Vertex F of ΔDEF has coordinates (−3, 7). If ΔDEF is reflected over the y-axis, what are the coordinates of vertex Fʹ?

Mathematics

1 answer:

docker41 [41]3 years ago

5 0

The reflection will be on the other side of the y-axis.
Its coordinates will be (3, 7).

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Include work pleaseeeee

Angelina_Jolie [31]

Okay so to represent juice we are going to use X, and to represent water we are going to use Y.

We also know that the first two starting equations are:

6x + y = 135

4x + 2y = 110

We can re-arrange the first equation so that it equals y (for now), so it is going to end up looking like this:

y = -6x +135

Now you can take that equation and plug into either one of the starting two equations. I chose the second equation. We just substitute what y equals in for y in the equation, so we have:

4x + 2(135 - 6x) = 110

Now solve

4x + 270 -12x = 110

-8x + 270 = 110

Subtract 270 from both sides

-8x = -160

Now divide by -8 on both sides

x = 20

We can now confirm that juice costs $20

Now lets plug that into the equation where we solved for y, to get the actual value of y.

y = 135 - 6(20)

y = 135 - 120

y = 15

The price of water costs $15

From this we can conclude that the cost of juice is $20 and the price of water is $15

5 0

2 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

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

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

*is* exact. If this modified DE is exact, then

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

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 :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$