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2 years ago

The endpoints of DE and D(-3, y) and x, 6). The midpoint of DE is M(4,2). What is the length of DE.

Mathematics

1 answer:

Natasha_Volkova [10]2 years ago

3 0

If the endpoints of DE where D(-3, y) and E(x, 6) with the midpoint M(4, 2), the length of DE is 16.12 units

The formula for calculating the midpoint is expressed as;

$M(x,y)=(\frac{x_1+x_1}{2}, \frac{y_1+y_2}{2})$

For the x-coordinate points:

X = -3+x/2

4 * 2 = -3 + x

8 = -3 + x

x = 8 + 3

x = 11

Get the value of y;

2 = y+6/2

y+6 = 2 * 2

y + 6 = 4

y = 4 - 6

y = -2

Hence the coordinates of D and E are D(-3, -2) and E(11, 6)

$D=\sqrt{(6-(-2))^2+(11-(-3))^2}\\D=\sqrt{8^2+14^2}\\D=\sqrt{64 + 196}\\D= \sqrt{260}\\D= 16.12units$

Hence the length of DE is 16.12 units

Learn more here; brainly.com/question/22624745

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It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

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$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

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$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

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Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

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