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2 years ago

Find the distance between the points given. (0.5) and (-5, 0)

Mathematics

1 answer:

Andrew [12]2 years ago

8 0

Answer:

I cant help unless i have workable units (0.5) is not a point

but im guessing that you were trying to say (0,5) in which case its

7.07106781187

its a triangle

5 on one side

5 on the other

solve for the hypotenuse

hypotenuse = $\sqrt{50}$

hypotenuse = 7.07106781187

I hope this helps!

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Find the function <img src="https://tex.z-dn.net/?f=%20y_%7B1%7D%20%28t%29" id="TexFormula1" title=" y_{1} (t)" alt=" y_{1} (t)"

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Characteristic equation:

$64r^2+48r+8=0$
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This gives a general solution of

$y_1=C_1e^{-1/2t}+C_2e^{-1/4t}$

With the given initial conditions, you get

$y_1(0)=1\implies 1=C_1+C_2$
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so that the first particular solution is

$y_1(t)=-e^{-1/2t}+2e^{-1/4t}$

With the second set of initial conditions, you would have

$y_2(0)=0\implies 0=C_1+C_2$
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so that the second particular solution is

$y_2(t)=-4e^{-1/2t}+4e^{-1/4t}$

The Wronskian of the two solution is

$W(y_1,y_2)=\begin{vmatrix}y_1&y_2\\{y_1}'&{y_2}'\end{vmatrix}=\begin{vmatrix}-e^{-1/2t}+2e^{-1/4t}&-4e^{-1/2t}+4e^{-1/4t}\\-\frac12e^{-1/2t}-\frac12e^{-1/4t}&2e^{-1/2t}-e^{-1/4t}\end{vmatrix}$
$W(y_1,y_2)=e^{-3/4t}$

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