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3 years ago

The elevation of a sunken ship is -120 feet. Your elevation is 5/8 of the ship's elevation. What is your elevation?

1 answer:

7 0

The question is basically asking <em>"What is 5/8 times -120?"</em>
<em>
</em>Well, 1/8 of -120 is going to be equivalent to -120 ÷ 8.
-120 ÷ 8 = -15.

5/8 of -120 is just going to be 5 eighths.
-15 × 5 = -75 feet

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CaHeK987 [17]

Answer:

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Step-by-step explanation:

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Then, we multiply 0.2 by 100 to get the percentage:

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3 years ago

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3 years ago

Read 2 more answers

A tank has the shape of a surface generated by revolving the parabolic segment y = x2 for 0 ≤ x ≤ 3 about the y-axis (measuremen

Darina [25.2K]

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,

$H=3^2=9\ ft$

So, the height of top of tank is, $y(3)=H=9\ ft$

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, $r=\sqrt y$

Now, area of cross section is, $A(y)=\pi (\sqrt y)^2$

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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2 years ago

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jolli1 [7]

Answer: Ian would have drove 91 miles using only 3 gallons of gas.

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2 years ago

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dimulka [17.4K]

$\\ \sf\longmapsto \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$