Answer:
B. b = 3a + 2
Step-by-step explanation:
We can write the equation in slope-intercept form as b = ma + c, where,
m = slope/rate of change
c = y-intercept/initial value
✔️Find m using any two given pair of values, say (2, 8) and (4, 14):
Rate of change (m) = change in b/change in a
m = (14 - 8)/(4 - 2)
m = 6/2
m = 3
✔️Find c by substituting (a, b) = (2, 8) and m = 3 into b = ma + c. Thus:
8 = 3(2) + c
8 = 6 + c
8 - 6 = c
2 = c
c = 2
✔️Write the equation by substituting m = 3 and c = 2 into b = ma + c. Thus:
b = 3a + 2
![\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.
I can help you. What is your question? Can you tell me?
Answer:
why
Step-by-step explanation:
Answer:
Each face of an octahedron is an equilateral triangle. An octahedron is a three-dimensional solid (polyhedron) that has eight faces.
Step-by-step explanation:
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