In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (Enter your answers as a comma-separated list of ordered pairs.) x = t^3 - 3t, y = t^2 - 4

(27.55, 7.22), (-11.3, 3.21).

Step-by-step explanation:

When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

.

This polynomial has roots such that , given by the following formulas:

In this question:

So

Then

So

Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

The first point is (27.55, 7.22)

t = -2.685

The second point is (-11.3, 3.21).

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry