CPhill

Anchor Bush and Putin in their seats.....but remember ....we can seat them in 2 ways.....Bush to Putin's left or vice-versa....

Any of 5 remaining people can be chosen to sit between them   = 5  ways

And the remaining people can be seated in  4! ways = 24 ways

So  2 * 5 * 24   =  240 arrangements

Using   R=u^2∗sin(2θ)/g      where R is the range = 20m , u = the intial velocity =  60m/2 , g = 9.8m/s    and θ  = the launch angle  .....we have....

20m = [3600m^2/s^2] sin(2θ) / [9.8m/s^2]

20  = [3600/9.8]sin(2θ)

sin (2θ)  = [20*9.8] / 3600

sin(2θ)  = .54444  = about 1.5605°

Here's the graph of the arrow's path : https://www.desmos.com/calculator/purfk8aeng

1.   [9, 12, 16]       the triangle is possible

The triangle   9-12-15  is a right triangle with the angle opposite the longest side = 90°

Therefore.......the angle opposite the longest side here will be > 90°  since 16 >15

Then, this trangle is obtuse  [we can prove this using the Law of Cosines]

2. (11,14, and 27)   this triangle is not possible

Remember that the sum of any two sides of a triangle must be greater than the remaining sde.....but....     11 +  14   = 25         and 25 <  27       ....so......no triangle is formed

3. (1.5,3.6, and 3.9)   

Remember that, if we multiply every side by 10, we will have a similar triangle to the one given....so.....this triangle will be    15 - 36 - 39

Note   15^2  + 36^2  = 39^2   ....so    this triangle is a right triangle

4. (2,3.7, and 4.1)      multiply by 10 again and we have    20 - 37 - 41

Note that   sqrt[20^ + 37^2]   =  about 42.05

So.....since  a right triangle is formed  with sides of 20, 37 and   ≈ 42.05

And the angle opposite the right triangle with a side of about 42.05 would be 90°

But this triangle has a third side which  is < 42.05

Thus.....the angle opposite the side 41 in a 20 - 37 - 41 triangle will be < 90°

So.....this triangle is acute because the other two angles will also be acute [we can prove this using the Law of Sines]

One more comment about this problem....we will always have integer solutions and equal triangles when the quantties involved form Pythagorean Triples.......specifically......

If the isoceles sides of the triangles  =   "c"  in the formula  a^2 + b^2  = c^2

And if we let the base of one  triangle  = two times the length of one of the legs in the triple.....then the height will be equal to the remaining number in the triple.

y = sin (n * arcsin x)   n > 0

y '  =  cos (n*arcsin x) * [ n  (1 - x^2)^(-1/2)]

y ''  = -sin(n*arcsin x) * [ n  (1 - x^2)^(-1/2)]* [ n  (1 - x^2)^(-1/2)]  + cos(n * arcsinx)* [n * x * (1 - x^2)^(-3/2) ] =

-sin(n*arcsin x)* [n^2 (1- x^2)^-1] + cos (n*arcsinx) * [ nx * (1- x^2)^(-3/2) ]  =

[n (1-x)^(-3/2)] * [ xcos(n*arcsinx  - nsin(n*arcsinx)(1-x^2)^(1/2) ]

So....

(1 - x^2) y ''  - xy '  + n^2y    =

(1-x^2) [n (1-x)^(-3/2)] [ xcos(n*arcsinx)- nsin(n*arcsinx)(1-x^2)^(1/2)] -

x [cos (n*arcsin x)] * [ n  (1 - x^2)^(-1/2)]  + n^2 [sin (n * arcsin x)]  =

[ n (1 -x)^(-1/2) ] [ xcos(n*arcsinx)  - nsin(n*arcsinx)(1-x^2)^(1/2) ] 

- x [cos (n*arcsin x)] * [ n  (1 - x^2)^(-1/2)]  + n^2 [sin (n * arcsin x)]   =

[ xcos(n*arcsinx)] *  [ n (1 -x)^(-1/2) ] -

[ n (1 -x)^(-1/2) ] * [ nsin(n*arcsinx)(1-x^2)^(1/2)] -

[x cos (n*arcsin x)] * [ n  (1 - x^2)^(-1/2)]  + n^2 [sin (n * arcsin x)]   =

n^2 [sin (n * arcsin x)]  -

 n * (1 -x)^(-1/2)  * [ nsin(n*arcsinx)] * (1-x^2)^(1/2)  =

n^2 [sin (n * arcsin x)]   -  [ n* n * (1 -x)^(-1/2) * (1 -x)^(1/2) * sin(n*arcsinx) ] =

n^2 [sin (n * arcsin x)]  - n^2 [ sin(n * arcsinx)]  =

0

I'm assuming that the numbers are supposed to be "60%"   and "12"   ....if so

12  = 60%   of the total items to be purchased.......so....using a proportion, we have

12/60  = N / 100   where N  is the total items  on the list

1/5 = N / 100      cross-multiply

100/5  = N   =   20   items to be purchased

Nice, Melody.....I always like these, too !!!!

[3 x 10^8 ]   / [6.682 x 10^-7]   

We can write this as :

[3 / 6.682 ]  x [ 10^8 / 10^-7]  ≈

.449 x 10^15  ≈

4.49  x 10^14

18 months =  1.5 years......so....

He didn't live with them for  1.5 / 7.5  = 1/5  = 20%  of his life

So......he must have lived with them the other 80%