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enot [183]
3 years ago
8

Circle D is shown. Line segments G D, F D, and E D are radii. Point H is on the other side of the circle. Angle G D F is 83 degr

ees and angle F D E is 66 degrees. What is the measure of Arc E F G?
Mathematics
2 answers:
VladimirAG [237]3 years ago
8 0

Answer:

149

Step-by-step explanation:

just got it right on the quiz

-BARSIC- [3]3 years ago
3 0

Answer:

149

Step-by-step explanation:

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Use the given angle measures to find the indicated
mina [271]

Answer:

x = -9°

m∠A = 37°

Step-by-step explanation:

(8-5x) + (10-3x) = 90

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18-8x = 90

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10-(-27) = 37°

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The Lin family bought a dozen bagels. They ate 1/4 of the bagels today and 5/12 of the bagels yesterday. What fraction of the ba
lina2011 [118]

Answer:

2/3

Step-by-step explanation:

To be able to find what the total fraction of bagels eaten, the fractions need to have common denominators. To do this, you multiply 1/4 by 3 in both the numerator and the denominator. Now that the two fractions have the same denominator, you can add the fraction's numerator to get the total fraction, 8/12. However, 8/12 can be simplified more, so you divide by 4 and end up with 2/3.

4 0
3 years ago
A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo
djverab [1.8K]

Answer:

a) y(t)=50000-49990e^{\frac{-2t}{25}}

b) 31690.7 g/L

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is \frac{dy}{dt}=R_{i}-R_{o}, where R_{i} is the rate of salt entering and R_{o} is the rate of salt going outside.

Then we have, R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}, and

R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}

So we obtain.  \frac{dy}{dt}=4000-\frac{2y}{25}, then

\frac{dy}{dt}+\frac{2y}{25}=4000, and using the integrating factor e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}, therefore  (\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}, we get   \frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}, after integrating both sides y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C, therefore y(t)= 50000 +Ce^{\frac{-2t}{25}}, to find C we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions y(0)=10, so 10= 50000+Ce^{\frac{-0*2}{25}}

10=50000+C\\C=10-50000=-49990

Finally we can write an expression for the amount of salt in the tank at any time t, it is y(t)=50000-49990e^{\frac{-2t}{25}}

b) The tank will overflow due Rin>Rout, at a rate of 80 L/min-40L/min=40L/min, due we have 500 L to overflow \frac{500L}{40L/min} =\frac{25}{2} min=t, so we can evualuate the expression of a) y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7, is the salt concentration when the tank overflows

4 0
3 years ago
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