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3 years ago

W(n)=n-1; find w-(4)

Mathematics

1 answer:

loris [4]3 years ago

6 0

Answer:5

Step-by-step explanation:

w(n)=4-1

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Identify the graph of 3x^2+y^2=9 for T(-1,3) and write an equation of the translated or rotated graph in general form.

Archy [21]

ANSWER

D. Ellipse;

$3{x}^{2} +{y}^{2} + 6x - 6y + 3= 0$

EXPLANATION

The given equation is

$3 {x}^{2} + {y}^{2} = 9$

Dividing through by 9 gives

$\frac{ {x}^{2} }{ 3} + \frac{ {y}^{2} }{9} = 1$

This is the equation of an ellipse centered at the origin.

If this ellipse has been translated, so that its center is now at (-1,3), then the equation of the translated ellipse becomes

$\frac{ {(x + 1) }^{2} }{ 3} + \frac{ {(y - 3)}^{2} }{9} = 1$

We multiply through by 9 to get,

$3 {(x + 1)}^{2} + {(y - 3)}^{2} = 9$

Expand to obtain;

$3( {x}^{2} + 2x + 1) + {y}^{2} - 6y + 9 = 9$

Expand to obtain;

$3{x}^{2} + 6x + 3+ {y}^{2} - 6y + 9 = 9$

Regroup and equate to zero to obtain;

$3{x}^{2} +{y}^{2} + 6x - 6y + 3= 0$

8 0

2 years ago

Read 2 more answers

PROBLEMAS QUE IMPLICAN UNA FUNCION DE 1ER GRADO

WINSTONCH [101]

anse is wter ixe 6 e8vd d Step-by-step explanation:

3 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 :

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

The modified DE,

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

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 :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

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

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

5 0

3 years ago

42=3(2-3h) can someone plz answer and show me how to do this

maria [59]

You have to use distributive property.

4 0

2 years ago

John is at a local bait shop; he wants to buy bait for his fishing trip. At the store, they are selling live bait for $12 a poun

galina1969 [7]

Answer:

$x\geq3$

$12x+7y\leq63$

$y\geq0$

Step-by-step explanation:

If we let x be the amount of live bait and y be the amount of natural bait, Then we can come up with the following inequalities;

We are told that John would like to get at least 3 pounds of live bait. At least 3 means 3 or more. Since x represents the amount of live bait, we have;

$x\geq3$

Moreover,we are informed that;

The store sells live bait for $12 a pound and natural bait for $7 a pound. x pounds of live bait would cost 12x while y pounds of natural bait would cost 7y. The total cost would thus be;

12x + 7y

but John only has a budget of $63. This implies that he can spend $63 at most, thus;

$12x+7y\leq63$

Finally we can have our last inequality as;

$y\geq0$

8 0

3 years ago