Answer:
sec²(x) - sec(x) + tan²(x) = (sec(x) - 1)(2sec(x) + 1)
Step-by-step explanation:
sec²(x) - sec(x) + tan²(x) =
= sec²(x) - sec(x) + [sec²(x) - 1]
= sec²(x) - sec(x) + [(sec(x) + 1)(sec(x) - 1)]
= sec(x)[sec(x) - 1] + [(sec(x) + 1)(sec(x) - 1)]
= (sec(x) - 1)(sec(x) + sec(x) + 1)
= (sec(x) - 1)(2sec(x) + 1)
Answer:
Step-by-step explanation:
x![\left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left \{ {{y=2} \atop {x=2}} \right. \int\limits^a_b {x} \, dx](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D%20%5Cright.%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx)
XE[_4.2,2.2]
FK+M=A-UK
FK-UK=A-M
(F-U)K=A-M
K=(A-M)/(F-U)
I think it’s C if you still need an answer but don’t trust me on that
D. data collection
Simply the act of collecting data from sample
