v=20√(273+t)
329 = 20*sqrt(273+t)
16.45 = sqrt(273+t)
273+t = 16.45^2
t- 16.45^2-273
t = -2.4 degrees celsius
Hope that helps? :)
The velocity of sound in air is given by the equation  , where v is the velocity in meters per second and t is the temperature in degrees Celsius. Find the temperature when the velocity is 329 meters per second by graphing the equation. Round the answer to the nearest degree. show your work
4 answers
Thanks , My answer was -2 , I guess I was close :)
Yes you were! Good job! Glad I could help. :)
v=20√(273+t)
329 = 20*sqrt(273+t)
16.45 = sqrt(273+t)
273+t = 16.45^2
t- 16.45^2-273
t = -2.4 degrees celsius
329 = 20*sqrt(273+t)
16.45 = sqrt(273+t)
273+t = 16.45^2
t- 16.45^2-273
t = -2.4 degrees celsius