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SpyIntel [72]
3 years ago
13

What is the length of a rectangle that has a width of 2.5 centimeters and an area of 20.5 square centimeters?

Mathematics
2 answers:
olchik [2.2K]3 years ago
7 0
Do 20.5 divided by 2.5 which equals
8.2
ira [324]3 years ago
6 0
For a rectangle,

area = length * width, so

length = area/width

length = (20.5 cm^2)/(2.5 cm) = 8.2 cm
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Which of the following are solutions to the equation below 3x^2+27x+54=0
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Answer:x=6

Step-by-step explanation:Step-1 : Multiply the coefficient of the first term by the constant 1 • 18 = 18

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Step-4 : Add up the first 2 terms, pulling out like factors :

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Step-5 : Add up the four terms of step 4 :

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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
nirvana33 [79]

Answer:

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}

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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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3 years ago
What is the rational number equivalent to 3.12​
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