The acceleration a(t) of a particle is -4a^2xi -a^2yj. The particle is initially at rest at the point (a, b). 1) show that the particle oscillates along the arc of the parabola 2ay^2 = b^2(x+a). 2) Find the velocity of the particle whenever it is at the points (a, -b) and (-a, 0).​

the answer is c

hope this !

where are the choices?

1/3 + 1/3 ≠ 2/6 because 1/3 alone = 2/6  

step by step explanation:

1/3 + 1/3 = 2/3       2/3=4/6

a, b, c, d are correct.

step-by-step explanation:

we have g(x) = 3 and are required to find which option is a translation of g(x).

now, 'translation' is a transformation which means to shift the function in any direction and the general form of translation is g(x) + k, where k is any real number.

so, according to the options we see that all are correct as they satisfy the general form of translation.

also, it can be seen from the graph below that the function g(x) = 3 is shifted from its position in all the four options.