Afireworks display is launched from a platform 10 feet above ground with an initial upward velocity of 70 feet per second. the height of the fireworks above ground after t seconds is given by the equation h = –16t2 + 70t + 10, where h is the height of the fireworks in feet and t is the time in seconds after they are launched. what is the maximum height of the fireworks display, to the nearest foot?

A. 87 feet

Step-by-step explanation:

1. You can find the t value of the vertex of the parabola as following:

2. Substitute values:

a=-16

b=70

Then:

3. Substitute the vaue obtained into the equation given in the problem. Therefore, you obtain the following result:

4. To the nearest foot:

h=87 feet

answer: t=3

step-by-step explanation:

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]