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3 years ago

Divide use partial quotients 6 divided by 258

Mathematics

1 answer:

Bond [772]3 years ago

5 0

43 I think...................

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Please help Please help

bekas [8.4K]

The answer is A. I looked it up for you on an surface area calculator.

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3 years ago

In a class of 30 students, 6 have a brother and 12 have a sister. There are 14 students

Elena-2011 [213]

30 students

14 have no siblings

that leaves 16 with siblings

6 have brothers and 12 have sisters which = 18 so 2 of the students have both a brother and a sister

2/30 students have a brother and a sister

1/15 chance

3 0

3 years ago

Safety rent a car rents an intermediate size car at a daily rate of $21.95 plus $.19 per mile. City rentals rents an intermediat

Katen [24]

So you can use

$21.95 + $.19x = $18.95 + $.21x

7 0

3 years ago

Usain Bolt ran the 2012 Olympic 100-meter race in 9.63 seconds. If he runs at this rate on a road with a speed limit of 25 miles

djverab [1.8K]

Answer:

3874.5342

Step-by-step explanation:

9.63 divide by 100 = 0.0963

0.0963x40 234= 3874.5342

6 0

3 years ago

Read 2 more answers

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