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Greeley [361]
3 years ago
13

Mr Bartley is taking the theater club

Mathematics
1 answer:
PtichkaEL [24]3 years ago
8 0
What is he doing behind stage
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I NEED POINTS

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20 pts awarded and brainliest chosen!!!!
g100num [7]

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D.  x <= -3 or   x >= 5

Step-by-step explanation:

3 | x-1 | > = 12

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x <= -3 or   x >= 5

4 0
3 years ago
A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and
goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}

Solve for S(t):

\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}

e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}

The left side is the derivative of a product:

\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}

Integrate both sides:

e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt

e^{t/800}S(t)=64e^{t/800}+C

S(t)=64+Ce^{-t/800}

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

0=64+C\impleis C=-64

and so (b) the amount of sugar in the tank at time t is

S(t)=64\left(1-e^{-t/800}\right)

As t\to\infty, the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

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3 years ago
Which of the following inequalities matches the graph (1 point)
Lilit [14]

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D

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(88÷1×2×5−9)+3×8 what does this make
cupoosta [38]

Answer: 6992

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5 0
2 years ago
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