CPhill

Nice, CoolStuff....!!!

See the image below

Let D = (0,0)

Let C = (24 sin60°, 24cos60°)   =  (12, 12√3)

Let A = (-12, 12√3)

And we can find B thusly

The distance from C to A   =  24

And, by symmetry, EC = 12

So triangle EBC is a 5-12-13 Pythagorean Right Triangle

So B =   (0,  12√3 + 5 )

Y is the midpoint of   DA   =  (-6, 6√3)

X is the midpoint of BC   =   (   [ 0 + 12] / 2 , [ 12√3 + 5 + 12√3] / 2 )  =

(6, 12√3 + 2.5)

So....the square of the distance from X to Y is given by :

(-6 - 6)^2 +  ( [ 12√3 + 2.5 ] - [6√3] )^2   =

(-12)^2   + [ 6√3 + 2.5 ]^2   =

144 + [ √108 + 2.5 ]^2

144  + 6.25 + 108 + 5√108 =

258.25 + 5√108  units  ≈ 310.21 ≈   (17.61)^2 units

See the graph here, NSS

https://www.desmos.com/calculator/0cqu6i2hjk

The x intercepts are  (-3, 0)   and (1, 0)

And the minimum of the function  is     -4     

Does this make sense ???

l b l   + b^3         b = -2

l - 2 l   +  (-2)^3    =

2  +   ( - 8)    =

-6

13^n    -  6^(n - 2)

Show it's true for n = 2

13^2 - 6^(2 - 2)   =

169 - 1  =  168

168 / 7   =  24

Assume that this is true for  n = k     where k ≥ 2

That is

13^k - 6^(k - 2)      is divisible by 7

Prove it's true for k + 1

That is

13^(k + 1) -  6^ (k + 1 - 2)   is divisible by 7       

[ note ....6^(k + 1 - 2)  = 6^(k - 2 + 1) ]

So we have 

13^(k+ 1)  - 6 ^( k - 2 + 1)

13^k * 13^1  -  6^(k-2) * 6^1

13 * 13^k -   6 * 6^(k - 2)

(6 + 7) 13^k - 6 * 6^(k - 2)

6 [ 13^k - 6^(k - 2) ]   + 7 * 13^k

And since we assumed that 13^k - 6^(k - 2) was divisible by 7, then the first term is divisible by 7, as well

And  the second term, 7*13^k,    is divisible by 7

g(x) is just f(x) shifted up 3 units

So

g(x)  =   f(x) + 3

Since the area of the trapezoid   ZWCD is 120

The height of this trapezoid is AB and the bases are ZD and WC

We have

120 = (1/2) AB  (ZD + WC)

120 = (1/2) 12 ( ZD + 6)     multiply through by 2

240 = 12 (ZD + 6)      divide both sides by 12

20 =  ZD + 6       ubtract 6 from both sides

14 = ZD

Now   angle DBW = angle BDZ

And angle ZQD = angle BQW

And since AZ = WC

Then  ZD = BW

So by AAS....triangle ZQD is congruent to triangle WQZ

And the altitude of trangle DQZ    = altitude of triangle BQW

And since twice these altitudes = AB....then each altitude = 1/2(AB) = 6

So....the area of triangle BQW   =

(1/2) BW * altitude of BQW   = 

(1/2) (14)(6) =

(1/2) (84) =

42 units^2

Thanks for that answer, PM....

The answers don't have any "space limits" ..........as long as you're giving good answers  [ or.....even attempting a difficult problem..... ] 

However......don't write a book.....the person asking the question will probably not appreciate extremely long answers

Thanks, GA......that's pretty interesting.....

Just a cursory examination of some lower powers of 2  seems to confirm this.....do you know the proof???

Let the side opposite the angle, θ, we are interested in = x

The hypotenuse = 25

So we have

sin θ = x / 25            differentiate both sides with respect to time, t

(d/dt ) sin θ   =     ( d / dt )  ( x /  25)

Remember that  on the left side

d / dt  = d / dθ * dθ / dt        where  d/dθ sinθ = cos θ         

And  on the right side

 d / dt   =  d / dx *  dx / dt          where   d/dx (x/25)  =  1/25

So we have....

 cos θ  * ( dθ / dt)   =  (1/25) * dx / dt            [ note that  dx/dt   = 2ft/s ]

dθ / dt    =     (1 / cos θ)  * (1/25) (2) 

dθ / dt  =   (2 / 25)  sec θ

When the ladder is 7 ft from the wall..... sec θ   =  

25 / √[ 25^2 - 7^2]  =    25 / √[625 - 49 ]  =  25 /  √576    =  25 / 24

So

dθ / dt   =      (2 / 25 )  * ( 25 / 24) rads / sec  =  1 / 12 rads/ sec ≈  4.77° / sec