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A menos que el enunciado diga algo como: busque todas las soluciones posibles. Pero si es un clásico resuelva, hasta donde dices fue exactamente lo que pensé.
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Eu imaginava que Stanford fosse extremamente difícil mas essa equação é muito fácil. Não precisa de tantos cálculos. Não estou me gabando porque só os humildes aprendem. É lógico que essa é uma das poucas questões para ser admito e não a única.
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Scusate chiedo una domanda che non c'entra col video. "Vorrei vedere la dimostrazione di come si "ricava" y = mx+q" NB. "Qual'è tutta la dimostrazione iniziale di somme e o moltiplicazioni di un eguaglianza di equazione che risolvendosi porta al risultato finale di y = mx+q" "Grafico escluso dalla dimostrazione" "È stata chiesta una risoluzione per espressione!!!!" Curiosità mia, perchè mi fu chiesta così come domanda al terzo superiore, feci il grafico era corretto ma il professore disse non ti ho chiesto quello la sapevi fare alle medie in primo superiore col grafico ma io te l'ho chiesta per espressione analitica fino a risultare y = mx+q e per il grafico lo conosce tutta la classe. Quindi per "espressione".
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Por acaso a pergunta dizia: "encontre todas as soluções possíveis"?... pois senão, de cara dá pra ver que "4" já satisfaz a resposta. Pra que complicar o que não precisa ser complicado?
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Why making it so incredibly complicated??? I must be missing something. Not clear if we are talking about complex numbers or just real.
Even if you consider complex numbers, if the "first" answer is 4, the other 3 answer will for sure be -4, 4i and -4i. This is very easy.
easy dear
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thanks
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1:41 C^4 = 4^4 (mismo exponente entonces C=4)
A menos que el enunciado diga algo como: busque todas las soluciones posibles.
Pero si es un clásico resuelva, hasta donde dices fue exactamente lo que pensé.
okay dear
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Ótimo exercício, parabéns pela didática
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thank you so much
Eu imaginava que Stanford fosse extremamente difícil mas essa equação é muito fácil. Não precisa de tantos cálculos. Não estou me gabando porque só os humildes aprendem. É lógico que essa é uma das poucas questões para ser admito e não a única.
This problem is cool. It's an elementary problem in imaginary numbers because it explores the Fundamental Theorem of Algebra. Congratulations.
thanks
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Scusate chiedo una domanda che non c'entra col video.
"Vorrei vedere la dimostrazione di come si "ricava" y = mx+q"
NB. "Qual'è tutta la dimostrazione iniziale di somme e o moltiplicazioni di un eguaglianza di equazione che risolvendosi porta al risultato finale di y = mx+q"
"Grafico escluso dalla dimostrazione"
"È stata chiesta una risoluzione per espressione!!!!"
Curiosità mia, perchè mi fu chiesta così come domanda al terzo superiore, feci il grafico era corretto ma il professore disse non ti ho chiesto quello la sapevi fare alle medie in primo superiore col grafico ma io te l'ho chiesta per espressione analitica fino a risultare y = mx+q e per il grafico lo conosce tutta la classe.
Quindi per "espressione".
Da 4/c•4/c=1,c*1.3
ok
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C =4
good
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My mind: FOUR
Video: NO😊
ok
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Por acaso a pergunta dizia: "encontre todas as soluções possíveis"?... pois senão, de cara dá pra ver que "4" já satisfaz a resposta.
Pra que complicar o que não precisa ser complicado?
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it has more solutions
A única dificuldade dessa questão é que não existe a alternativa TODAS AS OPÇÕES ESTÃO CORRETAS.
4
fantastic job
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PARA QUE TANTA CONTA , SO TIRAR MMC E FACILMENTE ACHA O VALOR DE C=4
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ok
16
ok
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check out the exact solution
4, solución real, en tres segundos
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yes
4 e -4, ninguém pediu soluções complexas
É o contrário, ninguém restringiu ao conjunto dos números Reais
@pedrocavalcante6455 Você tem razão, perdão.
nice
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As provas que apareceram
Je comprends pourquoi les étudiants sont tordus 😂
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why?
Errado
Whats the solution?
4
@steaky4305 ofcourse it is. Thankss!!
4,-4,4i,-4i
Four solutions
@@FaizanFayaz-qu3ek thanks
check out the video
C/4 = exp(ia), so (c/4)^4=exp(4ia), then 4a=2kπ giving a= kπ /2, k= 0,1,2,3 and that's it !
The op is avoiding complex algebra I'd Say...❤😂🎉
ok
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C=4
C=4