/sci/ - Science & Math

Lim n-> inf { [ (4n + 2) / ( 6n - 1) ] ^1 }
The ^1 can be ignored

Lim n-> inf { (4n + 2) / ( 6n - 1) }
This is the dimension [inf]/[inf] and you can use L'Hospital d/dn { 4n+2 } / d/dn { 6n -1 }
d/dn { 4n+2 } = 4
d/dn { 6n -1 } = 6

Lim n-> inf { 4 / 6 } = 4/6 = 2/3 = 0.66666666...

finish

>>5066009
Damn .... u allright ... xD

>>5066013
3/(x+2) +1/2

>>5065994
16*u^6 / ( 8*u^4*y + 30*u³*x )
16*u^3 / ( 8*u*y + 30*x )
8*u^3 / ( 4*u*y + 15*x )

i got

y*sqrt(48*y*x²) + x*sqrt(3*y³)
sqrt(48)*sqrt(y³)*sqrt(x²) + x*sqrt(3)*sqrt(y³)
sqrt(4)*sqrt(4)*sqrt(3)*sqrt(y³)*x + x*sqrt(3)*sqrt(y³)
2*2*sqrt(3)*sqrt(y³)*x + x*sqrt(3)*sqrt(y³)
x*sqrt(3)*sqrt(y³) * ( 4 +1 )
x*sqrt(3)*sqrt(y³)*5

finish

or one of this:

5*x*sqrt(3*y³)
5*x*sqrt(3)*sqrt(y³)
5*x*3^(1/2)*y^(3/2)
5*x*3^(0.5)*y^(1.5)

Ok another way to solve...

This is your function: f(x) = y = a*sqrt(b*(x-h))+k

4 = f(-3) = a*sqrt(b*(-3-h))+k <-- first look
3 = f(-2) = a*sqrt(b*(-2-h))+k
2 = f(1) = a*sqrt(b*(1-h))+k
1 = f(6) = a*sqrt(b*(6-h))+k

1.) 4 = f(-3) = a*sqrt(b*(-3-h))+k
This point is very interesting. Because a smaller valiue than -3 is not (real) possible. Like sqrt(0)
Inner the squareroot is: b*(-3-h). This is the smallest possible valiue.
b*(-3-h) = 0
when is this possible?
a) when b = 0 (0*(-3-h) = 0)
b) when h = -3 (-3-h = 0 => -3 = h)

a) is not possible, becaue than is every point k (a*sqrt(0*(x-h)) = a*sqrt(0) = a*0)
b) is the only possible value.

When sqrt(b*(-3-h)) = 0 is k = 4

f(x) = y = a*sqrt(b*(x-(-3)))+4 = f(x) = y = a*sqrt(b*(x+3))+4

splitting sqrt(b*(x+3)) = sqrt(b)*sqrt(x+3)

Now, we look what a and b can be.

4 = a*sqrt(b)*sqrt(-3+3)) +4 = a*sqrt(b)*sqrt(0) +4 = a*sqrt(b)*0 +4
3 = a*sqrt(b)*sqrt(-2+3)) +4 = a*sqrt(b)*sqrt(1) +4 = a*sqrt(b)*1 +4
2 = a*sqrt(b)*sqrt(1+3)) +4 = a*sqrt(b)*sqrt(4) +4 = a*sqrt(b)*2 +4
1 = a*sqrt(b)*sqrt(6+3)) +4 = a*sqrt(b)*sqrt(9) +4 = a*sqrt(b)*3 +4

substrate by 4 and dividate by 1, 2 and 3

0 = a*sqrt(b)*0
-1 = a*sqrt(b)*1 => -1 = a*sqrt(b)
-2 = a*sqrt(b)*2 => -1 = a*sqrt(b)
-3 = a*sqrt(b)*3 => -1 = a*sqrt(b)

=> a = -1/sqrt(b)

Fill in function:
f(x) = y = (-1/sqrt(b))*sqrt(b)*sqrt(x+3))+4 = -1*sqrt(x+3) = 4

a = -1
b = 1
h = -3
k = 4

Greetings from germany.

100 * k^8 = 1200
k^8 = 12
k = 8.root from 12

.5 = 0.5 = 1/2

24 / (1/2) = 24/1 * 2/1 = ( 24 * 2 ) / ( 1 * 1 ) = 48 / 1 = 48

x_1 + x_2 = 1
x_1 + x_2 = 2

x_1 = 1 - x_2
x_1 = 2 - x_2

1 - x_2 = 2 - x_2
1 = 2

wrong!

not solveable

x²/100

Ist nicht schwer ;)

w=7
d=28
=> x(d)=7

d-rounddown(d/w)*w

28-rounddown(28/7)*7
28-4*7
28-28
0 but it must be 7

My way:
d-(roundup(d/w)-1)*w or d-roundup(d/w)*w+w

28-roundup(28/7)*7+7
28-4*7+7
28-28+7
0+7
7 it is 7 :D

>>4427731

Id doesnt work!
Test with d=28
It will give you 0 to you!

Here ...

here

>>4424692

anticlockwise is mathematical positiv.
clockwise is mathematical negativ.

Ive mathematical positv calculate

Here

>>4422836

Here the calc

a*7*sqrt(3)/40

a = length of a side from the triangle

>>4422629


7x = 4x +36°
3x = 36°
x = 12°

>>4421987

Possible.
But. look >>4421828
3 Resitors -> 3 Resistors (as multiplikation) on the top of the line
and 3 * 2 resistors (as multiplikation) on the buttom.

4 Resistors -> 4 on the top and 4*3 on the buttom
5 Resistors -> 5 on the top and 5*4 on the buttom

n Resistors -> n on the top and n*(n-1) on the buttom

looks easyler ^^ >>4421799

>>4421897

yes. look:
>>4421799
1. line

further questions?

Here with 3 resitors