Aforced spring-mass system satisfies the ivpx′′+ 3x′+kx= sin(t), x(0) = 0, x′(0) = 0.find the minimum value of the spring constantksuch that the steady-state solution nevergoes above 0.1 meters. (that is, for large values oftthe mass oscillates no more than 0.1meters to either side of its equilibrium position.)

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation:

29933333.33

step-by-step explanation:

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116

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