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Stolb23 [73]
3 years ago
12

What is the approximate distance between the points (-3,-4) and (-8,1) on a coordinate grid?

Mathematics
1 answer:
VladimirAG [237]3 years ago
8 0

Answer:

5\sqrt{2}

Step-by-step explanation:

The distance between two points on a Cartesian coordinate grid is d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2 where d is the distance between (x_1,y_1)\rightarrow(-3,-4) and (x_2,y_2)\rightarrow(-8,1):

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\d=\sqrt{(-8-(-3))^2+(1-(-4))^2}\\\\d=\sqrt{(-8+3)^2+(1+4)^2}\\\\d=\sqrt{(-5)^2+5^2}\\\\d=\sqrt{25+25}\\\\d=\sqrt{50}\\\\d=5\sqrt{2}

Therefore, the distance between the two points is 5\sqrt{2} units.

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Hello There!

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The corresponding homogeneous ODE has characteristic equation r^2-r+1=0 with roots at r=\dfrac{1\pm\sqrt3}2, thus admitting the characteristic solution

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For the particular solution, assume one of the form

y_p=a\sin x+b\cos x

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(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x

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Then the general solution to this ODE is

\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}

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\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2

\implies y_c=C_1e^x+C_2e^{2x}

Assume a solution of the form

y_p=e^x(a\sin x+b\cos x)

{y_p}'=e^x((a+b)\cos x+(a-b)\sin x)

{y_p}''=2e^x(a\cos x-b\sin x)

Substituting into the ODE gives

2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x

-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x

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