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3 years ago

Need help with solving a 2x2 system of linear equations PLEASE HELP ME

Mathematics

1 answer:

Yakvenalex [24]3 years ago

7 0

$\bf \begin{array}{llll}
-5x-y=5\\
-5x+y=5\\
-------\\
-10x+0=10
\end{array}\implies -10x=10\implies x=-1$

so if x = -1, let's plug that in the second equation

-5(-1)+y = 5
5+y = 5
y = 0

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In the right triangle shown, <A= 30° and BC =6

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Answer:

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Step-by-step explanation:

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2 years ago

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2 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

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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Centripetal acceleration is given by:
g=v^2/r
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3 years ago