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Blizzard [7]
3 years ago
12

Solve the following system of equations x + y = 3 x + 2y = 6

Mathematics
1 answer:
lyudmila [28]3 years ago
7 0

In this question, you're solving the systems of equations to find what x and y equal to.

Solve the systems of equations:

x + y = 3

x + 2y = 6

Lets make the "y" variable the same by multiplying the top equation by -2.

-2(x + y = 3)

x + 2y = 6

----------------

-2x -2y = -6

x + 2y = 6

Solve:

-x =0

x = 0

Now, we know x = 0, now plug 0 into x in one of the equations and solve:

We'll plug it into the x + y = 3 equation.

0 + y = 3

y = 3

When you're done solving, you should get:

x = 0 and y = 3

Answer:

x = 0

y = 3

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

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Given:

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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

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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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3 years ago
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If the top side is 2.5, and the bottom is 5, then you divide to find the scaled factor since they correlate to each other.

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GREYUIT [131]
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