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3 years ago

10

Determine the solution to the equation below.

1 answer:

mezya [45]3 years ago

8 0

Answer:

excuses me I don't have point to ask questions so I write a thing instead of answering ❤

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Use Kelly’s steps to simplify this expression.

pantera1 [17]

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uranmaximum [27]

Given the differential equation

$5t^3 \frac{dx}{dt} +2x-2=0$

The solution is as follows:

$5t^3 \frac{dx}{dt} +2x-2=0 \\ \\ \Rightarrow5t^3 \frac{dx}{dt} =2-2x \\ \\ \Rightarrow \frac{5}{2-2x} dx= \frac{1}{t^3} dt \\ \\ \Rightarrow \int {\frac{5}{2-2x} } \, dx = \int {\frac{1}{t^3}} \, dt \\ \\ \Rightarrow- \frac{5}{2} \ln(2-2x)=- \frac{1}{2t^2} +A \\ \\ \Rightarrow\ln(2-2x)= \frac{1}{5t^2} +B\\ \\ \Rightarrow2-2x=Ce^{\frac{1}{5t^2}} \\ \\ \Rightarrow 2x=Ce^{\frac{1}{5t^2}}+2 \\ \\ \Rightarrow x=De^{\frac{1}{5t^2}}+1$

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3 years ago

The angle of elevation from a buoy in the water to the top of a lighthouse is 68degrees. If a buoy is 300 ft from the base of th

Katen [24]

Answer:

The height of the lighthouse is $742.5\ ft$

Step-by-step explanation:

Let

h -----> the height of the lighthouse

we know that

The tangent of angle of 68 degrees is equal to divide the height of the lighthouse by the horizontal distance from the buoy to the base of lighthouse

so

$tan(68\°)=\frac{h}{300}$

Solve for h

$h=(300)tan(68\°)=742.5\ ft$

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