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3 years ago

Please Help... Thanks!

Mathematics

1 answer:

Brut [27]3 years ago

6 0

Answer:

The system of linear inequalities to represent the graph is:

(1) y<=6 and

(2) y>=x

Step-by-step explanation:

The graph is:

1) The region below the horizontal line y=6: y<=6

2) The region above the oblique line that passes through the points:

Origin: O=(0,0)→P1=(0,0)=(x1,y1)→x1=0, y1=0

P2=(5,5)=(x2,y2)→x2=5, y2=5

Slope: m=(y2-y1)/(x2-x1)

m=(5-0)/(5-0)

m=(5)/(5)

m=1

Equation Point-slope:

y-y1=m(x-x1)

y-0=1(x-0)

y=1(x)

y=x

Above the line y=x: y>=x

Then the system of linear inequalities to represent the graph is:

(1) y<=6 and

(2) y>=x

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3 years ago

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

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

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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