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Neporo4naja [7]
2 years ago
14

Find the radian measure of 120 degrees.

Mathematics
1 answer:
Sedbober [7]2 years ago
7 0

Answer:

The radian :

\frac{2\pi}{3}

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solve the systems of linear equations by graphing y=-5/2x-7 x+2y=4 what is the solution to the system of linear equations A. (-4
Sati [7]
The appropriate choice is
  A. (-4.5, 4.25)

3 0
3 years ago
Read 2 more answers
F. How many miles could be driven with 18 gallons of<br> gas?
3241004551 [841]

Answer:

21.5 or 26.8

Step-by-step explanation:

a car with a 18 gallon gas tank can average 21.5 miles per gallon when driven in the stop and go traffic of the city and 26.8 miles per gallon when driven on the highway

8 0
3 years ago
an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a
viktelen [127]

Answer:

the rate of change of the water depth when the water depth is 10 ft is;  \mathbf{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft  is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the  rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at  a time t and r = radius of the cone shaped at a time t

Then the similar triangles  ΔOCD and ΔOAB is as follows:

\dfrac{h}{r}= \dfrac{20}{8}    ( similar triangle property)

\dfrac{h}{r}= \dfrac{5}{2}

\dfrac{h}{r}= 2.5

h = 2.5r

r = \dfrac{h}{2.5}

The volume of the water in the tank is represented by the equation:

V = \dfrac{1}{3} \pi r^2 h

V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h

V = \dfrac{1}{18.75} \pi \ h^3

The rate of change of the water depth  is :

\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

Since the water is drained  through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

\dfrac{dv}{dt}= - 4  \ ft^3/sec

Therefore,

-4 = \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

the rate of change of the water at depth h = 10 ft is:

-4 = \dfrac{ 100 \ \pi }{6.25}\  \dfrac{dh}{dt}

100 \pi \dfrac{dh}{dt}  = -4 \times 6.25

100  \pi \dfrac{dh}{dt}  = -25

\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi}

Thus, the rate of change of the water depth when the water depth is 10 ft is;  \mathtt{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

4 0
3 years ago
Help me this assignment is due soon.
AlekseyPX
So maybe it’s 5, it’s equivalent to the side with 6, just shorter
6 0
3 years ago
Find the area of the shaded region
Mrac [35]

Answer:

40.5

Step-by-step explanation:

Area of square is 9x9=81.

Area of triangle is 1/2(9)(9)=40.5

81-40.5=40.5

40.5 is the area of the shaded region.

4 0
2 years ago
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