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Blubberbär

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First name: Valentin

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721

Tuesday, March 22nd 2011, 6:23pm

1. Quadrieren
2. *d
3. durch 4Pi^2 teilen

=> m = (T^2 * d) / (4Pi^2)

Nightwish

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722

Tuesday, March 22nd 2011, 8:50pm

T = 2Pi * Wurzel (m/d) | durch 2Pi teilen

T / 2Pi = Wurzel (m/d) | quadrieren

(T / 2Pi) ^ 2 = m/d | mal d

m = [(T / 2Pi) ^ 2] * d

So sollte es passen

Blubberbär

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723

Tuesday, March 22nd 2011, 9:38pm

T = 2Pi * Wurzel (m/d) | durch 2Pi teilen

T / 2Pi = Wurzel (m/d) | quadrieren

(T / 2Pi) ^ 2 = m/d | mal d

m = [(T / 2Pi) ^ 2] * d

So sollte es passen

Und wo liegt nun der Unterschied zu meiner Antwort ;)

iBot

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724

Tuesday, March 22nd 2011, 9:46pm

Seins ist übersichtlicher und komplett zusammengefasst^^

Ich würde es auch so wie Nightwish machen. Erstmal die Wurzel alleine auf eine Seite bringen und dann den anderen Term in ne Klammer stecken und quadrieren. Das macht die Rechnung übersichtlicher und es entstehen keine Fehler, wenn man im Nachhinein noch versucht die Formel zusammen zufassen :P

fr33ze

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725

Tuesday, March 22nd 2011, 9:50pm

Vielleicht war bei ihm auch die Seite vorher noch angezeigt worden und er hat nicht drauf geachtet, das es schon jemand beantwortet hat, wollte nämlich auch gerade antworten bevor ich dann gesehn hab das noch ne Seite hintendran is^^
Gestern noch in Kiev, morgen schon in Wien
Zusammenhalten das ist unser Ziel
Wir sind aus München, wir sind die Bayern
Wir sind diejenigen die immer wieder feiern

Nightwish

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726

Tuesday, March 22nd 2011, 11:12pm

jop war auf der vorletzten Seite, habs aber auch so gelernt erstmal die Wurzel alleine zu stellen, finde ich auch einfacher und übersichtlicher so ;-)

Keltäää

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727

Wednesday, March 23rd 2011, 2:33pm

Euch beiden vielen Dank! :)
Ich mag Rauch lieber als Luft, wenn ich könnte würde ich auch im Schlaf rauchen, ich würd sogar im Sarg weiter rauchen

orientala

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728

Wednesday, March 30th 2011, 7:03pm

bin mir grad nicht sicher wie ich die klammern hier auflöse:

3x1 + (3x2 + 2)= 2x3 + 8 [Die zahlen hinter dem X bzw, den Xen sind nur zur Nummerierung]

iBot

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729

Wednesday, March 30th 2011, 7:08pm

Da vor der Klammer kein Malzeichen oder ein Minus steht, kannst du die Klammer einfach weglassen.
Also:
3x1 + 3x2 + 2= 2x3 + 8
3x1 + 3x2 - 2x3 = 6

orientala

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730

Wednesday, March 30th 2011, 7:14pm

und wenn da ein minus steht einfach die vorzeichen ändern von dem was in der klammer stand oder?

Blubberbär

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731

Wednesday, March 30th 2011, 7:50pm

Korrekt, kann man sich auch einfach selber überlegen:


1 - (2 + 3) = -4 = 1 - 2 - 3

Gruß

Rus

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732

Thursday, March 31st 2011, 5:51pm

Was bedeutet "Konsensrepublik" ? :D

Quoted

Original von DennisD
Kumpels is im Park mal der Tabak ausgegangen, also haben die ne Raupe geraucht,

:lol: :lol: :lol: :lol:

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Cpt.Chaos

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734

Thursday, March 31st 2011, 6:20pm

Servus,

Mein Bruder schreibt morgen Mathe und hat mich um Hilfe gebeten. Allerdings war ich in Mathe immer "mittelmäßig" und mittlerweile kann ich mich auch nicht mehr so ganz an das Zeug erinnern ^^

Vielleicht kann mir bzw. ihm ja einer von euch helfen, wäre nett :)

log7(3x-3)-log7(x-1) Die sieben soll natürlich leicht nach unten versetzt sein.
ln(1/a²)-ln(2a)-ln(1/a) / bedeutet geteilt

Danke

sebbmeistOr

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735

Thursday, March 31st 2011, 6:51pm

mh sind das gleichungen oder was ? ^^
wenn ja, wo ist die andere seite.

ansonsten bekommt man die logs bzw. ln weg, indem man die ganze gleichungung "hochnimmt" ...
und welche zahl man da als basis nimmt hängt eben vom log ab.
bei ln nimmt man z.b. e (eulersche zahl)
bei 10erlog eben 10 und bei siebner log 7...

also z.b. ln(x) = 3 ...
e^ln(x) = e^3 ...
(e^ln(x) = x ...)

also x =e^3 .



gruß seb

This post has been edited 1 times, last edit by "sebbmeistOr" (Mar 31st 2011, 7:04pm)


Cpt.Chaos

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736

Thursday, March 31st 2011, 7:16pm

Okay, vielen Dank :)

Kann diese Aufgabe vielleicht auch jemand lösen? :blush:

ax²+(a/2)x-1=0

thx

sebbmeistOr

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737

Thursday, March 31st 2011, 7:30pm

nach a oder nach x ?

wenn x ... dann binomische formel ;)



gruß seb

Cpt.Chaos

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738

Thursday, March 31st 2011, 7:32pm

nach a oder nach x ?

wenn x ... dann binomische formel ;)



gruß seb
"Für welche des Paramters a hat die Gleichung genau eine Lösung?" Hätte ich wohl dazu schreiben sollen. Kann mit der Formulierung allerdings schon nicht viel anfangen :D

sebbmeistOr

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739

Thursday, March 31st 2011, 7:35pm

jo dann binomische formel nach x ...
dann bekommt man ja normalerweise 2 ergebniss: x1,x2
und da das a auch in der binomformel drin is, sind x1 und x2 von a abhängig ...

jetzt schaut man wie a sein muss, dass es eben nur eine lösung gibt z.b. die gleiche x1(a) = x2(a) ;)



gruß seb

Cpt.Chaos

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740

Thursday, March 31st 2011, 8:16pm

jo dann binomische formel nach x ...
dann bekommt man ja normalerweise 2 ergebniss: x1,x2
und da das a auch in der binomformel drin is, sind x1 und x2 von a abhängig ...

jetzt schaut man wie a sein muss, dass es eben nur eine lösung gibt z.b. die gleiche x1(a) = x2(a) ;)



gruß seb
Kannst du das irgendwie genauer beschreiben bzw. direkt mal ausrechnen? Hört sich nach Faulheit an, ich weiß, aber irgendwie steh ich gerade ziemlich hart aufm Schlauch und komme nicht weiter... und das um diese Uhrzeit -.- Was man nicht alles als großer Bruder machen muss