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.

nothjing

step-by-step explanation:

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation:

the correct answer option is d. la = πrs.

step-by-step explanation:

we know that for a right cone, the lateral surface area is given by:

lateral area of right cone =

here the value under the square root is equal to slant height so replacing this value by in the formula to get:

lateral area of right cone =

therefore, the correct answer option is d.

b: y = 2x + 2

step-by-step explanation:

y = 2x + 1

we want a line that is parallel, so it will have the same slope

the line is in the form y= mx+b, so the slope is 2

we have a slope and the y intercept (0,2)

since x is 0 we know the y point which is the y intercept

we can use y= mx+b to write the equation of the line

m is the slope and b is the y intercept

y = 2x +2