T=2π/|b|. The period of an equation of the form y = a sin bx is T=2π/|b|.
In mathematics the curve that graphically represents the sine function and also that function itself is called sinusoid or sinusoid. It is a curve that describes a repetitive and smooth oscillation. It can be represented as y(x) = a sin (ωx+φ) where a is the amplitude, ω is the angular velocity with ω=2πf, (ωx+φ) is the oscillation phase, and φ the initial phase.
The period T of the sin function is T=1/f, from the equation ω=2πf we can clear f and substitute in T=1/f.
f=ω/2π
Substituting in T=1/f:
T=1/ω/2π -------> T = 2π/ω
For the example y = a sin bx, we have that a is the amplitude, b is ω and the initial phase φ = 0. So, we have that the period T of the function a sin bx is:
T=2π/|b|
The n-th term is given by
Then we can find the common ratio from the given terms.
The appropriate choice is -1.
Answer:
when multiply with exponents you add so
(3a^2b^7)(5a^3b^8)=15a^5b^13
Hope This Helps!!!
Answer:
A. 0.72
Step-by-step explanation:
When the ball hits the ground, h = 0.
0 = -16t² + 6t + 4
0 = 8t² − 3t − 2
Solve with quadratic formula:
t = [ 3 ± √(9 − 4(8)(-2)) ] / 16
t = (3 ± √73) / 16
t = 0.72
The answer to your question is 23
