The average daily temperature, t, in degrees Fahrenheit for a city as a function of the month of the year, m, can be modeled by the equation t=35cos(pi/6(m+3))+55, where m = 0 represents January 1, m = 1 represents February 1, m = 2 represents March 1, and so on. Which equation also models this situation?


A) t=-35sin(pi/6m)+55

B) t=-35sin(pi/6(m+3))+55

C) t=35sin(pi/6m)+55

D) t=35sin(pi/6(m+3))+55

The average daily temperature t in degrees Fahrenheit for a city as a function of the month of the year m can be modeled by the equation t35cospi6m355 where m 0 class=

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Answer:

Answer is A.

Step-by-step explanation:

For the equation: 35cos(pi/6(m+3)+55)

Simplify it by multiplying pi/6 in the bracket values:  

[tex]=35cos(\frac{m.pi}{6}+\frac{3.pi}{6}+55) \\=35cos(\frac{m.pi}{6}+55+\frac{pi}{2})[/tex]

Now suppose x = m.pi/6+55, while pi/2 = 90⁰

As per law cos(x+90⁰)=-sin(x)

so,

cos((m.pi/6)+55+90⁰=-sin((m.pi/6)+55)

Hence answer will be

[tex]=-35sin(\frac{m.pi}{6}+55)[/tex]

.

The law cos(x+90⁰)=-sin(x) can be proved by putting y=90⁰ in the following formula,

cos(x+y)=cos(x).cos(y)-sin(x).sin(y)

Answer:

A

Step-by-step explanation:

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