Consider a damped, forced mass/spring system. Let t denote time (in seconds) and let x(t) denote the position (in meters) of the mass at time t, with x = 0 corresponding to the equilibrium position. Suppose the mass m = 1 kg, the damping constant c = 3 N·s/m, the spring constant k = 2 N/m, the external force is F (t) = 20 cos(2t), the initial position x(0) = 1 m, and the initial velocity x′(0) = 2 m/s. a. Find the position function x(t). b. What part of the solution is the transient part and what part is the steady-state part? An RLC circuit contains in series a resistor R = 3 Ω, an inductor L = 1 H, and a capacitor C = 0.5 F. The current I(t) is provided by a source with emf E = 20cos(2t) Volts, where t is the time. Find the steady-state current Ip that develops after a long time (theoretically when t → [infinity]). For the differential equation y′′ + 4y = 2 tan(2x), solve its homogeneous part. Then use variation of parameters to find a particular solution yp.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

-22

times both sides by 2, times () by 8, add alile terms, move terms, add alike and caculate divide both sides by 3

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

12

step-by-step explanation:

12 + 5 = 17

12x2=24

24-7=17