A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point 1 foot above the equilibrium position with a downward velocity of 14 ft/s. 1. Determine the time (in s) at which the mass passes through the equilibrium position.
2. Find the time after the mass passes through the equilibrium position at which the mass attains its extreme displacement from the equilibrium position.
3. What is the position of the mass at this instant?

answerim :

n

step-by-step explanation:

im not sure but my dad knows

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.

a. (14,375+570)(5.3) b. 158,417$

step-by-step explanation:

a. you are basically just setting up the equation. f= amount of debt hence 14,375. g= how much she is borrowing from her parents weekly hence 570 and t= the interest on her debt. creating the equation.

b.   14,375+570= 14,945

      14,945 x 5.3 = 79,208.5

      79,208.5 x 2(years) = 158,417(answer)