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Natalka [10]
3 years ago
9

What percent of 48 is 54

Mathematics
1 answer:
lisov135 [29]3 years ago
4 0
\begin{array}{ccc}48&-&100\%\\54&-&p\%\end{array}\ \ \ |cross\ multiply\\\\48p=54\cdot100\\48p=5400\ \ \ \ |divide\ both\ sides\ by\ 48\\\boxed{p=112.5}

Answer: 54 is 112.5% of 48.
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D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x
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The solution depends on the value of k. To make things simple, assume k>0. The homogeneous part of the equation is

\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0

and has characteristic equation

r^2-16k=0\implies r=\pm4\sqrt k

which admits the characteristic solution y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}.

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be y_p=ae^{4x}+be^x. Then

\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x

So you have

16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x
(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x

This means

16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}
b(1-16k)=30\implies b=\dfrac{30}{1-16k}

and so the general solution would be

y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x
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Step-by-step explanation:

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4x=251
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