Respuesta :
f(x) = 4.5sin(πx/6) + 11.8
January, x = 0
February, x = 1
an so on
May = 4
f(4) = 4.5sin(4π/6) + 11.8
f(4) = 4.5sin(2π/3) + 11.8
We know that π = 180° so,
f(4) = 4.5sin(360°/3) + 11.8
f(4) = 4.5sin(120°) + 11.8
f(4) = 4.5sin(60° + 60°) + 11.8
We know that:
sin(a + b) = sin a . cos b + sin b . cos a
sin(60 + 60) = sin 60 . cos 60 + sin 60 . cos 60
sin(60 + 60) = 2.(sin 60 . cos 60)
sin(60 + 60) = 2.(√3/2 . 1/2)
sin(60 + 60) = 2.(√3/4)
sin(60 + 60) = √3/2
f(4) = 4.5*√3/2 + 11.8
f(4) = 2,25√3 + 11.8
√3 ≅ 1,73
f(4) = 2,25 . 1,73 + 11.8
f(4) = 13,7
The average daily maximum temperature in may is 13,7 °C
Answer:
Average daily maximum temperature of May is 15.7° C
Step-by-step explanation:
The average daily maximum temperature for Laura's hometown can be modeled by the function f(x) = [tex]4.5sin(\frac{\pi x}{6})+11.8[/tex]
Where x is the number of month.
We have to calculate the average daily maximum temperature in May.
Since x is the number of month so we will put x = 4 in the given function.
f(4) = [tex]4.5sin(\frac{4\pi }{6})+11.8[/tex]
f(4) = [tex]4.5sin(\frac{2\pi }{3})+11.8[/tex]
f(4) = [tex]4.5sin(\frac{360}{3})+11.8[/tex] [since 2π = 360°]
f(4) = [tex]4.5sin(120)+11.8[/tex]
f(4) = [tex]4.5sin(180-60)+11.8[/tex]
f(4) = 4.5sin(60°) + 11.8 [since sin(180-∅) = sin∅]
f(4) = [tex](4.5)(\frac{\sqrt{3}}{2})+11.8[/tex]
f(4) = 3.897 + 11.8 = 15.7° °C
The average daily maximum temperature of May is 15.7° C