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marshall27 [118]
3 years ago
9

Jacob is tiling his square walk with square tiles. He is able to fit 5 tiles up and 5 tiles across. If each tile has a length of

3b, what is the perimeter of the wall?
Mathematics
1 answer:
frozen [14]3 years ago
8 0

Answer:

The perimeter of the kitchen wall = 60 b

Step-by-step explanation:

The number if tiles used in the length side of the square walk = 5 tiles

The length of 1 tiles  = 3 b

So, the length of 5 tiles = 5 x ( Length of 1 tile)  

= 5 x (3 b)  =  15 b

So, the length of the square walk = 1 5 b

Perimeter of SQUARE  = 4 x SIDES

⇒ Perimeter of the wall  = 4 x (Side)

                                        = 4 x (15 b)  = 60 b

Hence, the perimeter of the kitchen wall = 60 b

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

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We have

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

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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}

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Notice that

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

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

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

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

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

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