6-Supondo que, em determinada região, a temperatura média semanal T(em °C) e a
quantidade de energia solar média semanal Q que atinge a região (em kcal/cm2
) possam ser expressas em função do tempo t, em semanas, por meio das funções:
Anexos:
dragon152:
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a) Quantidade de energia .
![Q(t)=400+200sen [ \pi ( \dfrac{t-11}{26} ) ]\Rightarrow Q(50)=400+200sen [ \pi ( \dfrac{50-11}{26} ) ] \\ \\ \\ Q(50)=400+200sen [ \pi ( \dfrac{39}{26} ) ] \Rightarrow Q(50)=400+200sen [ \pi ( \dfrac{3}{2} ) ] \\ \\ \\ Q(50)=400+200sen ( \dfrac{3 \pi }{2} )\Rightarrow Q(50)=400+200*(-1) \\ \\ \\ Q(50)=400-200\Rightarrow \boxed {Q(50)=200}](https://tex.z-dn.net/?f=Q%28t%29%3D400%2B200sen+%5B+%5Cpi+%28+%5Cdfrac%7Bt-11%7D%7B26%7D+%29+%5D%5CRightarrow+Q%2850%29%3D400%2B200sen+%5B+%5Cpi+%28+%5Cdfrac%7B50-11%7D%7B26%7D+%29+%5D+%5C%5C++%5C%5C++%5C%5C+Q%2850%29%3D400%2B200sen+%5B+%5Cpi+%28+%5Cdfrac%7B39%7D%7B26%7D+%29+%5D+%5CRightarrow+Q%2850%29%3D400%2B200sen+%5B+%5Cpi+%28+%5Cdfrac%7B3%7D%7B2%7D+%29+%5D+%5C%5C++%5C%5C++%5C%5C+Q%2850%29%3D400%2B200sen+++%28+%5Cdfrac%7B3+%5Cpi+%7D%7B2%7D+%29%5CRightarrow+Q%2850%29%3D400%2B200%2A%28-1%29+%5C%5C++%5C%5C++%5C%5C+Q%2850%29%3D400-200%5CRightarrow+%5Cboxed+%7BQ%2850%29%3D200%7D%0A%0A%0A "Q(t)=400+200sen [ \pi ( \dfrac{t-11}{26} ) ]\Rightarrow Q(50)=400+200sen [ \pi ( \dfrac{50-11}{26} ) ] \\ \\ \\ Q(50)=400+200sen [ \pi ( \dfrac{39}{26} ) ] \Rightarrow Q(50)=400+200sen [ \pi ( \dfrac{3}{2} ) ] \\ \\ \\ Q(50)=400+200sen ( \dfrac{3 \pi }{2} )\Rightarrow Q(50)=400+200*(-1) \\ \\ \\ Q(50)=400-200\Rightarrow \boxed {Q(50)=200}")
b) Temperatura média .
![T(t)=10+12sen[ \pi ( \dfrac{t-15}{26} )]\Rightarrow 10+12sen[ \pi ( \dfrac{t-15}{26} )]=22 \\ \\ \\ 12sen[ \pi ( \dfrac{t-15}{26} )]=12\Rightarrow sen[ \pi ( \dfrac{t-15}{26} )]=1\Rightarrow \pi ( \dfrac{t-15}{26} )= \dfrac{ \pi }{2} \\ \\ \\ \dfrac{t-15}{26}= \dfrac{ 1 }{2}\Rightarrow t-15= \dfrac{26}{2} \Rightarrow t-15=13 \\ \\ \\ t=15+13 \Rightarrow \boxed{t=28}](https://tex.z-dn.net/?f=T%28t%29%3D10%2B12sen%5B+%5Cpi+%28+%5Cdfrac%7Bt-15%7D%7B26%7D+%29%5D%5CRightarrow+10%2B12sen%5B+%5Cpi+%28+%5Cdfrac%7Bt-15%7D%7B26%7D+%29%5D%3D22+%5C%5C++%5C%5C++%5C%5C+12sen%5B+%5Cpi+%28+%5Cdfrac%7Bt-15%7D%7B26%7D+%29%5D%3D12%5CRightarrow+sen%5B+%5Cpi+%28+%5Cdfrac%7Bt-15%7D%7B26%7D+%29%5D%3D1%5CRightarrow++%5Cpi+%28+%5Cdfrac%7Bt-15%7D%7B26%7D+%29%3D+%5Cdfrac%7B+%5Cpi+%7D%7B2%7D++%5C%5C++%5C%5C++%5C%5C++%5Cdfrac%7Bt-15%7D%7B26%7D%3D+%5Cdfrac%7B+1+%7D%7B2%7D%5CRightarrow+t-15%3D+%5Cdfrac%7B26%7D%7B2%7D+%5CRightarrow+t-15%3D13+%5C%5C++%5C%5C++%5C%5C+t%3D15%2B13+%5CRightarrow+%5Cboxed%7Bt%3D28%7D%0A "T(t)=10+12sen[ \pi ( \dfrac{t-15}{26} )]\Rightarrow 10+12sen[ \pi ( \dfrac{t-15}{26} )]=22 \\ \\ \\ 12sen[ \pi ( \dfrac{t-15}{26} )]=12\Rightarrow sen[ \pi ( \dfrac{t-15}{26} )]=1\Rightarrow \pi ( \dfrac{t-15}{26} )= \dfrac{ \pi }{2} \\ \\ \\ \dfrac{t-15}{26}= \dfrac{ 1 }{2}\Rightarrow t-15= \dfrac{26}{2} \Rightarrow t-15=13 \\ \\ \\ t=15+13 \Rightarrow \boxed{t=28}")
Resposta : semana 28
a) Quantidade de energia .
b) Temperatura média .
Resposta : semana 28
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