MATH SOLVE

3 months ago

Q:
# Which function has a period equal to half the period of the function in y = -3sin(2/3x - 2π) + 2?a. y = 3cos(2/3x - π) + 2 b.y = -3/2cos(2/3x - 2π) + 2 c.y = -3cos(2/3x - 2π) + 2d.y = 3cos(4/3x - 2π) + 2

Accepted Solution

A:

Answer:D. [tex]y=3\cos(\frac{4}{3}x-2\pi)+2[/tex]Step-by-step explanation:The given function is:[tex]y=-3\sin(\frac{2}{3}x-2\pi)+2[/tex]This function is of the form:[tex]y=A\sin(Bx-C)+D[/tex], where [tex]B=\frac{2}{3}[/tex]The period is given by:[tex]T=\frac{2\pi}{B}[/tex][tex]T=\frac{2\pi}{\frac{2}{3}}=3\pi[/tex]Half of this period is [tex]\frac{3\pi}{2}[/tex].The function that has a period of [tex]\frac{3\pi}{2}[/tex] is [tex]y=3\cos(\frac{4}{3}x-2\pi)+2[/tex]