Asound wave is given the following equation:

y= 8 sin(512πt) where t= time in seconds.

how many cycles will occur between t=2 and t=3.5 seconds?

384

Step-by-step explanation:

Standard equation of a sine wave is:

y = A sin(2πf x + B) + C

where A is the amplitude, f is the frequency, B is the phase shift, and C is the vertical offset.

In this case:

2πf = 512π

f = 256

This means there are 256 cycles per second.  The number of cycles between t = 2 and t = 3.5 is:

(3.5 − 2) × 256

384

(0,,,0)

step-by-step explanation:

the table of x and y values is shown in the attached pic(part of question)

we have to decide according that graph

given points are (0, 0) (–1, 0), (2, 0)

when we graph these points on graph they show an increasing line as shown in the attached pic

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.