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2 years ago

(3x^2 + 2x -2) + ( -2x^2 + 5x+5) My answer 5x^2 + 7× +7 Am i right

Mathematics

1 answer:

sertanlavr [38]2 years ago

4 0

$(3x^2 + 2x -2) + ( -2x^2 + 5x+5)=\\
3x^2 + 2x -2 -2x^2 + 5x+5=\\
x^2+7x+3$

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Find the distance between the pair of coordinates: (-4, -3) and (-8, 7). A 10.77 B 3.74 C 04 D 12.65 Question

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2 years ago

Read 2 more answers

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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2 years ago

Help! Please! !!!!!!!!!!!!

asambeis [7]

Answer:

Step-by-step explanation: what is your question

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2 years ago