If y = cos(kt), then its first two derivatives are
y' = -k sin(kt)
y'' = -k² cos(kt)
Substituting y and y'' into 49y'' = -16y gives
-49k² cos(kt) = -15 cos(kt)
⇒ 49k² = 15
⇒ k² = 15/49
⇒ k = ±√15/7
Note that both values of k give the same solution y = cos(√15/7 t) since cosine is even.
Answer:
7z^3+6z^2-27z+22
Step-by-step explanation:
The given question is of subtraction:
So,
(7z^3+42z^2-15z+1)- (36z^2+12z-21)
=7z^3+42z^2-15z+1-36z^2-12z+21
Combining alike terms to simplify
=7z^3+42z^2-36z^2-15z-12z+1+21
=7z^3+6z^2-27z+22
So the answer in simplified form is:
7z^3+6z^2-27z+22 ..
Answer:
Step-by-step explanation:
Sample space, The counting Principle and, a table can be used
22.5 i think but i have no idea good luck
