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3 years ago

7

Mercury and venus are the sun than earth is.

2 answers:

Leno4ka [110]3 years ago

3 0

closer to, shorter, longer. (T=A^1.5

just took the assigniment.

damaskus [11]3 years ago

3 0

Answer:

So, their orbital periods are than Earth's orbit. The further a planet is from the sun, the its orbit is.

Step-by-step explanation:

Even though Venus is farther from the sun than mercury, Venus's surface is hotter than Mercury's because Venus has a carbon dioxide atmosphere that traps the sun's heat.

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Would the answer be 12.5

matrenka [14]

Yes hope this helps man!!

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3 years ago

Hai công ty A và B cùng kinh doanh một mặt hàng. Xác suất để công ty A thua lỗ là 0,3; xác suất để công ty B thua lỗ là 0,2 và x

DENIUS [597]

Answer:

Ask in English then I can help u

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3 years ago

The median of 50 50 54 56 60 62 63 64 79 72 72

miss Akunina [59]

Answer:

62

Step-by-step explanation:

50,50,54,56,60,62,63,64,79,72,72,

median ids the number at the center

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3 0

3 years ago

How do i do rise over run if the slope is not a fraction

Furkat [3]

Ok for example let’s use 2 as the slope. You are going to put the 2 over 1 so it will look like this 2/1. And that would be your slope!

7 0

3 years ago

Read 2 more answers

Consider the following initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1 Let ∂f ∂x = (x + y)2 = x2 + 2xy + y2

IRISSAK [1]

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

Suppose the ODE has a solution of the form $F(x,y)=C$ , with total differential

$\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

This ODE is exact if the mixed partial derivatives are equal, i.e.

$\dfrac{\partial^2F}{\partial y\partial x}=\dfrac{\partial^2F}{\partial x\partial y}$

We have

$\dfrac{\partial F}{\partial x}=(x+y)^2\implies\dfrac{\partial^2F}{\partial y\partial x}=2(x+y)$

$\dfrac{\partial F}{\partial y}=2xy+x^2-2\implies\dfrac{\partial^2F}{\partial x\partial y}=2y+2x=2(x+y)$

so the ODE is indeed exact.

Integrating both sides of

$\dfrac{\partial F}{\partial x}=(x+y)^2$

with respect to $x$ gives

$F(x,y)=\dfrac{(x+y)^3}3+g(y)$

Differentiating both sides with respect to $y$ gives

$\dfrac{\partial F}{\partial y}=2xy+x^2-2=(x+y)^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies x^2+2xy-2=x^2+2xy+y^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=-y^2-2$

$\implies g(y)=-\dfrac{y^3}3-2y+C$

$\implies F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y+C$

so the general solution to the ODE is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=C$

Given that $y(1)=1$ , we find

$\dfrac{(1+1)^3}3-\dfrac{1^3}3-2=C\implies C=\dfrac13$

so that the solution to the IVP is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=\dfrac13$

$\implies\boxed{(x+y)^3-y^3-6y=1}$

5 0

3 years ago