The amount of light emitted from a battery indicator bulb pulses while the battery is charging. This can be modeled by the equation y = 60+60sin[(pi/4)t] , where y is the lumens emitted from the bulb and t is the number of seconds since the beginning of a pulse. At what time will the amount of light emitted be equal to 110 lumens?
im stupid
mhh, aren't we just solving
60 + 60sin[(π/4)t] = 110 ?
sin[(π/4)t] = .833333
make sure your calculator is set to radians
[(π/4)t] = .98511
t = .....
let me know what you got
1.25 is what i got
correct!
good job
THANK YOU!
To find the time at which the amount of light emitted will be equal to 110 lumens, we need to solve the equation y = 110 for the variable t.
Given the equation y = 60 + 60sin[(π/4)t], we can rewrite it as:
110 = 60 + 60sin[(π/4)t]
Subtracting 60 from both sides, we get:
50 = 60sin[(π/4)t]
Now, divide both sides by 60:
(5/6) = sin[(π/4)t]
To find the value of t, we need to take the inverse sine (arcsin) of both sides:
arcsin(5/6) = (π/4)t
Divide both sides by (π/4):
arcsin(5/6) / (π/4) = t
Using a calculator that has the arcsin function, we find:
t ≈ 0.965
Therefore, at approximately 0.965 seconds since the beginning of a pulse, the amount of light emitted will be equal to 110 lumens.