All Answers 358501 Answers

Happy B-day hoping you have a great Birthday that you are another year old and _{one more year closer to being dead} But besides that Hoping your going to have a great day!

1. We will have to fit the pieces into 3 compartments like this:

If we try to place a piece between these compartments, it will create an unfillable area.

Within each of these compartments, there are only 2 ways to fit the pieces:

So the total number of possibilities   =   2 * 2 * 2   =   2³  =  8

They are:

1.  AAA

2.  AAB

3.  ABA

4.  ABB

5.  BAA

6.  BAB

7.  BBA

8.  BBB

And here is a picture of them:

LOL that is a good joke man and you scored it like your post made the question answered XD 

Holy cow that is funny!!! 

What is this? Mad Libs?

Sorry about that pushyyyyyyy!

Happy Birthday! 

THX, Guest.....impressive   !!!

Thanks :)

Nice, Pushy!!!

To  get the su of  a + b + c

(a + b + c)^2   =   25      [ as Pushy found ]

Take the square root

a + b + c  =   5

4x^2  + 5x + 2        factor out the 4

4(x^2 + (5/4)x + 1/2)

Take (1/2)  of 5/4  = (1/2)(5/4)  =  5/8.....square it  =  25/64.....add and subtract this within the parentheses

4 (x^2 + (5/4)x + 25/64 + 1/2 - 25/64)  simplify

4( x^2 + (5/4)x + 25/64  + 32/64  - 25/64)

4( x^2 + (5/4)x + 25/64  +  7/64)        factor the first three terms in the parenthses

4 [ (x + (5/8) )^2  + 7/64  ]       distribute the 4  over the terms inside the parentheses

4(x + (5/8) )^2  +  28/64

4 (x + (5/8) )^2 +  7/16

I totally Agree with Nickolas 

You didnt Cphills is like this

SpaceSpace  

See

Hear is a answer from Cphill 

P(< 15 from Box A)  = (14/20)  =   7/10      (1)

P(even or > 25 from Box B)   = 

P(even from Box B) + P(>25 from Box B)  - P(even and > 25 from Box B)  =

(10/20) + (5/20) - (3/20)  = 12/20  = 3/5     (2)

So.....the probability of  (1) and (2)   =   7/10  *  3/5  =    21 /  50

See I hope this helps! 

Derivative:

diff( 4*x^2+5*x+2, x )

Integral:

int( 4*x^2+5*x+2, x )

Limit:

limit( 4*x^2+5*x+2, x=0 )

I dont know if this helps but hope it does the calculator puleed it up that is one hear!

Thanks :)

Here is a relatively involved computer code that lists all the integers that meet the 3 conditions in the question. I'm not sure if this can be figured out through combinations and permutations:

 n=0;p=0;cycle:a= (10000+n);b=int(a/10000);c=int(a/1000);d=c%10;e=int(a/100);f=e%10;g=int(a/10);h=g%10;i=int(a/10);j=a%10;n=n+1;if(a%3==0, goto next, goto cycle);next:if(b!=d and b!=f and b!=h and b!=j and d!=f and d!=h and d!=j and f!=h and f!=j and h!=j and d%2==1 and d!=1, goto loop,goto cycle);loop:p=p+1;printa," ",;if(n<87863, goto cycle, 0);print"Total = ",p

Partial List:

13026  13029  13047  13056  13059  13062  13065  13068  13074  13086  13089  13092  13095  13098  13206  13209  13245  13248  13254  13257  13260  13269  13275  13278  13284  13287  13290  13296  13407   23016  23019  23046  23049  23058  23061  23064  23067  23076  23079  23085  23091  23094  23097  23106  23109  23145  23148  23154  23157  23160  23169  23175  23178  23184  23187  23190  23196  23406    35016  35019  35028  35046  35049  35061  35064  35067  35076  35079  35082  35091  35094  35097  35106  35109  35124  35127  35142  35148  35160  35169  35172  35178  35184  35187  35190  35196  35208    43017  43026  43029  43056  43059  43062  43065  43068  43071  43086  43089  43092  43095  43098  43107  43125  43128  43152  43158  43167  43170  43176  43179  43182  43185  43197  43206  43209  43215    53016  53019  53028  53046  53049  53061  53064  53067  53076  53079  53082  53091  53094  53097  53106  53109  53124  53127  53142  53148  53160  53169  53172  53178  53184  53187  53190  53196  53208    63012  63015  63018  63021  63024  63027  63042  63045  63048  63051  63054  63057  63072  63075  63078  63081  63084  63087  63102  63105  63108  63120  63129  63147  63150  63159  63174  63180    73014  73026  73029  73041  73056  73059  73062  73065  73068  73086  73089  73092  73095  73098  73104  73125  73128  73140  73146  73149  73152  73158  73164  73182  73185  73194  73206  73209  73215    83016  83019  83025  83046  83049  83052  83061  83064  83067  83076  83079  83091  83094  83097  83106  83109  83124  83127  83142  83145  83154  83157  83160  83169  83172  83175  83190  83196  83205   93012  93015  93018  93021  93024  93027  93042  93045  93048  93051  93054  93057  93072  93075  93078  93081  93084  93087  93102  93105  93108  93120  93126  93147  93150  93156  93162  93165  >> Total =  4476

Congratulations! 

So, we know that a^2 + b^2 + c^2 = 13.

Also, ab + bc + ca = 6.

And we want to find a + b + c. So, we sqaure a + b + c.

That would be a^2 + ab + ac + ba + b^2 + bc + ca + cb + c^2.

We group the squares together to get a^2 + b^2 + c^2, which is 13.

Then we group the ab or bc or ac together.

That gets us 2(ab + bc + ac) which is 2*6 which is 12.

Then we add to get 13 + 12 = 25.

Thanks Cphill 

Nickolas just keep that in mind Cubeman! 

THX Nicholas

Good job, Nick!!!

Well number 1 does look like a good calculator question!   so lets start 

1. 500/2=250   why because 250+250  or 250x2=  500

2.  well to find a average   we first add those all up so 

4+5+23+1+78 = 111  Now the next step is to say "How many numbers do we have to add"  Well the answer is 5 so nowe the second step is

111/5= 22.2

3.  I belive it is     

One Million two Hundred Thousand one hundred twnety-four! 

THx 

Hi Cubeman Welcome to the Forum! 

The square root of 16 is 4!