CPhill

That's a pretty neat method, Alan....I haven't seen it before.....!!!

Top  left figure....we have  a hexagon....it it composed of 12 equal right triangles.....

each  has  legs  of  45/2  and 45 / [2√3] units.....each triangle has an area  of  (1/2) (product of the legs)...so we have

Area  =

12 * (1/2) * (45/2) * (45 / [2√3 ] =

6 * 45^2 / [4√3] =

2025√3 / 2 units^2  ≈

1753.7  units^2

The next three figures are  equilateral triangles.....their areas  =  [√3/4] * side^2

I'll calculate the area of the one with a side length of   20√3....you should be able to do the other two

Area  = [√3/4 ] [20√3]^2   =  [√3/ 4] [ 1200]  = 300√3  units^2 ≈ 519.6 units^2

The last two figures are also equilateral triangles  with circumradiuses  of 14 and 28...the area of either  is given by  3* (1/2)(circumradius)^2 * √3/2

The area of the  one with the circumradius of 14  =

3*(1/2) (14)^2 * √3/2  =

(3/4)(196)√3  =

147√3  units^2 ≈

254.6 units^2

You should be able to calculate the area of the one with the circumradius of 28 based on this

P(short)  =  1  - P(tall)  =  1  - .15   =  .85  = 85%

Eh....I'm pretty sjure that you could have done the same thing, Coldplay....

( pq^3)^3 (4p^2q)^2

________________                using  (a^m)^n   =  a^(m * n)....so we have

     (2pq^2)^3

(p^3 * q^9) (4^2 * p^4 * q^2 )

_______________________

  ( 2^3 * p^3 * q^6)

( p^3 * q^9 * 16  * p^4  * q^2)

________________________         using  a^m * a^n  =  a^(m + n)

   (8 * p^3 * q^6)

16 *p^7 * q^11

_____________           using    a^m / a^n   =  a^(m - n)

     8 * p^3 * q^6 

2 * p^4 * q^5          just as Coldplay found   !!!!!

Last  one

x  ≤ -4   this means  ( -inf, 4 ]

or          

x  ≥  5    this means  [ 5, inf )

So  we  have

(-inf, 4 ] or [5, inf )

Second one

y  + 3x ≥ 2    ⇒  y  ≥ -3x + 2

y + 4 ≥ (1/2)x   ⇒   y ≥ (1/2)x - 4

Look at the graph  on the bottom left...note that   (2,0)  is a point in the shaded region

So...replacing these into the inequalities, we have that

0 ≥  -3(2) + 2              and          0 ≥ (1/2)(2) - 4

0 ≥ -6 + 2                                    0 ≥ 1 - 4

0 ≥ -4   true                                 0 ≥  -3    true

So.....this graph is the correct one

For the first one...let's pick a point in the shaded region...I'll choose  (-1, 6)

This point should satisfy all of the inequalities  in any answer

Look at the third  answer....we have that

6 - 6  ≤ -11 (-1)         6  + 3(-1) ≥  -4           -2(-1)  + 3(6)   > 15                       

0  ≤ 11   true             6  - 3  ≥ -4                     2    +  18   >   15

                                     3  ≥  -4   true                20  >  15     true

So.....this answer is the correct one

x = (1/4)y + a               y  = (1/4)x + b

4x = y + 4a                  4y = x + 4b

y = (4x - 4a)                x  = (4y - 4b)

y = 4 ( x - a)                x = 4 ( y - b)

2 = 4( 1 - a)                 1  = 4 (2 - b)

(2/4)  = 1 - a               (1/4)  = 2 - b

(1/2)  = 1  - a                b = 2 - 1/4

a  = 1  -1/2                   b = 8/4 - 1/4

a  = 1/2                         b = 7/4

So...a +  b  =  1/2 + 7/4  =  2/4 + 7/4   =  9/4   

Thanks Daisy!!!!....here's another way

j  = C / k      where C is the proportionality constant

And we know that

16 = C / 21      multiply both sides by 21

336  = C

So  when  k = 14  we have that

j = 336 / 14  =   24