Answer:
Least number of bus require for trip = 5 buses (Approx)
Step-by-step explanation:
Given:
Total number of classes = 9
Number of student in each class = 25
Number of teacher = 4
Number of chaperones = Double of teacher
Bus hold = 45 people
Find:
Least number of bus require for trip
Computation:
Total number of student = 9 × 25
Total number of student = 225
Number of chaperones = 4 × 2
Number of chaperones = 8
Total people = 225 + 8 + 4
Total people = 237
Least number of bus require for trip = Total people / Bus hold
Least number of bus require for trip = 237 / 45
Least number of bus require for trip = 5.266
Least number of bus require for trip = 5 buses (Approx)
If this is a parabolic motion equation, then it is a negative parabola, which looks like a hill (instead of a positive parabola that opens like a cup). Your equation would be h(t)= -16t^2 + 20t +3. That's the equation for an initial velocity of 20 ft/s thrown from an initial height of 3 ft. And the -16t^2 is the antiderivative of the gravitational pull. Anyway, if you're looking for the maximum height and you don't know calculus, then you have to complete the square to get this into vertex form. The vertex will be the highest point on the graph, which is consequently also the max height of the ball. When you do this, you get a vertex of (5/8, 9.25). The 9.25 is the max height of the ball.
If you have a calculator, you can use the inverse cos " cos⁻¹ " key. Many upper level books write that as Arccos so as not to confuse you with a negative exponent. We do that because if you take the cosine of any angle, and then the inverse cosine of any angle, it leads you back to the original angle. Think of it as a big "UNDO" button.
To one decimal place, that result is 61.6 degrees.
![\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ a_n=2-5(n-1)\implies a_n=\stackrel{\stackrel{a_1}{\downarrow }}{2}+(n-1)(\stackrel{\stackrel{d}{\downarrow }}{-5})](https://tex.z-dn.net/?f=%5Cbf%20n%5E%7Bth%7D%5Ctextit%7B%20term%20of%20an%20arithmetic%20sequence%7D%20%5C%5C%5C%5C%20a_n%3Da_1%2B%28n-1%29d%5Cqquad%20%5Cbegin%7Bcases%7D%20n%3Dn%5E%7Bth%7D%5C%20term%5C%5C%20a_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%20d%3D%5Ctextit%7Bcommon%20difference%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20a_n%3D2-5%28n-1%29%5Cimplies%20a_n%3D%5Cstackrel%7B%5Cstackrel%7Ba_1%7D%7B%5Cdownarrow%20%7D%7D%7B2%7D%2B%28n-1%29%28%5Cstackrel%7B%5Cstackrel%7Bd%7D%7B%5Cdownarrow%20%7D%7D%7B-5%7D%29)
so, we know the first term is 2, whilst the common difference is -5, therefore, that means, to get the next term, we subtract 5, or we "add -5" to the current term.
just a quick note on notation:
![\bf \stackrel{\stackrel{\textit{current term}}{\downarrow }}{a_n}\qquad \qquad \stackrel{\stackrel{\textit{the term before it}}{\downarrow }}{a_{n-1}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current term}}{a_5}\qquad \quad \stackrel{\textit{term before it}}{a_{5-1}\implies a_4}~\hspace{5em}\stackrel{\textit{current term}}{a_{12}}\qquad \quad \stackrel{\textit{term before it}}{a_{12-1}\implies a_{11}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bcurrent%20term%7D%7D%7B%5Cdownarrow%20%7D%7D%7Ba_n%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bthe%20term%20before%20it%7D%7D%7B%5Cdownarrow%20%7D%7D%7Ba_%7Bn-1%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bcurrent%20term%7D%7D%7Ba_5%7D%5Cqquad%20%5Cquad%20%5Cstackrel%7B%5Ctextit%7Bterm%20before%20it%7D%7D%7Ba_%7B5-1%7D%5Cimplies%20a_4%7D~%5Chspace%7B5em%7D%5Cstackrel%7B%5Ctextit%7Bcurrent%20term%7D%7D%7Ba_%7B12%7D%7D%5Cqquad%20%5Cquad%20%5Cstackrel%7B%5Ctextit%7Bterm%20before%20it%7D%7D%7Ba_%7B12-1%7D%5Cimplies%20a_%7B11%7D%7D)
Answer:
d
Step-by-step explanation:
1+1^2=4
2+1=3^2=9
3+1=4^2=16
^(times by self)
