Answer: 4/15 of the movie
Step-by-step explanation:
In the morning, Charlie watched ¹/₃ of a movie. He then watched some more at night.
At the end of the day, he still had ²/₅ of the movie to watch.
The proportions of an item must always add up to one. This means that if you added the 1/3 to the 2/5 and then to the fraction Charlie watch at night, they should add up to one.
¹/₃ + ²/₅ + x = 1
x = 1 - ¹/₃ - ²/₅
Answer:
Tn = 2Tn-1 - Tn-2
Step-by-step explanation:
Before we can generate the recursive sequence, we need to find the nth term of the given sequence.
nth term of an AP is given as:
Tn = a+(n-1)d
If a17 = -40
T17 = a+(17-1)d = -40
a+16d = -40 ...(1)
If a28 = -73
T28 = a+(28-1)d = -73
a+27d = -73 ...(2)
Solving both equations simultaneously using elimination method.
Subtracting 1 from 2 we have:
27d - 16d = -73-(-40)
11d = -73+40
11d = -33
d = -3
Substituting d = -3 into 1
a+16(-3) = -40
a - 48 = -40
a = -40+48
a = 8
Given a = 8, d = -3, the nth term of the sequence will be
Tn = 8+(n-1) (-3)
Tn = 8+(-3n+3)
Tn = 8-3n+3
Tn = 11-3n
Given Tn = 11-3n and d = -3
Tn-1 = Tn - d... (3)
Tn-1 = 11-3n +3
Tn-1 = 14-3n
Tn-2 = Tn-2d...(4)
Tn-2 = 11-3n-2(-3)
Tn-2 = 11-3n+6
Tn-2 = 17-3n
From 3, d = Tn - Tn-1
From 4, d = (Tn - Tn-2)/2
Equating both common difference
(Tn - Tn-2)/2 = Tn - Tn-1
Tn - Tn-2 = 2(Tn - Tn-1)
Tn - Tn-2 = 2Tn-2Tn-1
2Tn-Tn = 2Tn-1 - Tn-2
Tn = 2Tn-1 - Tn-2
The recursive formula will be
Tn = 2Tn-1 - Tn-2
Hmm if I'm not mistaken, is just an "ordinary" annuity, thus
![\bf \qquad \qquad \textit{Future Value of an ordinary annuity} \\\\ A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right] \\\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Cqquad%20%5Ctextit%7BFuture%20Value%20of%20an%20ordinary%20annuity%7D%0A%5C%5C%5C%5C%0AA%3Dpymnt%5Cleft%5B%20%5Ccfrac%7B%5Cleft%28%201%2B%5Cfrac%7Br%7D%7Bn%7D%20%5Cright%29%5E%7Bnt%7D-1%7D%7B%5Cfrac%7Br%7D%7Bn%7D%7D%20%5Cright%5D%0A%5C%5C%5C%5C%5C%5C)
![\bf \begin{cases} A= \begin{array}{llll} \textit{original amount}\\ \textit{already compounded} \end{array} & \begin{array}{llll} \end{array}\\ pymnt=\textit{periodic payments}\to &400\\ r=rate\to 10\%\to \frac{10}{100}\to &0.1\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, so once} \end{array}\to &1\\ t=years\to &14 \end{cases} \\\\\\ A=400\left[ \cfrac{\left( 1+\frac{0.1}{1} \right)^{1\cdot 14}-1}{\frac{0.1}{1}} \right] ](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0AA%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Boriginal%20amount%7D%5C%5C%0A%5Ctextit%7Balready%20compounded%7D%0A%5Cend%7Barray%7D%20%26%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%0A%5Cend%7Barray%7D%5C%5C%0Apymnt%3D%5Ctextit%7Bperiodic%20payments%7D%5Cto%20%26400%5C%5C%0Ar%3Drate%5Cto%2010%5C%25%5Cto%20%5Cfrac%7B10%7D%7B100%7D%5Cto%20%260.1%5C%5C%0An%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%0A%5Ctextit%7Bannually%2C%20so%20once%7D%0A%5Cend%7Barray%7D%5Cto%20%261%5C%5C%0A%0At%3Dyears%5Cto%20%2614%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D400%5Cleft%5B%20%5Ccfrac%7B%5Cleft%28%201%2B%5Cfrac%7B0.1%7D%7B1%7D%20%5Cright%29%5E%7B1%5Ccdot%2014%7D-1%7D%7B%5Cfrac%7B0.1%7D%7B1%7D%7D%20%5Cright%5D%0A)
Answer:
sum of all side
8+22+6+8+8+17+5+5+13+4
100 is the area of this figure
Step-by-step explanation:
please mark me brainleat
