admin

16 Questions
1410 Answers

+4

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Neue Datenschutzrichtlinie, Mai 2018

Hallo,

Diese Webseite wurde aktualisiert mit kleineren Anpassungen an die DS-GVO.
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admin 27.05.2018

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New Privacy Policy, May 2018

Hello,

this website was updated today because of some legal changes here in Europe. ("GDPR").
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admin 27.05.2018

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New version is online for two weeks now, any problems?

Hi,

Any problems? You can use the "Report a Problem" link in the footer, or send me an email to britnex+bugs@gmail.com
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admin 12.09.2015

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New website is online.

Hi,

Any problems? You can use the "Report a Problem" link in the footer, or send me an email to britnex+bugs@gmail.com , thank you!
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Melody ●●●●●●●●●●●●●●●●

admin 30.08.2015

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Neue Version der Webseite ist online

Hi,

Probleme bitte per Email an britnex+bugs@gmail.com melden, Danke!
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admin 30.08.2015

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+3146

three times ten to the power of negative 2

help what is 3x10to the power of negative 2

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admin 26.01.2012

+3146

0

[input](-9)(-8)/(-2)[/input]

admin19.02.2012

+3146

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[input]4=(5x-5)/(7x)-(6x^2-50x+16)/(7x)[/input]

4=((5x-5)-(6x^2-50x+16))/(7x)
4=(5x-5-6x^2+50x-16)/(7x)
4=(-6x^2+55x-21)/(7x)
28x=-6x^2+55x-21
28x-55x= -6x^2-21
-27x = -6x^2-21
-6x^2+27x-21=0
-6*(x^2-27x/6+21/6)=0

-6*(x^2-27x/6+21/6) is zero, if (x^2-27x/6+21/6) is zero:

x _1,2 = -p/2 +- sqrt( (p/2)^2 - q )
p = -27/6
q = 21/6
x1 = -(-27/6)/2 + sqrt( ((-27/6)/2)^2 - 21/6 )
[input]-(-27/6)/2 + sqrt( ((-27/6)/2)^2 - 21/6 )[/input]
x1 = 3.5 = 7/2

x2 = -(-27/6)/2 - sqrt( ((-27/6)/2)^2 - 21/6 )
[input]-(-27/6)/2 - sqrt( ((-27/6)/2)^2 - 21/6 )[/input]
x2 = 1

admin19.02.2012

+3146

0

[input]100+19%[/input]
[input]100-19%[/input]
[input]100*19%[/input]

admin19.02.2012

+3146

0

2x+3y=1

-6x-9y=-1 | /-2
2x+3y=1/2

1=2x+3y=1/2
false

[input]solve( 2x+3y=1, -6x-9y=-1 )[/input]

admin18.02.2012

+3146

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[input]798*(1+0.031/4)^(21*4)[/input]

admin18.02.2012

+3146

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[input]-6(2y+3)+6=-12[/input]

admin18.02.2012

+3146

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[input]2/1-2/3[/input]

admin18.02.2012

+3146

0

m^2 + 8m + 3 =0
m^2 + 8m = -3

m^2 + 8m + 16 = -3 + 16
(m + 4)^2 + 16 = -3+16
(m + 4)^2 = -3+16
(m + 4)^2 = 13

(m + 4) = ±sqrt(13)
m = ±sqrt(13) - 4

m ₁ = -sqrt(13) - 4
m ₂ = sqrt(13) - 4

[input]x^2 + 8x + 3 =0[/input]

admin18.02.2012

+3146

0

isaac3211:
y''+y'-6y=e^(-3x)?

first the complementary solution for: y''+y'-6y=0
y(x)=e^(λ*x)

e^(λ*x) d^2/dx^2 + e^(λ*x) d/dx - 6*e^(λ*x) = 0

substitute:
e^(λ*x) d^2/dx^2 with λ^2*e^(λ*x)
e^(λ*x) d/dx with λ*e^(λ*x)

λ^2*e^(λ*x)+λ*e^(λ*x)-6*e^(λ*x) = 0

(λ^2+λ-6)*e^(λ*x) = 0
solve:
λ^2+λ-6=0
λ=-3 and λ=2

so:
y1(x) = c1*e^(-3*x)
y2(x) = c2*e^(2*x)
(c1,c2 = constant)
=> the complementary solution y(x):
y(x) = y1(x)+y2(x) = c1*e^(-3*x) + c2*e^(2*x)

[input]diff(diff(y(x),x),x)+diff(y(x),x)-6*y(x)=0[/input]

ok, looks good. now to the particular solution: y''+y'-6y=e^(-3x)
yp(x) = x*a1*e^(-3x)

yp(x) d/dx = a1*e^(-3x)*-3*a1*e^(-3x)*x
[input]diff( x*a1*e^(-3x), x)[/input]

yp(x) d^2/dx^2 = a1*( -6*e^(-3x)+9*e^(-3x)*x )
[input]diff(diff( x*a1*e^(-3x), x),x)[/input]

now insert both diff'd yp(x) and yp(x) into y''+y'-6y=e^(-3x)
a1*( -6*e^(-3x)+9*e^(-3x)*x ) + a1*e^(-3x)*-3*a1*e^(-3x)*x - 6*x*a1*e^(-3x) = e^(-3x)
-5*a1*e^(-3x) = e^(-3x)
a1 = -1/5
yp(x) = -(1/5)*e^(-3x)*x

y(x) = c1*e^(-3*x) + c2*e^(2*x) + y(p)
y(x) = c1*e^(-3x) + c2*e^(2x) - (1/5)*e^(-3x)*x

admin18.02.2012

+3146

0

[input]5(4^x)=6[/input]

admin17.02.2012