April shoots an arrow upward at a speed of 80 feet per second from a platform 25 feet high. the pathway of the arrow can be represented by the equation h=-16t squared+80t+25, where h is the highest and t is the time in seconds. what is the maximum height of the arrow.

The answer is: The maximum height of the arrow is 125 feet.

To solve this problem, we need to know the motion of the arrow is described by a quadratic equation (parabola). Also, we are given a parabola with a negative coefficient. So, to solve this problem, we need to remember the following:

- To know if the parabola is opening upward or downward, we need to look the sign of the coefficient of the quadratic term.

- We need to remember that the highest or lowest point of a parabola is given by the y-coordinate of the vertex.

We can calculate the vertex using the following formula:

Then, to find the y-coordinate of the vertex, we need to substitute the x value into the parabola equation.

So, we are given the pathway of the arrow represented by the quadratic equation:

Where,

So, calculating the vertex coordinates in order to find the maximum height of the arrow (highest point), we have:

Then, substituting "t" into the equation of the parabola, we have:

Therefore, we have that the vertex coordinates are (2.5,125) being the y-coordinate the highest point of the parabola, which is equal to the maximum height of the arrow.

Hence, the maximum height of the arrow is 125 feet.

Have a nice day!

Have a nice day!

- 2 + 10x³ - 21x² + 19x - 4

step-by-step explanation:

each term in the second factor is multiplied by each term in the first factor

x²(- 2x² + 4x - 1) - 3x(- 2x² + 4x - 1) + 4(- 2x² + 4x - 1) ← distribute

= - 2 + 4x³ - x² + 6x³ - 12x² + 3x -8x² + 16x - 4

collect like terms

= - 2 + 10x³ - 21x² + 19x - 4 ← in standard form