For a better understanding of the explanation provided here kindly go through the file attached.
Since, the weight attached is already at the lowest point at time, t=0, therefore, the equation will have a -9 as it's "amplitude" and it will be a Cosine function. This is because in cosine function, the function has the value of the amplitude at t=0.
Now, we know that the total angle in radians covered by a cosine in a given period is
and the period given in the question is t=3 seconds. Therefore, the angular velocity,
of the mentioned system will be:

Combining all the above information, we see that the equation which models the distance, d, of the weight from its equilibrium after t seconds will be:

Thus, Option B is the correct option. The attached diagram is the graph of the option B and we can see clearly that at t=3, the weight indeed returns to it's original position.
No enough info
you can just say that five brought there own lunch while 19 didnt
Step-by-step explanation:
going around the field one time is walking all 4 sides.
so, 4×13 m
going around the field 5 times is walking all 4 sides 5 times.
that means
5×4×13 m = 20×13 = 260 m
The thousand place
Of 6.589 is 0.009 because after the decimal it goes tenths hundredth then thousands