since the period of your sine curve should be 12
2π/k = 12
k = 6/π
your equation probably should have been:
f(x)=15.3sin(πx/6)+44.1 instead of what you typed.
since Jan ----> x = 0
June ----> x = 5
so just sub in x = 5 and evaluate
Let me know what you got.
The average daily minimum temperature for Pablo’s hometown can be modeled by the function f(x)=15.3sin(πx6)+44.1 , where f(x) is the temperature in °F and x is the month.
x = 0 corresponds to January.
What is the average daily minimum temperature in June?
Round to the nearest tenth of a degree if needed.
Use 3.14 for π .
really stuck could someone help out??
4 answers
i did but im still stuck
stuck how? Just replace x with 5 and evaluate!
too bad you could not be bothered to show us what you did.
f(x) = 15.3sin(πx/6)+44.1
f(5) = 15.3sin(5π/6)+44.1
= 15.3(1/2)+44.1
...
too bad you could not be bothered to show us what you did.
f(x) = 15.3sin(πx/6)+44.1
f(5) = 15.3sin(5π/6)+44.1
= 15.3(1/2)+44.1
...
Okay so I worked on the same problem and I got this answer, rounded to the nearest tenth like the problem says; 10.9
Here's my work:
15.3 sin (3.14(5)/6)+44.1
15.3 sin (15.7/6)+44.1
15.3 sin(2.616666667)+44.1
I5.3 sin(46.71666667)
I got=10.91958335
Unfortunately, I can't put in screenshots here :|
Hopefully my answer is correct!
Here's my work:
15.3 sin (3.14(5)/6)+44.1
15.3 sin (15.7/6)+44.1
15.3 sin(2.616666667)+44.1
I5.3 sin(46.71666667)
I got=10.91958335
Unfortunately, I can't put in screenshots here :|
Hopefully my answer is correct!