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allochka39001 [22]

3 years ago

5

Fencing costs $10 per foot. Frances had a rectangular garden that is 4 feet wide and 5 feet long. How much money will Frances ne

ed to spend to put fence around the garden?

1 answer:

kicyunya [14]3 years ago

4 0

Answer:

$180

Step-by-step explanation:

1. Find the perimeter: 2w + 2L

2(4) = 8 + 2(5) = 10

8 + 10 = 18 feet of fencing

2. multiply the cost per foot (10) by the number of feet (18)

10*18 = 180

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Answer: Option 'B' and 'E' are correct.

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(X^2+y^2+x)dx+xydy=0<br> Solve for general solution

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

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

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

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

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

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

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

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

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

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

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

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So the solution to the ODE is

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

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Step-by-step explanation:

Median means middle number

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3 years ago

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