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Cerrena [4.2K]
3 years ago
10

Line XY bisects and is perpendicular to AB and CD. CD = 10 and XY = 24. What is the length of AP?

Mathematics
1 answer:
myrzilka [38]3 years ago
8 0

Answer:

<u>AP = 13</u>

Step-by-step explanation:

See the attached figure.

CD = 10 and XY = 24

Line XY bisects and is perpendicular to AB and CD

∴ XY bisects at Y

∴ CY = 0.5 CD = 0.5 * 10 = 5

And from the geometry of the shape

P is the midpoint of XY

∴ XP = YP = 0.5 XY = 0.5 * 24 = 12

XY is perpendicular to CD

∴ ΔPYC is a right triangle at Y

∴ PC (hypotenuse) = \sqrt{PY^2+YC^2} =\sqrt{12^2+5^2}=\sqrt{144+25} =\sqrt{169} =13

From the figure AP = PC = 13

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A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo
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a) y(t)=50000-49990e^{\frac{-2t}{25}}

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Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is \frac{dy}{dt}=R_{i}-R_{o}, where R_{i} is the rate of salt entering and R_{o} is the rate of salt going outside.

Then we have, R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}, and

R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}

So we obtain.  \frac{dy}{dt}=4000-\frac{2y}{25}, then

\frac{dy}{dt}+\frac{2y}{25}=4000, and using the integrating factor e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}, therefore  (\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}, we get   \frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}, after integrating both sides y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C, therefore y(t)= 50000 +Ce^{\frac{-2t}{25}}, to find C we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions y(0)=10, so 10= 50000+Ce^{\frac{-0*2}{25}}

10=50000+C\\C=10-50000=-49990

Finally we can write an expression for the amount of salt in the tank at any time t, it is y(t)=50000-49990e^{\frac{-2t}{25}}

b) The tank will overflow due Rin>Rout, at a rate of 80 L/min-40L/min=40L/min, due we have 500 L to overflow \frac{500L}{40L/min} =\frac{25}{2} min=t, so we can evualuate the expression of a) y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7, is the salt concentration when the tank overflows

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