Through: (1,5), perp. to y=-1/10x + 3 pls help

1 answer:

8 0

First to get the equation you knew to understand one thing about perpendicular lines. The slope of the line is the opposite reciprocal of the perpendicular lines or the new slope is m = 10.

Then you use the formula
y = mx + b
you plug in your values from the point and the new slope.
(1,5) with new slope m
5= 10(1)+b
5-10=b
-5 = b
then make your new equation
y = 10x -5
that's your line that goes through point (1,5) and is perpendicular to the line given

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Answer:

Differential equation

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Step-by-step explanation:

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$\frac{dy}{dt} =ky(1-y)$

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$\frac{dy}{y(1-y)}=k\cdot dt \\\\\int \frac{dx}{y(1-y)} =k \int dt \\\\-ln(1-\frac{1}{y} )+C_0=kt\\\\1-\frac{1}{y} =Ce^{-kt}\\\\\frac{1}{y} =1-Ce^{-kt}\\\\y=\frac{1}{1-Ce^{-kt}}$

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$y(0)=0.2\\y(3)=0.4\\\\y(0)=0.2=\frac{1}{1-Ce^0}\\\\1-C=1/0.2\\\\C=1-1/0.2= -4\\\\\\y(3)=0.4=\frac{1}{1+4e^{-3k}} \\\\1+4e^{-3k}=1/0.4\\\\e^{-3k}=(2.5-1)/4=0.375\\\\k=ln(0.375)/(-3)=0.327\\\\\\y=\frac{1}{1+4e^{-0.327t}}$