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Alexeev081 [22]
3 years ago
11

the length of a rectangle is 9 cm and the diagonal of the rectangle is 11 cm. Find the width of the rectangle​

Mathematics
1 answer:
ss7ja [257]3 years ago
7 0

Answer: The width  is the square root of 40 or 6.32

Step-by-step explanation:

9^2 + b^2 = 11^2  where b squared is the width  

81 + b^2 = 121

-81              -81

  b^2 = 40

b= 6.32

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Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat
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Answer:

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For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

\frac{dV}{dt}= \frac{100 L}{min}

Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}

And replacing we got:

\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}

And that represent 0.127 \frac{cm}{min}

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

V = \pi r^2 h

For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

\frac{dV}{dt}= \frac{100 L}{min}

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}

And solving for \frac{dh}{dt} we got:

\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}

We need to convert the rate given into m^3/min and we got:

Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}

And replacing we got:

\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}

And that represent 0.127 \frac{cm}{min}

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4 years ago
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<u>Step-by-step explanation:</u>

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Tristan is building steps up to a deck that is 156 5 6 yards above the ground. The steps must all be the same height. If he want
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Answer:

The height of each step is 1957 yard

Step-by-step explanation:

we are given

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now, we can use formula

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now, we can plug these values

and we get

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