Question

The average daily maximum temperature for Laura’s hometown can be modeled by the function f(x)=4.5sin(πx/6)+11.8 , where f(x) is the temperature in °C and x is the month.
x = 0 corresponds to January.


What is the average daily maximum temperature in May?


Round to the nearest tenth of a degree if needed.


Use 3.14 for π .

Answer 1

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 2

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) = $4.5sin(\frac{\pi x}{6})+11.8$

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) = $4.5sin(\frac{4\pi }{6})+11.8$

f(4) = $4.5sin(\frac{2\pi }{3})+11.8$

f(4) = $4.5sin(\frac{360}{3})+11.8$ [since 2π = 360°]

f(4) = $4.5sin(120)+11.8$

f(4) = $4.5sin(180-60)+11.8$

f(4) = 4.5sin(60°) + 11.8 [since sin(180-∅) = sin∅]

f(4) = $(4.5)(\frac{\sqrt{3}}{2})+11.8$

f(4) = 3.897 + 11.8 = 15.7° °C

The average daily maximum temperature of May is 15.7° C