A force of 540 newtons stretches a spring 3 meters. A mass of 45 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 6 m/s. Find the equation of motion. x(t) = m
A mass of 1 slug is attached to a spring whose constant is 5 lb/ft. Initially, the mass is released 1 foot below the equilibrium position with a downward velocity of 9 ft/s, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to 2 times the instantaneous velocity.
Find the equation of motion if the mass is driven by an external force equal to f(t) = 20 cos 2t + 5 sin 21.
x(t) = h
Solve the given initial-value problem.
D^2x dt^2 + 4x = -6 sin 2t + 3 cos 2t, x(0) = -1, x'(0) = 1
X(t) = -cos(2t) - sin(2t) / 3 + 2t cos(2t) + tsin()2t_ 3
Solve the given initial-value problem.
xy"+y'=x, y{ 1) = 3, y\ 1) = - 1 /2
y(x) = .

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.

a reflection across the x-axis , followed by the by a reflection across the y-axis , and then a scale dilation of 0.5 .

d 25 pi/6

step-by-step explanation:

the diameter is 10 inches

the radius is 1/2 of the diameter or 5 inches

the area of the circle is

a = pi r^2

  = pi 5^2

  = 25 pi

the slice is 60 degrees of the entire circle.

60 degrees out of 360   60/360 = 1/6

the area of the slice is 1/6 of the circle

a slice = 1/6 a circle

a slice = 1/6 * 25 pi

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