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Sliva [168]
3 years ago
6

The SOLT missionaries are throwing a birthday party. There is obviously cake. Mr. Daniel only wants 1/2 of a piece. Mr. Richie e

ats half of Mr. Daniel's leftover. Ms. Laura eats half of the remaining piece. Ms. Kimberly eats half of Ms. Laura's leftover. Seven more missionaries keep sharing the piece of cake following the same pattern. If Ms. Sarah is the last of the seven, how big is the piece that Ms. Sarah eats?
Mathematics
1 answer:
marishachu [46]3 years ago
3 0

Answer:

1/ 112

Step-by-step explanation:

1-½ = ½ ( Mr. Daniel's leftover)

½×½= ¼ ( Mr. Richie)

½×¼= 1/8 ( Ms Laura)

½×1/8 = 1/16 ( Ms. Kimberly)

1/7 × 1/16 = 1/ 112 (the last turn)

so, Ms. Sarah eats 1/ 112 of a piece.

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(x^2y+e^x)dx-x^2dy=0
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It looks like the differential equation is

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

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\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x

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\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0

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\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

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

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\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}

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Then the general solution to the DE is

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

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