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ExtremeBDS [4]
1 year ago
11

Which measure is of an angle that is coterminal with a 425° angle?

Mathematics
2 answers:
olga55 [171]1 year ago
7 0

Answer:

B. 425° – (840n)°, for any integer n

velikii [3]1 year ago
3 0

The measure of angle that is coterminal with 425° is 425° + (1440n)°

<h3>What are coterminal angles?</h3>

Coterminal angles are angles in multiples of the angles in standard positions examples(360°).

It is mostly gotten by adding  360° or multiples of 360 to the angle.

One of the unique behavior of coterminal angles is that their sine, cosine and tangent are equal.

Analysis:

From the options, the only one with an addition of 360 or multiples of 360 is 425 + 1440n

If we put n = 1, the coterminal angle gotten is 1865.

if we find the sine of 425°, we get 0.9063 which is same as the sine of 1865°.

In conclusion, the coterminal angle to 425° is 425° + (1440n)°.

Learn more about coterminal angle: brainly.com/question/23093580

#SPJ1

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2 years ago
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In the recent year 18.2% of all registered doctors were females. If there were 57,000 females registered doctors that year what
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Your equation is

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2 years ago
Wally Beige has painted a rectangular mural that is 7 ft tall and 11 ft wide. He plans to paint a border of equal width all the
4vir4ik [10]

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Width = 4ft

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Area of the rectangular mural = Length × Breadth

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3 years ago
Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g
AysviL [449]

Answer:

The quantity of salt at time t is m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} }), where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}

Where:

c - The salt concentration in the tank, as well at the exit of the tank, measured in \frac{pd}{gal}.

\frac{dc}{dt} - Concentration rate of change in the tank, measured in \frac{pd}{min}.

V - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

V \cdot \frac{dc}{dt} + 6\cdot c = 3

60\cdot \frac{dc}{dt}  + 6\cdot c = 3

\frac{dc}{dt} + \frac{1}{10}\cdot c = 3

This equation is solved as follows:

e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }

\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }

e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt

e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C

c = 30 + C\cdot e^{-\frac{t}{10} }

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c_{o} = \frac{10\,pd}{60\,gal}

c_{o} = 0.167\,\frac{pd}{gal}

Now, the integration constant is:

0.167 = 30 + C

C = -29.833

The solution of the differential equation is:

c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }

Now, the quantity of salt at time t is:

m_{salt} = V_{tank}\cdot c(t)

m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })

Where t is measured in minutes.

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