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3 years ago

What is the degree measure of an arc 4(pi) ft. long in a circle of radius 10 ft.?

2 answers:

3 0

Hey there!

$20*pi/360=4*pi/x solve for x x=360*4*pi/(20*pi)=72 degrees$

Your correct answer would be. . .

$\boxed{\boxed{72}}$

Hope this helps.
~Jurgen

Aloiza [94]3 years ago

3 0

Answer:

Hey there!

20*pi/360=4*pi/x solve for x x=360*4*pi/(20*pi)=72 degrees

Your correct answer would be. . .

Hope this helps.

Step-by-step explanation:

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Answer:

22.5

Step-by-step explanation:

If you divide y/x each ticket costs $2.5, so multiply 2.5x9 and that's your answer. :)

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2 years ago

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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 $0.127 \frac{cm}{min}$

5 0

3 years ago

Total Points: 50

Sveta_85 [38]

Answer:

1. slope

2.Positive

3.up and down

4. Left and right

5. 1/1 or over 1 up 1

Step-by-step explanation:

7 0

3 years ago

If we wanted to prove quadrilateral MNOP was a parallelogram, which method below would not work?

rjkz [21]

Answer:

The correct option is 4.

4) Doing two distance formulas to show that adjacent sides are not the same length.

Step-by-step explanation:

Parallelogram is a quadrilateral which has opposite sides equals and parallel. Example of a parallelogram are rhombus, rectangle, square etc.

We can prove that a quadrilateral MNOP is a parallelogram. If we find the slopes of all four sides and compare those of the opposite ends, same slopes would indicate the opposite sides are parallel, hence the quarilateral is a parallelogram. We can also find the distance of two opposing sides, and slopes of twp opposing sides to determine whether it is a parallelogram or not. The most difficult approach is that diagonals bisect each other at same point.

However, using only two distance formulas will not give us enough information to determine whether a side is parallel or not.

7 0

3 years ago