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This computer code confirms the above theoretical calculations and gives exactly same answer of 420 integers.

a=1;b=1;c=1;d=1;e=1;f=1;g=1;p=0; cycle:if(a*b*c*d*e*f*g==10000, goto loop, goto next); loop:print a*1000000+b*100000+c*10000+d*1000+e*100+f*10+g," ",;p=p+1; next:g++;if(g<10, goto cycle, 0);g=1;f++;if(f<10, goto cycle, 0);g=1;f=1;e++;if(e<10, goto cycle,0);g=1;f=1;e=1;d++;if(d<10, goto cycle,0);g=1;f=1;e=1;d=1;c++;if(c<10, goto cycle,0);g=1;f=1;e=1;d=1;c=1;b++;if(b<10, goto cycle,0);g=1;f=1;e=1;d=1;c=1;b=1;a++;if(a<10, goto cycle, 0);print"Total = ",p

CPhill, that's a very intuitive guess!

However, there's one factor you haven't taken into account. Try to see if you can find it :)

Here's my take.....

The ant never gets to the end of the string

Each second, the distance  that the ant covers is   t  inches

And each second, the length of the string  (in inches)  =  12  + 12 t

So.....the ratio of the distance that the ant has traveled in t seconds  to the string length at that time is just

        t

________

12 t  +  12

As  t approaches some infinite number of seconds, this ratio   converges  to

       t                   1

_______  =      ____

   12 t                  12

Therefore, the ant can never travel much more than 1/12  the string length  as t approaches infinity

THX, ZZZZZZ and BasicMaths.....!!!!!!

Guest, here are some easy identities for simplifying quadratics.

First, if a quadratic is in the form a^2 - b^2, then the form of the factored equation would be (a+b)(a-b). This is called the difference of squares.

Second, if in the normal quadratic form of ax ^2 + bx + c, then one must find a number that adds up to b, and multiplies to c. If a is a coefficient of a > 1, then one must find a number that adds to b and multiplies to ac. 
 

Hope this helps!

BasicMaths

Each dice has a 1/6 chance to roll a six which means there are 6^5 possible rolls that 5 dice could have

if we isolate the roles that have a six, that means we are dividing the equation by 6

because only a 6th of them have a six, which brings it down to 6^4 possible roles

then we isolate the next six which would be the same thing because only a 6th of them have a six

so it would be 6^3 possible roles

next we need to turn 6^3 possible roles in to a fraction which involes divididing it from 6^5

which would be a 1/6^2 chance or 1/32 chance of roling a pair of sixes

I should mention the reason why I know this is because I was bored and wanted to know the

chances of getting a yatzee on the first roll (which is 1/6^4 by the way)

Thank you, so much.

first things first, you need two equations

the two equations are 4x+12y=60 and x+y=6

then you want to simplify like dividing the first equation by 4

so it would be x+3y=15

now you want to subtract the other equation from it to get rid

of x so now it would be 0x+2y=9

then we simplify again so 2y/2=9/2

and it would be y=4.5

now we replace y with 4.5 in one of the equations (I did the second one

but you'll get the same answer if you do the first) so it would be x+4.5=6

than subtract 4.5 from both sides and your done and that leaves you with

1.5 hours runing and 4.5 hours biking

Thx Cphill

Nice, Gadfly   !!!!

Let

a, b  =  √10 

c = 2  

The circumradius can be found as

  (abc) / √((a + b + c)(b + c - a)(c + a - b)(a + b - c))

So we have

( 10*2) / √   [ ( 2√10 + 2) ( 2) ( 2 ) ( 2√10 - 2)  ]  =

20  /  √ [ (40 - 4) (4) ] = 

20 / √ [ 36 * 4]  =

20 / √144 =

20 /12  =

5/3   units

x^2/4 − y^2/2 = 1

x^2 = 4(y+2)        (1)

Multiply the first equation through by  4  and we get that

x^2 -  2y^2  =  4

x^2  = 2y^2 + 4      (2)

Equate (1) and (2)   and we have that

4(y + 2)   =  2y^2 + 4

2 (y + 2)  = y^2 + 2

2y + 4  =  y^2 + 2

y^2 - 2y - 2  =  0      complete the square on y

y^2 - 2y + 1   - 2  - 1   = 0

( y - 1)^2  - 3  =  0

(y - 1)^2  =  3

y - 1  = ±√3

y = 1 ±√3

y = 1 + √3      or  y  = 1 -√3

And

x^2  = 4 ( y + 2)

x = ±2√[ y + 2]

So the solutions are

(2√[ 3 - √3] , 1 -√3 )  

(-2√[ 3 - √3] , 1 -√3 )

(2√[ 3 + √3] , 1 +√3 )

(-2√[ 3 + √3] , 1 +√3 )      

sin x +  cos x  =   1/5         squrare both sides

sin^2x +  2sinxcosx   + cos^2x  =  1/25

(sin^2x +cos^2x  )  + 2sin x cos x   =1/25

1   +  sin (2x)  =  1/25

sin (2x )  =   1/25  - 1

sin (2x)  =   -24/25

Let  2x  =  A

So

arcsin (-24/25)   =   A   ≈  -1.288 rads  =     pi + 1.288 rads ≈  4.43 rads

So  

2x   = 4.43 rads

x  =   2.215 rads

So      tan (2.215)  ≈ -1.33  

Go online to this link and see the solution there: https://nrich.maths.org/13124/solution

 Circumradius
The circumcircle of a triangle is a circle that passes through all of the triangle's vertices, and the circumradius of a triangle is the radius of the triangle's circumcircle. Circumcenter (center of circumcircle) is the point where the perpendicular bisectors of a triangle intersect.

R = { a b c } / { 4 r s } = { 3.16 * 3.16 * 2 } / { 4 * 0.721 * 4.162 } = 1.667, where r =inradius, s =semiperimeter.

To simplify polynomial fractions, you can factor and find canceling x-intercepts.

\(x^2-25 \over x^2+10x + 25\)can be factored in both the numerator and denominator. \(x^2 -25 \) becomes \((x+5)(x-5)\), and 

\(x^2+10x+25\) becomes \((x+5)(x+5)\).

From there you can get \((x+5)(x-5) \over (x+5)(x+5)\).

You can then eliminate one \(x+5\) from each part of the fraction, resulting in \(x-5 \over x+5\).

If no obvious factors are there, you can always use the quadratic formula to find them when the highest power is 2.

Hope this helps!

"Find all n satisfying 1 + 3 + 5 + 7 + ... + 125 = n + (n + 2) + (n + 4) + (n + 6) + ... + 209."

L = 1+3+5+7+...+125 = 3969

R = n+(n+2)+(n+4)+(n+6)+...+209

Write the last term of R as n+2N

R = n*(N+1) + N*(N+1)

Now the last term n + 2N = 209, so N = (209 - n)/2, hence:

R = (n + 209)(211 - n)/4

Equate L and R:

(n + 209)(211 - n)/4 = 3969

I'll leave you to solve this quadratic to find the two values of n.

Yes!. It would be the 19th term of the sequence as you can see in the following:

2 , 2 , 4 , 6 , 10 , 16 , 26 , 42 , 68 , 110 , 178 , 288 , 466 , 754 , 1220 , 1974 , 3194 , 5168 , 8362 , 13530 , = 20 terms.

The circumradius is 30*3/2 = 45.

I did not get that.

This is a combination question.

You have to work out the prob of 0 + prob of 1 + prob of 2

That should get you started. 

this should get you started!

Thank you, Melody !

Thanks for enlightening me    

well in the 5 times, there are 30 faces 

Each roll has atleast one "6" in it, you want to find the probability "If we get a 6 at most in 2 of the rolls only right?" 

so in 2 times, there are two "6", 

and in 2 rolls there are 6*2=12 possibility 

2/12= 1/6