CPhill

Ant, I assume we know the length of the medians....if so....

Call one median AD   and the other BE

Then,,,,we have two right triangles  ADC and BEC   

So.....this sets up a system of equations using the Pythagorean Theorem

[ (1/2)AC]^2 + [BC]^2  =  [BE]^2     ⇒   (1/4)AC^2  + BC^2  = BE^2    (1)

[ (AC]^2  +  [ [(1/2)BC]^2  = [ AD]^2  ⇒   AC^2  + (1/4)BC^2  = AD^2  (2)   

Multiply (1) by  -4       and we have

-AC^2   - 4BC^2  = - 4BE^2

AC^2  + (1/4)BC^2  = AD^2          add these

(-15/4)BC^2  =  -4BE^2  + AD^2

Since we assume that we know BE and AD, we can solve this for BC   and then use either (1) or (2) to find AC

Then.....just use thePythagorean Theorem again to find  AB

\(\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}\)

Note that

(  1)^3 - ( ∛3)^3  =   ( 1 - ∛3 ) ( 1 + ∛3 + (∛3)^2)  =     ( 1 - ∛3 )   ( 1 - ∛3 + ∛9)

So  this implies that if we multiply the numerator and  denominator by  ( 1 - ∛3)  we get

2∛9 ( 1 - ∛3)                         2 ∛9 - 2∛9 *∛3        2∛9 - 2∛27          2 [ ∛9 - 3 ]

__________________  = ______________  =  ___________  = ___________ =

( 1 - ∛3 + ∛9) ( 1 - ∛3)         (1)^3  - (∛3)^3            1  -  3                    -2

- [ ∛9 - 3 ]     =

3 - ∛9      [  =   3  - 3^(2/3) ]

(b) How many ordered triples (x,y,z) of integers are there such that \sqrt{x^2 + y^2 + z^2} = 7? Does the question have a geometric interpretation?

I  get the following  triples

6, 3, 2    

But each of these can take on pos/ neg   values for each integer   so there are 2^3  = 8 possibilities  and each of these can be arranged in 3! ways  = 6 ways....so  8 * 6  = 48  possibilities

And

7 0  0       the "7" can be pos/negative   = 2 possibilities  and can be in any 3 positions  = 2 * 3  = 6 possibilities

So.....(If I haven't missed any)  we have 48 + 6   =  54  points [ just as you found ]

This is a sphere centered at  (0, 0, 0)   with a radius of 7

Here's a (not so good) image : 

THX, hectictar.......very nice  !!!!

4x^2 - 12x +  a

We can write

(2x - m)^2  =   4x^2  - 4mx + m^2

This implies that 4m  = 12....so  m = 3     and m^2  = 9   = "a"

So we have

4x^2 - 12x + 9   =   (2x- 3)^2

The area of the smaller circle  = pi *r^2

The area of the dartboard itself is  pi (r + r)^2  =   pi (2r)^2   =   4 pi * r^2

So   the area between the circles  is   4pir^2 - pi*r^2   =  3 pi*r^2

And the blue area is 1/4 of this  = (3/4) pi * r^2

So.....the ratio of the blue area to the total area of the dartboard is  (3/4) / 4  =  3/16

If  60% of the darts miss the board, then 40% must hit the board = 200 * .40 =  80

And (3/16)  of these will land in the blue area  = (3/16) (80)  =   3 (80/16)  = 3 * 5   =   15 darts

(1-2i)(1-i)(1+i)(1+2i)   =

(1 + 2i) (1 - 2i) (1+ i) (1 - i) =

(1 - 4i^2) ( 1 - i^2)              [ remember that i^2  = - 1  ]

(1 - 4(-1) )  ( 1 - (-1) )  =

( 1 + 4)  (1 + 1)  =

5  *   2    =

10

The sum can be represented by :

a + (a + d)  + ( 2a + d) + ( 3a + 2d) + (5a + 3d) + (8a + 5d) + (13a + 8d) + (21a + 13d) + (34a + 21d) + (55d + 34d)  =   143a + 88d

And we know that   the  7th term  = 6, so

13a + 8d  =  6         multiply through by 11

143a + 88d  =  66

So....the sum is 66

sin A   =   2cosA

sin^2A  + cos^2A  = 1

[ 2cosA]^2 + cos^2A  = 1

4cos^2 A  + cos^2A  = 1

5cos^2A  =  1

cosA^2 A  =   1 /5

cosA =  ±√[1/5]

But....since this is a right triangle, the cosine could only be negative if angle A was obtuse which is impossble

So

cos A  = √[1/5]

10!  =  3628800

11!  = 39916800