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**Agnishom**- Replies: 1

I am the only registered member who has the highest number of threads in the guest book.

**Agnishom**- Replies: 1

I have enabled autocorrect. You will now see my spellings improve but my grammar will stop making sense

Nice page.

Now, how do I use this to get the time-independent equation?

**Agnishom**- Replies: 3

How many solutions of x! = nPr are there such that n and r are different?

What reason is that?

Welcome to reality.

What os your favorite color?

The reaction force between m and M is 5kg*5m/s^2

To balance the friction, you must give a force of 5kg*5m/s^2*0.4

You did not get my email, PatternMan?

bobbym wrote:

The site already has one.

yes, and the name of that program is bobbym.

Now, has this guest posted here before at all?

You cannot teach a man anything, you can only help him find it within himself......bob bundy

Brilliant is a competition site, rather than education. For that purpose, I wil recommend mathopolis, mathisfun and expii.

Ask me if you need a refferal code for expii

I do realise it now, bob. Thanks

Hi iluvdogs;

Who is your favorite dog?

Hi;

What is your favorite color?

That is a trivial result

only when

Are you saying that if x - y = 1, then x^2 - y^2 is a prime?

Solution

Let us call the curves c1, c2 and c3

It is evident that:

c1 = (cos t, sin t, 0)

c2 = (cos t, 0, sin t)

c3 = (0, sin t, cos t)

[I am considering the circles to be centered at the origin and of radius 1. This is just to make the algebra simple]

You wanted me to prove that c1 intersects c2 making an angle of pi/2 and c2 intersects c3 making an angle of pi/2 and c3 intersects c1 making an angle of pi/2

Consider the intersection of c1 and c2. At this point, c1(t) = c2(t)

Or, (cos t, sin t, 0) = (cos t, 0, sin t)

Solving, you get t = 0

Now, c1(0) = c2(0) = (1, 0, 0)

Differentiating the curves,

c1' = (-sin t, cos t, 0)

c2' = (-sin t, 0, cos t)

At t=0, c1'(0) = (0, 1, 0) and c2'(0) = (0, 0, 1)

Let x1 be the tangent of c1 at the intersection point

We have, x1= c1(0) + t(c1'(0)) = (1, 0, 0) + t(0, 1, 0) = (1, t, 0)

Let x2 be the tangent of c2 at the intersection point

We have, x2= c2(0) + t(c2'(0)) = (1, 0, 0) + t(0, 0, 1) = (1, 0, t)

By the angle between c1 and c2, we obviously mean the angle between x1 and x2.

Let that angle be θ

x1 . x2 = (1, t, 0) . (1, 0, t) = 1*1 + t*0 + t*0 = 1

|x1| = Sqrt[1+t^2]

|x2| = Sqrt[1+t^2]

At the intersection point, t=0. So,

Similarly, you can go on proving the same thing for the two other intersection points (but that has been left to the reader as an exercise )

QED

(1/12)y^4 = (1/2)y^2 - y + (1/2)C[1]y^2 + C[2]y - C[3]

Is that the answer?

Context Free Code:

```
CF::Size = [s 225]
startshape leggo
shape leggo {
CIRCLE[s 36]
loop 180[r 2]
vein[s .6 hue 65..180 sat .3 b .8 x 12 y 12 ]
loop 206[r 1.75]
myveins[hue 160..250 sat .3 b .7 x -40 y -40] //Comment out
this line for some trippy shizz
loop 20 [r 14..20]
yomama[s .8 hue 0 sat .7 b .7 a -.5 x (-71..-68 )y
(-71..-68)]
mycircle[]
}
shape yomama
rule .99{
CIRCLE[]
yomama[s .995 x -.1...5 y -.1...5 r -10..10]
}
rule .01{
CIRCLE[]
yomama[s .995 x 0...5 y 0 ...5]
yomama[s .6 r -60..60]
}
rule .05{
CIRCLE[]
yomama[s .995 x 0...5 y 0 ...5]
yomama[s .6 r -60..60 a -.8]
}
shape myveins{
vein[]
loop 10 [r 18]
vein[s .2 r -85]
}
shape vein
rule .99{
CIRCLE[a -.2]
vein[s .99 x 0...5 y 0 ...5]
}
rule .0775{
CIRCLE[a -.2]
vein[s .99 x 0...5 y 0 ...5]
vein[s .6 r -60..60]
}
path mycircle{
myrad = 100
MOVETO(0,(-myrad))
ARCTO(0,(myrad),myrad)
ARCTO(0,(-myrad),myrad)
STROKE(1)[]
}
```

by DackAttack

Welcome to reality.

What is your favorite color?

**Agnishom**- Replies: 1

Somebody teach me iterated expectations please?

Step 2: Test whether A is divisible by any prime number less than K. If yes A is not a prime number. If not, A is prime number.

What happens if A is divisible by K?

I can fit this polynomial on the data

How about calling Math or Maths just M?