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motikmotik
3 years ago
10

An airplane descends 2.8 miles to an elevation of 5.85 miles. Find the elevation of the plane before an airplane descends 2.2 mi

les to an elevation of 5.85 miles. Find the elevation of the plane before its descent.
descent.
Mathematics
2 answers:
Natali5045456 [20]3 years ago
7 0

Answer:

8.05 miles

Step-by-step explanation:

If it descended 2.2 miles to 5.85 miles then that means it was at 5.85 + 2.2 miles before its descent.

5.85 + 2.2 = 8.05

Advocard [28]3 years ago
4 0

Answer:

8.05 miles

Step-by-step explanation:

5.85 + 2.2 = 8.05

5.85 + 2.8 = 8.65 (not the answer, but as an example ;b

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Answer:

100\pi\int\limits^9_0 {(\sqrt y)^2(14-y)} \, dy ft-lbs.

Step-by-step explanation:

Given:

The shape of the tank is obtained by revolving y=x^2 about y axis in the interval 0\leq x\leq 3.

Density of the fluid in the tank, D=100\ lbs/ft^3

Let the initial height of the fluid be 'y' feet from the bottom.

The bottom of the tank is, y(0)=0^2=0

Now, the height has to be raised to a height 5 feet above the top of the tank.

The height of top of the tank is obtained by plugging in x=3 in the parabolic equation . This gives,

H=3^2=9\ ft

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Now, 5 ft above 'H' means H+5=9+5=14

Therefore, the increase in height of the top surface of the fluid in the tank is given as:

\Delta y=(14-y) ft

Now, area of cross section of the tank is given as:

A(y)=\pi r^2\\r\to radius\ of\ the\ cross\ section

Radius is the distance of a point on the parabola from the y axis. This is nothing but the x-coordinate of the point.

We have, y=x^2

So, x=\sqrt y

Therefore, radius, r=\sqrt y

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Work done in pumping the contents to 5 feet above is given as:

W=D\int\limits^{y(3)}_{y(0)} {A(y)(\Delta y)} \, dy

Plug in all the values. This gives,

W=100\int\limits^9_0 {\pi (\sqrt y)^2(14-y)} \, dy\\\\W=100\pi\int\limits^9_0 { (\sqrt y)^2(14-y)} \, dy\textrm{ ft-lbs}

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Step-by-step explanation:

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