Given the parametric equations, x = 2t + 5 and y = 3t^2 - 1, find the point on the graph at time, t = 4. explain how you got your answer.

Parametric equations,   at t = 4

The point on the graph at time, t = 4 is (13,47).

Given:

The expression t = 4 means that 4 units of time have come off the clock (possibly 4 seconds). Plug this value into each equation given for x and y

In short, if t = 4 then it leads to x = 13 and y = 47 at the same time. These x and y values pair up to get the final answer (13,47).

Therefore, the point on the graph at time, t = 4 is (13,47).

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(13, 47)

The expression t = 4 means that 4 units of time have come off the clock (possibly 4 seconds). Plug this value into each equation given for x and y

x = 2*t + 5 = 2*4 + 5 = 8 + 5 = 13

y = 3*t^2 - 1 = 3*4^2 - 1 = 3*16 - 1 = 48 - 1 = 47

In short, if t = 4 then it leads to x = 13 and y = 47 at the same time. These x and y values pair up to get the final answer (13,47)

step-by-step explanation:

the answer is d

the answer is in the first quadrant, which make the y-int positive.

since the dots are going down, therefore the correlation is negative

i hope it is right.