A16 pound weight attached to a spring exhibits simple harmonic motion. determine the equation of motion if the spring constant is 8 lb/ft and if the weight is released 6 inches below the equilibrium position with a downward velocity of 1 ft/s. a(t)-4 sin(4t)-2 cos(4t) ar(t)sin(4t) 2 cos(4t) r(t) (4t)o() az (t) -7sin (4t)cos (4t cos save

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.

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