A) y=5x+5<br> B)y=x+5<br> C)y=5x+25<br> D)y=x+125

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 $0.127 \frac{cm}{min}$

5 0

3 years ago

Which relationship describes angles 1 and 2?

galina1969 [7]

Angle 1 and 2 are vertical angles

Hope this helps :)

8 0

3 years ago

Read 2 more answers

What is the value of x?

grandymaker [24]

i got 5.14cm i don´t know how much you are supposed to round though

36/42= 0.857142857 x 6=5.142857143

6 0

3 years ago

You can jog around your block twice and the park once in 10 minutes. You can jog around your block twice and the park 3 times in

exis [7]

2x + y = 10
2x + 3y = 22

Please consider marking this answer as Brainliest to help me advance, and let me know if you need any clarifications.

4 0

3 years ago

Qaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Kipish [7]

Answer:

32.8 miles

Step-by-step explanation:

Amy is driving to Seattle. Suppose that the remaining distance to drive (in miles) is a linear function of her driving time (in minutes). When graphed, the function gives a line with a slope of -0.95. See the figure below. Amy has 48 miles remaining after 31 minutes of driving. How many miles will be remaining after 47 minutes of driving?

Answer: The general equation of a line is given as y = mx + c, where m is the slope of the line and c is the intercept on the y axis. Given that the slope is -0.95, substituting in the general equation :

y = -0.95x + c

Amy has 48 miles remaining after 31 minutes of driving, to find c, we substitute y = 48 and x = 31. Therefore:

48 = -0.95(31) + c

c = 48 + 0.95(31)

c = 48 + 29.45

c = 77.45

The equation of the line is

y = -0.95x + 77.45

After 47 minutes of driving, the miles remaining can be gotten by substituting x = 47 and finding y.

y = -0.95(47) + 77.45

y = -44.65 + 77.45

y = 32.8 miles

3 0

3 years ago

A) y=5x+5 B)y=x+5 C)y=5x+25 D)y=x+125