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kati45 [8]
3 years ago
8

on friday 1/6 of band practice was spent trying on uniforms. The band spent 1/4 practice on marching. the remaining practice tim

e was spent playing music.
Mathematics
2 answers:
AlladinOne [14]3 years ago
4 0

Answer: 7/12 of the band practice time is spent on playing music.

Step-by-step explanation:

The data we have is:

On friday, the band spent 1/6 of the time in triyinh uniforms and 1/4 of the time on practice marching.

Then the amount of time that the band did not practice music is:

1/6 + 1/4 of their total time:

we know that 12 is a common multiple of both, 6 and 4, so we can write this as:

2/12 + 3/12 = 5/12

Then the total time that the band practices music is:

1 - 5/12 = 7/12

Ahat [919]3 years ago
3 0

Answer:

7/12 of band practice was spent on playing music



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Assuming we're dealing with \frac{x+2}{-5} < 4, which is the same as writing (x+2)/(-5) < 4, then the answer is choice C.

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\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

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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} }

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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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The solution of the differential equation is:

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

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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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