

some thing big

Thanks, CD.......that property of the radius meeting a tangent at right angles comes in handy, at times....!!!

Genius CPhill strikes again! Thanks a million...you made it so much easier to understand! Now, I'll know how to do this in the future. Muchas Gracias, Senor CPhill! Lol ^_^ 

Good job, gretzu......B is correct.....!!!

Angle CDO is a right angle, because a radius drawn to a tangent always meets that tangent at right angles.

And by the Pythagorean Theorem, OD^2 + CD^2  = CO^2

But, CO  = (BC + OB)   and OB = OD  because they are radii

So.......by substitution.....

OD^2  + CD^2  = CO^2

OD^2 + CD^2 = (BC + OD)^2

OD^2  + 30^2  = ( 13 + OD)^2      simplify

OD^2 + 900  = 169 + 26OD + OD^2     subtract OD^2  from both sides

900 = 169 + 26OD      subtract 169 from both sides

731  = 26OD   divide both sides by 26

731/ 26  = OD   = about 28.1

But OD is a radius ...... and the diameter is twice this   = 56.2 =  "B"

What exactly are asking help with?