![\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{}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ h=-16t^2+\stackrel{\stackrel{v_o}{\downarrow }}{65}t](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%7B%7D%7B%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h_o%3D%5Cstackrel%7B%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~%5Cdotfill%5C%5C%5C%5C%20h%3D-16t%5E2%2B%5Cstackrel%7B%5Cstackrel%7Bv_o%7D%7B%5Cdownarrow%20%7D%7D%7B65%7Dt)
now, take a look at the picture below, so for 2) and 3) is the vertex of this quadratic equation, 2) is the y-coordinate and 3) the x-coordinate.
Answer:
5, -5
Step-by-step explanation:
well 5-5 is 0 so that's it
:))))
Answer: 52
Explanation:
You need to replace the x-values with 8 to solve this. So, the equation would look like this. f(8)= 7 x 8 - 5
7 times 8 is 56. 56 minus four is 52 which is your answer.
If you needed to graph this then it would look like this... (8, 52)
The 8 is in the input because it is the x-value (we replace x with 8) and 52 is output/range/y-value because it is the answer.
4/6=0.6666666667
0.6666666667×39=26 millimeters
Step-by-step explanation:
Find the slope of JK by the formula :y2-y1/(x2-x1) whose value is 4/5
similarly, calculate the slope of GH it's value is also 4/5
Find slope of GK now by same formula y2-y1/( x2-x1 ).the value will be -5/3
Similarly calculate slope of HJ and you will get its value to be -5/3
since parallel lines have equal slopes JK and GH are parallel and GK and HJ are parallel. Since opposite sides are parallel GHJk is parallelogram
