Please help can give brainliest
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Weird. A period appears above this... huh.
Answer:
[Th]e cube has a greater value.
Step-by-step explanation:
What the word problem really wants us to get [is ]the question of 'Which is greater, A=6a^2 when 'a' [is] 18 or A=4r^2 when r = 9? And here's how to solve that.
Starting with the[ c]ube we have A=6a^2. A bit t[o]o simple, right?
A=6(18)^2 Substitute numbers.
A=6(324) Solve ex[p]onents.
A=1944 Mult[i]ply.
So w[e] know that the cube is 1944 meters cube[d ] in area. But what about the more [f]ormidable sphere? Fo[r] it we need a slightly m[o]re co[m]plicated formula, A=4r^2. However, instead of using the real pi I will be rounding to 3.14, since we have no calculator so anything more would take way too long and fry your[ bra]in.
A=4(3.14)(9)^2 Subst[i]tute numbers.
A=4(3.14)(81) Solve expone[n]ts.
A=12.56(81) Multip[ly].
A=1017.36 Multiply again[.]
Now, since I'm sure all of us can count, we know that 1944 is greater than 1017.36. Or in other words, the cube is bigger than the sphere.
And PLEASE don't copy this guys. Make your own iteration. Change it up!
Take the homogeneous part and find the roots to the characteristic equation:
This means the characteristic solution is
.
Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form
. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.
With
and 
, you're looking for a particular solution of the form 
. The functions 
satisfy
where
is the Wronskian determinant of the two characteristic solutions.
So you have
So you end up with a solution
but since
is already accounted for in the characteristic solution, the particular solution is then
so that the general solution is
This means the characteristic solution is
Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form
With
where
So you have
So you end up with a solution
but since
so that the general solution is