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y = x^2  +  a

y = ax

Setting the y's equal 

x^2 + a   = ax       rearrange as

x^2  - ax  +  a   = 0

For us to have real solutions.....the  discriminant must be ≥ 0

So

a^2  - 4a  ≥  0

a ( a - 4) ≥  0

This will  be true on these intervals

a ≥ 4       and      a ≤ 0

So

(-infinity , 0 ]   U   [4, infinity )

Let DAC  = A      EBD  = B     ACE  = C    BDA  = D     CEB   = E

Starting a the top left and  moving clock-wise, call the  interior angles  of the  interior  irregular pentagon 

  F , G, H, I , J 

The sum of these interior angles =   540°

And the sum of the angles of the five triangles abutting  this irregular pentagon =  5 * 180  = 900°

So.....the    base  angles  on the sides of the  triangles  abutting this irregular pentagon  lead to this equation

2(180 - F) + 2(180 - G)  + 2(180 - H)  + 2 (180 - I) + 2(180 - J)  +  A + B + C + D + E  = 900

Simplify

1800 -  2 ( F + G + H + I + J)  +  A + B + C + D + E  =  900

1800 - 2(540)  +  A + B + C + D + E  = 900

1800 - 1080  + A + B + C + D + E =  900

720  + A + B + C + D + E   = 900

A + B + C + D + E   =  900  -720   =   180°

Thanks, but sorry, that isn't right.

THX, OlympusHero!!!!

(a + bi)^2   =  b +  ai

a^2  + 2abi  +  (bi)^2  = b + ai

a^2 + 2abi  - b^2   =   b + ai

Equating terms  we have that

2abi  =  ai

2b  =  1

b = 1/2

And

a^2 - b^2  =  b

a^2  -  (1/4)  = (1/2)

a^2 =  3/4

a = √3 / 2

So

(a,b)    =  ( √3 / 2 ,  1 / 2  )

x^2  + 2bx  +  (a - b)   =  0 

If this has one real solution then  the discriminant must  =  0 ......so.....

(2b)^2  - 4(a - b)   =  0

4b^2  - 4a + 4b   =  0

b^2  -  a  +  b   = 0

a  =  b^2 + b

Note  that   the orginal  equation  reduces to

x^2  +2bx + b^2  =  0      which  factors  to

(x + b)^2  =  0        which only has  one real solution......x  = -b

sin (10°) sin (50°) sin(70°)  =

sin(70°)sin(50°)sin(10°)  =         [ because   sin A   = cos (90 - A)  we can write  ]

cos(20°) cos(40°)cos(80°)

Note that

sin (2A)  =  2 sinA cos A       solve for   cos A     and we have that

cos A =   sin 2A

              _______

               2 sin A

So

cos (20°)   =    sin (40°)

                       _________

                         2 sin (20°)

cos(40°)  =   sin (80°)

                   _________

                      2sin(40°)

cos(80°)  =   sin (160°)

                    __________

                     2sin(80°)

Multiplying these together, we have that

sin (40°)    *     sin ( 80°)      *    sin (160°)

_______        _________        __________  =  

2sin(20°)         2 sin(40°)           2sin(80°)

sin (160°)

_________   =                     [  note  that sin 20°  =  sin (180 - 20°)  = sin 160°]

8 sin(20°)

1

__

 8

BTW......this  is  the application of something known as " Morrie's Law"

The name is due to the famous physicist Richard Feynman, who used to refer to the identity under that name. Feynman picked that name because he learned it during his childhood from a boy with the name Morrie Jacobs and afterwards remembered it for all of his life.

There  are   2^6  =   64 possibilites

He will be less than  two milles from his starting point only when  3 heads  and 3 tails  occur  in any sequence of 6 flips

So......  

TTTTTT  =    1  way

TTTTTH  =    6 ways

TTTTHH  =  choose  tails to occur in any four of six positions  =     C(6,4)  = 15 ways

HHHHHH  = 1 way

HHHHHT  =  6 ways

HHHHTT = choose heads to occur  in any four of six positions  C(6, 4)  =  15 ways

So......the probability =

2 [  1 +  6   + 15 ]               22               11

______________   =        ___  =        ___ 

           64                           32               16

Sorry Melody: I thought I had found a mistake, which on second thought, is NOT a mistake !!!. I was taking "combinations of the 4 numbers, instead of their permutation", which you have it accurately counted. Shows you how much I know about combinations and permutations! I don't really know where the difference of 2520 - 1440 =1080 is !!.

By the way, I was able to list 1/3 of 1440 = 480 permutations, and here is the list. See if you can find any mistakes or duplicates in it:

(112233, 112234, 112235, 112236, 112243, 112244, 112245, 112246, 112253, 112254, 112255, 112256, 112263, 112264, 112265, 112266, 112332, 112342, 112352, 112362, 112432, 112442, 112452, 112462, 112532, 112542, 112552, 112562, 112632, 112642, 112652, 112662, 113223, 113224, 113225, 113226, 113232, 113242, 113252, 113262, 113322, 113422, 113522, 113622, 114223, 114224, 114225, 114226, 114232, 114242, 114252, 114262, 114322, 114422, 114522, 114622, 115223, 115224, 115225, 115226, 115232, 115242, 115252, 115262, 115322, 115422, 115522, 115622, 116223, 116224, 116225, 116226, 116232, 116242, 116252, 116262, 116322, 116422, 116522, 116622, 121233, 121234, 121235, 121236, 121243, 121244, 121245, 121246, 121253, 121254, 121255, 121256, 121263, 121264, 121265, 121266, 121332, 121342, 121352, 121362, 121432, 121442, 121452, 121462, 121532, 121542, 121552, 121562, 121632, 121642, 121652, 121662, 122133, 122134, 122135, 122136, 122143, 122144, 122145, 122146, 122153, 122154, 122155, 122156, 122163, 122164, 122165, 122166, 122331, 122341, 122351, 122361, 122431, 122441, 122451, 122461, 122531, 122541, 122551, 122561, 122631, 122641, 122651, 122661, 123213, 123214, 123215, 123216, 123321, 123421, 123521, 123621, 124213, 124214, 124215, 124216, 124321, 124421, 124521, 124621, 125213, 125214, 125215, 125216, 125321, 125421, 125521, 125621, 126213, 126214, 126215, 126216, 126321, 126421, 126521, 126621, 131232, 131242, 131252, 131262, 131322, 131422, 131522, 131622, 132132, 132142, 132152, 132162, 133212, 134212, 135212, 136212, 141232, 141242, 141252, 141262, 141322, 141422, 141522, 141622, 142132, 142142, 142152, 142162, 143212, 144212, 145212, 146212, 151232, 151242, 151252, 151262, 151322, 151422, 151522, 151622, 152132, 152142, 152152, 152162, 153212, 154212, 155212, 156212, 161232, 161242, 161252, 161262, 161322, 161422, 161522, 161622, 162132, 162142, 162152, 162162, 163212, 164212, 165212, 166212, 211233, 211234, 211235, 211236, 211243, 211244, 211245, 211246, 211253, 211254, 211255, 211256, 211263, 211264, 211265, 211266, 211332, 211342, 211352, 211362, 211432, 211442, 211452, 211462, 211532, 211542, 211552, 211562, 211632, 211642, 211652, 211662, 212133, 212134, 212135, 212136, 212143, 212144, 212145, 212146, 212153, 212154, 212155, 212156, 212163, 212164, 212165, 212166, 212331, 212341, 212351, 212361, 212431, 212441, 212451, 212461, 212531, 212541, 212551, 212561, 212631, 212641, 212651, 212661, 213123, 213124, 213125, 213126, 213312, 213412, 213512, 213612, 214123, 214124, 214125, 214126, 214312, 214412, 214512, 214612, 215123, 215124, 215125, 215126, 215312, 215412, 215512, 215612, 216123, 216124, 216125, 216126, 216312, 216412, 216512, 216612, 221133, 221134, 221135, 221136, 221143, 221144, 221145, 221146, 221153, 221154, 221155, 221156, 221163, 221164, 221165, 221166, 221331, 221341, 221351, 221361, 221431, 221441, 221451, 221461, 221531, 221541, 221551, 221561, 221631, 221641, 221651, 221661, 223113, 223114, 223115, 223116, 223131, 223141, 223151, 223161, 223311, 223411, 223511, 223611, 224113, 224114, 224115, 224116, 224131, 224141, 224151, 224161, 224311, 224411, 224511, 224611, 225113, 225114, 225115, 225116, 225131, 225141, 225151, 225161, 225311, 225411, 225511, 225611, 226113, 226114, 226115, 226116, 226131, 226141, 226151, 226161, 226311, 226411, 226511, 226611, 231231, 231241, 231251, 231261, 232131, 232141, 232151, 232161, 232311, 232411, 232511, 232611, 233121, 234121, 235121, 236121, 241231, 241241, 241251, 241261, 242131, 242141, 242151, 242161, 242311, 242411, 242511, 242611, 243121, 244121, 245121, 246121, 251231, 251241, 251251, 251261, 252131, 252141, 252151, 252161, 252311, 252411, 252511, 252611, 253121, 254121, 255121, 256121, 261231, 261241, 261251, 261261, 262131, 262141, 262151, 262161, 262311, 262411, 262511, 262611, 263121, 264121, 265121, 266121) Total = 480
 

3. How many ways are there to put 4 balls in 4 boxes if the balls are not distinguishable but the boxes are?

Call  k the number of balls

Call  n  the number of boxes

The number of ways  [ with no restrictions....a box may  be empty ]  =

C ( k + n - 1 , n - 1)   =  C ( 4 + 4 - 1 , 4 - 1)   =  C ( 7, 3)   =  35 ways

1. How many ways are there to put 4 balls in 4 boxes if the balls are distinguishable and the boxes are distinguishable?

Call the  number of balls  =  k

Call the number of boxes   =  n

The number  of  ways to  distribute  these  [ with no restrictions.....a box may be empty   ]  =

n^k  =  4⁴  =  256 ways

2. How many ways are there to put 4 balls in 4 boxes if the balls are not distinguishable and neither are the boxes?

This is just the number  of  ways to partition  the  number 4

So  we have

4  0    0

3  1    0

2   2   0

2   1   1

The boxes are indistinguishable  so   any   one of these distributions  is also indistinguishable...in other words

4 0  0      is the same as  0 4 0     or   0 0 4

So....4 ways

We have  3 possibilities

All  10    were   heads =     only 1 way

9  were heads =  C (10,9)  = 10 ways

8 were heads  = C(10,8)  = 45 ways

So the total  number of possible sequences =   1 + 10  +  45   =   56

Diagonal of red square =   5√2  =  side of second largest square

5√2 * √2  =  5 * 2   = 10    =  side of largest square

So  the area of the largest square  =   10^2  =   100 units^2

(192, 301, 300, 300, 300, 300, 300, 300, 300, 300) >>>>>>Total Number = 2893
(0, 301, 600, 900, 1200, 1500, 1800, 2100, 2400, 2700)>>Total Sum = 13501

As you can see, zero appears a fewer number of times (192).

1   2     3     4       5      6       7       8           9

LCM  =

5 * 7 *  2^3  *  3^2 =    2520 

Let D  =  David's sheep     H  =Harry's sheep   P  = Peter's sheep   S   =Sam's sheep

We have these equations

D + 3  = H  - 1 ⇒      D + 4   = H                (1)

D + H + P = S          (2)

P - 3  = 3D     ⇒   3D + 3   = P          (3)

And the last statement means that

D + H + P + S  = 44   (4)

Using (2)  and (4)   we have that

[D + H + P]  = S     so  subbing this into (4)  we have that

S  +  S    =  44

2S  =44

S = 22    = Sam's sheep

So

D + H + P =  22

And subbing  (1)  and (3)  we have that

D +  (D + 4)  + (3D + 3)  =  22

5D + 7   =22

5D  = 15

D  = 3    =  David's sheep 

Hi Melody: Here is a short computer code which is very brief version of the full code. It calculates 16 combinations for for 1 of 6 numbers. To write a "full conditional statement", you have to repeat the "if condition" 8 x 6 x 2 =96 times !!!! So, I simply took a "shortcut" and arrived at 1440 as explained below the code:

a=0;b=1;c=1;d=1;e=1;f=1;g=1;p=0; cycle:n=b*100000+c*10000+d*1000+e*100+f*10+g; if(b==1 and c==1 and d==2 and e==2  and f!=1 and f!=2 and g!=1 and g!=2, goto loop, goto next);loop:printn," ",;p=p+1; next:g++;if(g<7, goto cycle, 0);g=1;f++;if(f<7, goto cycle, 0);g=1;f=1;e++;if(e<7, goto cycle,0);g=1;f=1;e=1;d++;if(d<7, goto cycle,0);g=1;f=1;e=1;d=1;c++;if(c<7, goto cycle,0);g=1;f=1;e=1;d=1;c=1;b++;if(b<7, goto cycle,0);g=1;f=1;e=1;d=1;c=1;b=1;a++;if(a==0, goto cycle,0);print"Total = ",p

OUTPUT: 112233  112234  112235  112236  112243  112244  112245  112246  112253  112254  112255  112256  112263  112264  112265  112266  Total =  16
This is only for 1 number. Then you cycle throug 6 numbers =16 x 6 = 96. and this is only one go for 15 combinations:6 C 4 = 15. And 15 x 96 =1440. This is how I arrived at 1440.

P.S. I found your mistake in this calculation:   x and y are different. 4*3*6!/2!2! = 12*180 =   2160 ways

The mistake is : when x, y are different, you only have 6 choices not 12 as follows: (3,4), (3,5), (3,6), (4,5), (4,6), (5,6) making your calculation =6 * 6! / 2!2! =1080. Then add it to your first calculation and we get:

1080 + 360 = 1440 which agrees with the computer output.

a) By stars-and-bars, the number of ways is C(52,4) = 270725.

b) By stars-and-bars, the number of ways is C(52,5) = 2598960.

22343234532^2 = 499220129351957259024

22343234533 * 22343234531 = 499220129351957259023

SUBTRACT  one from the other and the answer is   1

Compute (22343234532 X 22343234532)-(22343234533 X 22343234531).      

The final answer is 1     

Adding commas to make the numbers easier to read, we obtain    

(22,343,234,532 X 22,343,234,532) – (22,343,234,533 X 22,343,234,531).    

For convenence, allow N to stand for 22,343,234,532    

Now the expression becomes    [N • N]  –  ( [N + 1] • [N – 1] )

Multiply it out                               N² – (N² – 1)      

This equals                                                1        

_.

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