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1 year ago

PLSSSSSSSSSSSSSSS HELP pls

Mathematics

1 answer:

sertanlavr [38]1 year ago

7 0

${\qquad\qquad\huge\underline{{\sf Answer}}}$

Let's solve ~

$\qquad \sf \dashrightarrow \: 10x + 30 = 90$

[ according to given figure ]

$\qquad \sf \dashrightarrow \: 10x = 90 - 30$

$\qquad \sf \dashrightarrow \: 10x = 60$

$\qquad \sf \dashrightarrow \: x = 60 \div 10$

$\qquad \sf \dashrightarrow \: x = 6 \degree$

Correct choice is D

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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

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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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