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anyanavicka [17]

2 years ago

15

How do you round 9.246 to the nearest tenth?

1 answer:

Flauer [41]2 years ago

3 0

You look at the hundreths place (4) and if it is 5 or more you round the tenths place up. so the number is rounded to 9.2

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Mathematics help needed. <br><br> Which of the following statements are true regarding functions?

Evgen [1.6K]

A and c. No horizontal line

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2 years ago

Use a standard inch ruler to answer this question

Elina [12.6K]

Huh?

You forgot something.

8 0

2 years ago

A Ladder that is 32 ft long leans against a building. The angle of elevation of the ladder is 70 degrees. To the nearest tenth o

kolezko [41]

See the picture attached to better understand the problem

we know that
in the right triangle ABC
sin 70°=opposite side angle 70°/hypotenuse

in this problem
opposite side angle 70°=AB
hypotenuse=AC----> 32 ft
sin 70°=AB/32------> AB=32*sin 70°-----> AB=30.07 ----> AB=30.1 ft

4 0

2 years ago

Read 2 more answers

Ashley ate Mexican food 4 out of 7 days on her vacation. If her vacation was 35 days, how many days did she eat Mexican food?

lina2011 [118]

The number of days Ashley ate Mexican food while on vacation is 20 days.

<h3>How many days did Ashley eat mexican food?</h3>

The first step is to determine the fraction of the number of days Ashley atae Mexican food. In order to do this, divide 4 by 7

4/7.

The second step is to multiply 4/7 by 35

4/7 x 35= 20 days

To learn more about multiplication, please check: brainly.com/question/3385014

8 0

2 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 $0.127 \frac{cm}{min}$

5 0

3 years ago