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disa [49]
3 years ago
12

6(y+1.5)=18 what does y equals?

Mathematics
2 answers:
Mashcka [7]3 years ago
5 0

Answer:

y = 1.5

Step-by-step explanation:

1) Divide both sides by 6.

y + 1.5 =  \frac{18}{6}

2) Simplify 18/6 to 3.

y + 1.5 = 3

3) Subtract 1.5 from both sides.

y = 3 - 1.5

4) Simplify 3 - 1.5 to 1.5.

y = 1.5

Therefor, the answer is, y = 1.5.

leva [86]3 years ago
3 0

Answer:

Step-by-step explanation:

6y+9=18

6y=2

y=1/3

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3 years ago
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}

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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3 years ago
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