Seja f (u,v,h,p) = 
cos(h) tg(p) .
O valor aproximado de f ( -2 , 2 , π/4 , π/3 ) , é :
a) 1,22
b) 1,00
c) 1,73
d) 1,41
e) 0,02
Soluções para a tarefa
Respondido por
4
Oi Roger
f(u,v,h,p) = e^(u+v) * cos(h) * tg(p)
u = -2, v = 2, h = π/4 , p = π/3
e^(u+v) = e^(-2+2) = e^0 = 1
cos(h) = cos(π/4) = √2/2
tg(p) = tg(π//3) = √3
f(u,v,h,p) = e^(u+v) * cos(h) * tg(p) = 1*√2/2*√3 = √6/2 = 1.22 (A)
f(u,v,h,p) = e^(u+v) * cos(h) * tg(p)
u = -2, v = 2, h = π/4 , p = π/3
e^(u+v) = e^(-2+2) = e^0 = 1
cos(h) = cos(π/4) = √2/2
tg(p) = tg(π//3) = √3
f(u,v,h,p) = e^(u+v) * cos(h) * tg(p) = 1*√2/2*√3 = √6/2 = 1.22 (A)
Usuário anônimo:
Muito obrigado ^^ !
disponha
Respondido por
9
Boa tarde Roger!
Solução!
Vamos substituir os valores na função!
![Lembrando!\\\\\\
cos(\dfrac{ \pi }{4})=cos 45\º= \frac{ \sqrt{2} }{2}\\\\\\\
Tg (\frac{ \pi }{3})=Tg60\º= \sqrt{3}](https://tex.z-dn.net/?f=Lembrando%21%5C%5C%5C%5C%5C%5C%0A+cos%28%5Cdfrac%7B+%5Cpi+%7D%7B4%7D%29%3Dcos+45%5C%C2%BA%3D+%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C%0ATg+%28%5Cfrac%7B+%5Cpi+%7D%7B3%7D%29%3DTg60%5C%C2%BA%3D+%5Csqrt%7B3%7D+++ "Lembrando!\\\\\\
cos(\dfrac{ \pi }{4})=cos 45\º= \frac{ \sqrt{2} }{2}\\\\\\\
Tg (\frac{ \pi }{3})=Tg60\º= \sqrt{3}")
Retomando a função e substituindo os dados.
![f(-2,2, \frac{ \pi}{4} , \frac{ \pi }{3} )=e^{-2+2}.( \frac{ \sqrt{2} }{2} ).( \sqrt{3})\\\\\\\
f(-2,2, \frac{ \pi}{4} , \frac{ \pi }{3} )=e^{0}.( 0,707)}{2} ).( 1,732)\\\\\\\
f(-2,2, \frac{ \pi}{4} , \frac{ \pi }{3} )=1.( 0,707)}{2} ).( 1,732)\\\\\\\
f(-2,2, \frac{ \pi}{4} , \frac{ \pi }{3} )=1,224524\\\\\\
f(-2,2, \frac{ \pi}{4} , \frac{ \pi }{3} )=1,22\\\\\\\\\\\\\\
\boxed{Resposta:1,22~~\boxed{Alternativa~~A}}](https://tex.z-dn.net/?f=+f%28-2%2C2%2C+%5Cfrac%7B+%5Cpi%7D%7B4%7D+%2C+%5Cfrac%7B+%5Cpi+%7D%7B3%7D+%29%3De%5E%7B-2%2B2%7D.%28+%5Cfrac%7B++%5Csqrt%7B2%7D++%7D%7B2%7D+%29.%28++%5Csqrt%7B3%7D%29%5C%5C%5C%5C%5C%5C%5C%0Af%28-2%2C2%2C+%5Cfrac%7B+%5Cpi%7D%7B4%7D+%2C+%5Cfrac%7B+%5Cpi+%7D%7B3%7D+%29%3De%5E%7B0%7D.%28+0%2C707%29%7D%7B2%7D+%29.%28+1%2C732%29%5C%5C%5C%5C%5C%5C%5C%0Af%28-2%2C2%2C+%5Cfrac%7B+%5Cpi%7D%7B4%7D+%2C+%5Cfrac%7B+%5Cpi+%7D%7B3%7D+%29%3D1.%28+0%2C707%29%7D%7B2%7D+%29.%28+1%2C732%29%5C%5C%5C%5C%5C%5C%5C%0Af%28-2%2C2%2C+%5Cfrac%7B+%5Cpi%7D%7B4%7D+%2C+%5Cfrac%7B+%5Cpi+%7D%7B3%7D+%29%3D1%2C224524%5C%5C%5C%5C%5C%5C%0Af%28-2%2C2%2C+%5Cfrac%7B+%5Cpi%7D%7B4%7D+%2C+%5Cfrac%7B+%5Cpi+%7D%7B3%7D+%29%3D1%2C22%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BResposta%3A1%2C22%7E%7E%5Cboxed%7BAlternativa%7E%7EA%7D%7D "f(-2,2, \frac{ \pi}{4} , \frac{ \pi }{3} )=e^{-2+2}.( \frac{ \sqrt{2} }{2} ).( \sqrt{3})\\\\\\\
f(-2,2, \frac{ \pi}{4} , \frac{ \pi }{3} )=e^{0}.( 0,707)}{2} ).( 1,732)\\\\\\\
f(-2,2, \frac{ \pi}{4} , \frac{ \pi }{3} )=1.( 0,707)}{2} ).( 1,732)\\\\\\\
f(-2,2, \frac{ \pi}{4} , \frac{ \pi }{3} )=1,224524\\\\\\
f(-2,2, \frac{ \pi}{4} , \frac{ \pi }{3} )=1,22\\\\\\\\\\\\\\
\boxed{Resposta:1,22~~\boxed{Alternativa~~A}}")
Boa tarde!
Bons estudos!
Solução!
Vamos substituir os valores na função!
Retomando a função e substituindo os dados.
Boa tarde!
Bons estudos!
Qualquer dúvida é só mandar uma mensagem.
Muito obrigado ^^ !
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