Solve the following (a) a 4 pound weight stretches a spring 6 inches dowward to a position y-0. set up an initial value problem (differential equation and initial conditions) fr the positio y(t) ( t secoods later) ? fthe spring has been lifted back up the 6 inches and is released. (b) compute the laplace transform clys) wbe y ls the solut lon of the initial value problem: 2"+ty'+6y 0,withy0)-2 andyio)-2

dude   i don't even know. i've been trying for hours.

step-by-step explanation:

c.

step-by-step explanation:

idn

step-by-step explanation:

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.