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Anika [276]
3 years ago
15

What is the area of a rectangle that is 4 x 4 1/2 ?

Mathematics
2 answers:
Yakvenalex [24]3 years ago
6 0
The area of the rectangle is 18
Klio2033 [76]3 years ago
5 0
4 x 4 1/2 = 8 4/8 = 8 2/4 = 8 1/2
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(X^2+y^2+x)dx+xydy=0<br> Solve for general solution
aksik [14]

Check if the equation is exact, which happens for ODEs of the form

M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0

if \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}.

We have

M(x,y)=x^2+y^2+x\implies\dfrac{\partial M}{\partial y}=2y

N(x,y)=xy\implies\dfrac{\partial N}{\partial x}=y

so the ODE is not quite exact, but we can find an integrating factor \mu(x,y) so that

\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0

<em>is</em> exact, which would require

\dfrac{\partial(\mu M)}{\partial y}=\dfrac{\partial(\mu N)}{\partial x}\implies \dfrac{\partial\mu}{\partial y}M+\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}

\implies\mu\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=M\dfrac{\partial\mu}{\partial y}-N\dfrac{\partial\mu}{\partial x}

Notice that

\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=y-2y=-y

is independent of <em>x</em>, and dividing this by N(x,y)=xy gives an expression independent of <em>y</em>. If we assume \mu=\mu(x) is a function of <em>x</em> alone, then \frac{\partial\mu}{\partial y}=0, and the partial differential equation above gives

-\mu y=-xy\dfrac{\mathrm d\mu}{\mathrm dx}

which is separable and we can solve for \mu easily.

-\mu=-x\dfrac{\mathrm d\mu}{\mathrm dx}

\dfrac{\mathrm d\mu}\mu=\dfrac{\mathrm dx}x

\ln|\mu|=\ln|x|

\implies \mu=x

So, multiply the original ODE by <em>x</em> on both sides:

(x^3+xy^2+x^2)\,\mathrm dx+x^2y\,\mathrm dy=0

Now

\dfrac{\partial(x^3+xy^2+x^2)}{\partial y}=2xy

\dfrac{\partial(x^2y)}{\partial x}=2xy

so the modified ODE is exact.

Now we look for a solution of the form F(x,y)=C, with differential

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

The solution <em>F</em> satisfies

\dfrac{\partial F}{\partial x}=x^3+xy^2+x^2

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

Integrating both sides of the first equation with respect to <em>x</em> gives

F(x,y)=\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+f(y)

Differentiating both sides with respect to <em>y</em> gives

\dfrac{\partial F}{\partial y}=x^2y+\dfrac{\mathrm df}{\mathrm dy}=x^2y

\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C

So the solution to the ODE is

F(x,y)=C\iff \dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+C=C

\implies\boxed{\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3=C}

5 0
3 years ago
3x + 5x + 8y Simplified
sattari [20]

Answer:

8x + 8y

Step-by-step explanation:

Simplify


Hope this helps. Have a great day. :)

-wfz

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3 years ago
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X squared plus x y - 12 y squared
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(x-4y) (x+3y)
Grouping :)
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7/6 divided by 2 in simplest form
iris [78.8K]
Here is your question: 7/6 divided by 2.

= 7/6÷2 = 7/6 * 1/2 =.

Multiply and you get = 7/12.
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