Answer:
66
Step-by-step explanation:
It's given that
Tn = 5n + 16
The 10th term is
T10 = 5(10) + 16
= 66
Dive your answer by three so 9.60/3 = 3.20 or £3.20
Answer:
<BCO = <BAO = 20degrees
Step-by-step explanation:
If <ABC measures 100 and is inscribed in a circle O. find <BAO and <BCO
To get <BAO and <BCO, we need to get <AOC first.
From the figure, it can be seen that triangle ABC is an isosceles trinagle. Hence;
<BAC + <BCA + 100 = 180
Since <BAC = <BCA
<BAC + <BAC = 180 - 100
2<BAC = 80
<BAC = 80/2
<BAC = 40
Also;
<BAO = <BCO and <BAO = <BAC/2
<BAO = 40/2 = <BCO
Hence <BCO = <BAO = 20degrees
180 ÷3 = 60 ° (angles on a straight line equals to 180 ° )
a+b+60 =180°
a+ b= 180 ° - 60° = 120°
(a+b) are 2 variables
so 120 ÷2 =60 °
therefore a and b =60 °
c +b+ 60° = 180° ( co - interior angles are supplementary angles )
c+60° +60° =180°
c +120° =180°
c =180°-120°
c=60 °
d=60° (alternate angles are equal )
or
c+b+d=180°
60° +60° + d = 180°
d=180°-120°
d=60°
Hello, please consider the following.
![\displaystyle \begin{aligned} \int\limits^x {5sin(5t)sin(t)} \, dt &= -\int\limits^x {5sin(5t)} \, d(cos(t))\\ \\&=-[5sin(5t)cos(t)]+ \int\limits^x {25cos(5t)cos(t)} \, dt\\\\&=-5sin(5x)cos(x)+ \int\limits^x {25cos(5t)} \, d(sin(t))\\ \\&=-5sin(5x)cos(x)+[25cos(5t)sin(t)]+ \int\limits^x {25sin(5t)sin(t)} \, dt\\\\&=-5sin(5x)cos(x)+25cos(5x)sin(x)+ \int\limits^x {(25*5)sin(5t)sin(t)} \, dt\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20%5Cint%5Climits%5Ex%20%7B5sin%285t%29sin%28t%29%7D%20%5C%2C%20dt%20%26%3D%20-%5Cint%5Climits%5Ex%20%7B5sin%285t%29%7D%20%5C%2C%20d%28cos%28t%29%29%5C%5C%20%5C%5C%26%3D-%5B5sin%285t%29cos%28t%29%5D%2B%20%5Cint%5Climits%5Ex%20%7B25cos%285t%29cos%28t%29%7D%20%5C%2C%20dt%5C%5C%5C%5C%26%3D-5sin%285x%29cos%28x%29%2B%20%5Cint%5Climits%5Ex%20%7B25cos%285t%29%7D%20%5C%2C%20d%28sin%28t%29%29%5C%5C%20%5C%5C%26%3D-5sin%285x%29cos%28x%29%2B%5B25cos%285t%29sin%28t%29%5D%2B%20%5Cint%5Climits%5Ex%20%7B25sin%285t%29sin%28t%29%7D%20%5C%2C%20dt%5C%5C%5C%5C%26%3D-5sin%285x%29cos%28x%29%2B25cos%285x%29sin%28x%29%2B%20%5Cint%5Climits%5Ex%20%7B%2825%2A5%29sin%285t%29sin%28t%29%7D%20%5C%2C%20dt%5Cend%7Baligned%7D)
And we can recognise the same integral, so.
And then,
Thanks
