PLEASE HELP ME WITH THIS QUESTION I DONT WANT TO FAIL AND I FEEL AS IF IM WRONG!
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 .
this is what my Algebra teacher taught me to do
2π/B=2π/π0/6
cross out the π and I'm left with 2,0, and 6. She's telling me to multiply them so i did and got 0.
This is the part thats bothering me cuz i believe im wrong!
f(x)=3.8cos(πx20)+2.2
cos(x) has a minimum value of -1
so, f(x) has a minimum value of
2.2 + 3.8(-1) = -1.6
The temperature of a liquid during an experiment can be modeled by the function f(x)=3.8cos(πx20)+2.2 , where f(x) is the temperature in °C and x is the number of minutes into the experiment.
What is the lowest temperature the liquid reached during the experiment?
Round to the nearest tenth of a degree if needed.
Use 3.14 for π .
Enter your answer in the box.
SORRY I POSTED THE WRONG QUESTION BUT CAN YOU HELP MEWITH THAT ONE TOO PLEASE
f(x)=15.3sin(π*x*6)+44.1
For January, x = 0
f(0) = 15.3sin(π*0*6) + 44.1
= 15.3sin(0) + 44.1
sin(0) = 0
f(0) = 44.1
Now, for June, x = 5
f(5) = 15.3sin(π*5*6) + 44.1
= 15.3sin(30π) + 44.1
= 15.3sin(15*2π) + 44.1
= 15.3sin(2π) + 44.1
= 44.1
Are you sure you've got the correct equation? As it is now, the temperature would be 44.1 degrees for each month.
wait hold on thank you for explaining the first one to me but i meant to post this question ---> The temperature of a liquid during an experiment can be modeled by the function f(x)=3.8cos(πx20)+2.2 , where f(x) is the temperature in °C and x is the number of minutes into the experiment.
What is the lowest temperature the liquid reached during the experiment?
Round to the nearest tenth of a degree if needed.
Use 3.14 for π .
Enter your answer in the box.
and i got the answer 0 for that question which i belive im wrong. I looked at the recordings that my teacher sent and they didnt help at all. sorry im so stressed i dont want to fail.
It seems like you are trying to find the average daily minimum temperature in June, given the function f(x) = 15.3sin(πx/6) + 44.1, where x represents the month.
To find the average daily minimum temperature in June, you need to substitute x = 6 into the function and solve for f(x).
Let's plug in x = 6 into the function:
f(6) = 15.3sin(π(6)/6) + 44.1
f(6) = 15.3sin(π) + 44.1
Now, we know that sin(π) is equal to 0, so we can simplify the equation further:
f(6) = 15.3(0) + 44.1
f(6) = 0 + 44.1
f(6) = 44.1
Therefore, the average daily minimum temperature in June is approximately 44.1°F.
It seems like you made some errors in your calculations. When you plugged in x = 6 into the function, you should have obtained f(6) = 15.3sin(π) + 44.1, rather than f(6) = 15.3sin(π(0)/6) + 44.1. Remember that sin(π) is equal to 0, so that's why the temperature simplifies to f(6) = 44.1.
I hope this explanation helps you understand how to approach this problem correctly. Let me know if you have any further questions!