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Zielflug [23.3K]
4 years ago
5

Camryn practices the trumpet every 11th day and the flute every 3rd day. Camryn practiced both the trumpet and the flute today.

How many days until Camryn practices the trumpet and flute again in the same day?
Mathematics
1 answer:
QveST [7]4 years ago
4 0

Answer: 33 days.


Step-by-step explanation:

Given :- Camryn practices the trumpet every 11th day and the flute every 3rd day.

Camryn practiced both the trumpet and the flute today.  

To find the number of days until Camryn practices the trumpet and flute again in the same day, we need to find the least common multiple (L.C.M.) of 11 and 3.

As they both are co-prime numbers thus its L.C.M.= 11×3=33

Therefore, Camryn practices the trumpet and flute again in the same day after 33 days.

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3 years ago
Identify the coordinates of the point (3,-2), translated 5 units left and 6
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Answer:

(- 2, 4 )

Step-by-step explanation:

A translation of 5 units left means subtract 5 from the original x- coordinate.

A translation of 6 units up means add 6 to the original y- coordinate.

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4 years ago
A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of
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Let A(t) denote the amount of salt in the tank at time t.

Salt flows in at a rate of

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and flows out at a rate of

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(in case you're unsure about the denominator: the tank starts off with 200 gal of solution, and each minute solution flows in at a rate of 3 gal/min and thus the tank gains (3 gal/min) * (1 min) = 3 gal. At the same time, solution flows out at a rate of 2 gal/min and thus the tank loses 2 gal, giving a net change in volume of (3 - 2)*t = t gal)

Then the net rate of salt flow is given by the ODE,

\dfrac{\mathrm dA(t)}{\mathrm dt}-\dfrac{2A(t)}{200+t}=3

Multiply both sides by (200+t)^{-2}:

(200+t)^{-1}\dfrac{\mathrm dA(t)}{\mathrm dt}-2(200+t)^{-3}A()=3(200+t)^{-2}

\implies\dfrac{\mathrm d}{\mathrm dt}\bigg((200+t)^{-2}A(t)\bigg)=3(200+t)^{-2}

Integrating both sides and solving for A(t) gives

(200+t)^{-2}A(t)=-\dfrac3{200+t}+C

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The tank starts off with 100 lb of salt in solution, so A(0)=100 and we find

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and so

A(t)=-2(200+t)+\dfrac{(200+t)^2}{80}=\dfrac{(200+t)(40+t)}{80}

The tank will begin to overflow once the volume of solution reaches 500 gal; this happens when

500=200+t\implies t=300

or 300 minutes or 5 hours after solution starts flowing. At this point, the tank will contain

A(300)=2125

or 2125 lb of salt.

Theoretically, the amount of salt in the tank will increase forever, since A(t)\to\infty as t\to\infty.

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4 years ago
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Step-by-step explanation:

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