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WITCHER [35]
3 years ago
6

2- 5x=-13 how do u solve this

Mathematics
1 answer:
kenny6666 [7]3 years ago
6 0
2 - 5x = -13
-2          -2
---------------
-5x = -15
/-5     /-5
-------------
x = 3

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

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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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Samantha has 58 ounces of orange juice that she pours into glasses holding 7 1/4 ounces. How many glasses of orange juice can sh
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It can't be A. since if you only look at managers, you are missing all the sales executives.

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