Consider a vibrating quarter-circular membrane, 0 < r < a, 0 < θ < π/2, with u = 0 on the entire boundary. [Hint: You may assume without derivation that λ > 0 and that product solutions u(r, θ, t) = φ(r, θ)h(t) = f(r)g(θ)h(t) satisfy ∇2φ + λφ = 0 dh dt = −λkh d2g dθ2 = −μg r d dr r df dr + (λr2 − μ)f = 0.]

Solve the initial value problem if u(r, θ, 0) = g(r, θ), ∂u ∂t (r, θ, 0) = 0.

i think the answer would be c

5 + 8 +7 + 4 + 0 +1 = 25

step-by-step explanation:

answer: the proof is mentioned below.

step-by-step explanation:

since, given:   lines bc = ef

prove:   lines ac > ef

          statements                                           reason

1. bc=ef                                           1. given

2. ac= ab+bc                                 2. because b is the point between the  

                                                            line segment ac

3.   ac > bc                                       3. because if a line segment is equal to

                                                          the sum of two line segments then it is

                                                          always greater then each of these

                                                          individual line segments

4. ac > ef                                         4. by equality property.

hello, it is c -4

hope i you have a good day