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3 years ago

11

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1 answer:

nevsk [136]3 years ago

8 0

Slope =2/3
70,80) (10,40)
(80-40)/(70-10) = 40/60
4/6
2/3

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Can anyone help me with this?

Artemon [7]

7) 5+4 = 9

4+5=9

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5+6 = 11

9) 6+7 = 13

7+6 = 13

10) 7+8 = 15

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11) 8+9 = 17

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Which statement describes a positive effect of selective breeding?

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B, because people do selective breeding to get something they want out of it. Which decreases genetic diversity, genetic diversity meaning there being less and less traits and characteristics and we wont something perfect not exactly something with traits that we dont like. However dont get me wrong, nothings perfect. But yeah.

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2 years ago

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the perimeter of a rectangle is 15x+17y if the length is 7/2x +7y then find the width of the rectangle

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2 years ago

Solve the equation. Check your solution. −4x − 8x + 17 = 23

Novay_Z [31]

-4x -8x +17 = 23
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3 years ago

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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,

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

7 0

3 years ago