Answer:
- The sequence is an Arthemtic Progression
An=A1+(n-1)d
A1 is first term, An is nth term, n is number of term, and d is common difference
therefore
A4=35, A1= -17
A4=A1+(4-1)d
35= -17+3d
35+17=3d
52=3d
52/3=3d/3
14=d
common diffrence(d)=14
- The general solution is given by
An= -17+(n-1)14
An= -17+14n-14
An= -31+14n
<u>An= 14n-31</u>
A14 term, means n=14
From An=A1+(n-1)d
A14= -17+(14-1)14
= -17+(13×14)
= -17+182
= 165.
<u>Therfore, the 14th term is 165.</u>
2. A sequence has a CR of 4/5 and its eighth term (a8) is (393216/3125). What is its general equation? Its 3rd term?
<u>solution</u>
common ratio(r)=4/5
eighth term(G8)=393216/3125
From Gn= G1r^(n-1)
G8 means n=8
G8=G1r^(n-1)
393216/3125=G1(4/5)^(8-1)
393216/3125=G1(4/5)^7
G1=(393216/3125)/(4/5)^7
G1=600
<u>The first term is given by G1=600</u>
The General equation is given by
The General equation is given by Gn= 600(4/5)^(n-1)
3rd term (G3)
G3= G1(4/5)^(3-1) where n=3,
=600(4/5)^2
=600(16/25)
=384
<u>Therefore, the 3rd term is given by G3= </u><u>3</u><u>8</u><u>4</u><u>.</u>
<u>I</u><u> </u><u>h</u><u>a</u><u>v</u><u>e</u><u> </u><u>m</u><u>a</u><u>d</u><u>e</u><u> </u><u>s</u><u>o</u><u>m</u><u>e</u><u> </u><u>C</u><u>o</u><u>r</u><u>r</u><u>e</u><u>c</u><u>t</u><u>i</u><u>o</u><u>n</u><u>s</u><u> </u><u>i</u><u> </u><u>m</u><u>e</u><u>s</u><u>s</u><u>e</u><u>d</u><u> </u><u>u</u><u>p</u><u> </u><u>s</u><u>o</u><u>m</u><u>e</u><u>w</u><u>h</u><u>e</u><u>r</u><u>e</u><u>.</u>
Answer:
b
Step-by-step explanation:
Answer:
x = 9/4
Step-by-step explanation:
f(x) = 4x - 9
(first substitute f(x) with 0)
0 = 4× - 9
(Move four)
-4x = -9
=> x = -9/-4
= 9/4
![\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.