C = 2 * 3.14 * 4 = 25.1 to nearest tenth ( 1 decimal place)
The answers to the questions are
- The mathematical model is given as s = −16t2 + 1054
- The height after 4.5 seconds is 730 feet
- the time it would take to strike the ground is 8.11 seconds.
<h3>How to solve for the position of the object</h3>
The mathematical model of this problem would be written as
s = −16t2 + v0t + s0
s0 = 1054
then we would have
s = −16t2 + 1054
b. after 4.5 seconds the height is going to be
s = −16t2 + 1054
= −16(4.5)² + 1054
= -16 * 20.25 + 1054
= 730
C. the time that it takes to strike the ground
s = −16t² + 1054
= 16t² = 1054
t² = 1054/16
= 65.88
t = √65.88
t = 8.11
Hence the time it would take to strike the ground is 8.11 seconds.
Read more on velocity here:
brainly.com/question/4931057
#SPJ1
Set the argument (x+1) equal to zero.
I will assume that we are trying to find the slope between (-9,2) and (9, 2)
The slope's equation ⇒ 
<u>Let's set the variables</u>:
(x1, y1) --> (-9, 2)
(x2, y2) --> (9, 2)
<u>Now let's plug them in:</u>
<u>Thus the slope of the line is 0.</u>
Hope that helps!
![z=2x^4+8x \\ dz=(8x^3+8)\ dx=8(x^3+1)\ dx \\ \int\limits^0_{-1} {(2x^4+8x)^3*(4x^3+4)} \, dx =4* \int\limits^0_{-6} { \dfrac{z^3}{2}} \, dz \\ = 2* \left[\begin{array}{ccc} \dfrac{z^4}{4}\end{array}\right] ^0_{-6}= \left[\begin{array}{ccc} \dfrac{z^4}{2}\end{array}\right] ^0_{-6}=0-648=-648](https://tex.z-dn.net/?f=%20z%3D2x%5E4%2B8x%20%5C%5C%0Adz%3D%288x%5E3%2B8%29%5C%20dx%3D8%28x%5E3%2B1%29%5C%20dx%20%5C%5C%0A%0A%20%5Cint%5Climits%5E0_%7B-1%7D%20%7B%282x%5E4%2B8x%29%5E3%2A%284x%5E3%2B4%29%7D%20%5C%2C%20dx%20%3D4%2A%20%5Cint%5Climits%5E0_%7B-6%7D%20%7B%20%5Cdfrac%7Bz%5E3%7D%7B2%7D%7D%20%5C%2C%20dz%20%5C%5C%0A%3D%202%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cdfrac%7Bz%5E4%7D%7B4%7D%5Cend%7Barray%7D%5Cright%5D%20%5E0_%7B-6%7D%3D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cdfrac%7Bz%5E4%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20%5E0_%7B-6%7D%3D0-648%3D-648)