![\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0\qquad \textit{from the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20~~~~~~%5Ctextit%7Bin%20feet%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-16t%5E2%2Bv_ot%2Bh_o%20%5Cend%7Barray%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Cstackrel%7B64%7D%7B%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h_o%3D%5Cstackrel%7B0%5Cqquad%20%5Ctextit%7Bfrom%20the%20ground%7D%7D%7B%5Ctextit%7Binitial%20height%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Bheight%20of%20the%20object%20at%20%22t%22%20seconds%7D%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)

Check the picture below, it hits the ground at 0 feet, where it came from, the ground, and when it came back down.
Answer:
three (3)
Step-by-step explanation:
4x is the first term
-4 is the second term
3y is the third term
Answer:
Step-by-step explanation:
<u>Incenter is the intersection of angle bisectors.</u>
- ∠MJP = ∠OJP, ∠MKP = ∠NKP, ∠NLP = ∠OLP
<u>First find the value of x:</u>
- 7x - 6 = 5x + 4
- 7x - 5x = 4 + 6
- 2x = 10
- x = 5
<u>Find the angle MJP:</u>
<u>We know sum of interior angles of a triangle is 180°. Using this find the missing angle measure:</u>
- 2*m∠MJP + 2*m∠NJP + 2*m∠MKP = 180°
- m∠MJP + m∠NJP + m∠MKP = 90°
- 29° + 26° + m∠MKP = 90°
- m∠MKP = 90° - 55°
- m∠MKP = 35°
Answer:
x = -2
Step-by-step explanation:
C . 250 cm^2
cause its the same