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3 years ago

Solve the problem.

Mathematics

2 answers:

Aleonysh [2.5K]3 years ago

4 0

Perimiter=2(L+W)
P=2(100+120)
P=2(220)
P=440

0.1 per foot
440*0.10=44

$44

RideAnS [48]3 years ago

3 0

<span>I know the answer it is really easy and I hope you figure it out and when you do contact me and we will compare answers.</span>

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57.03 + 2.08=

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Which is bigger; 0.125 or 0.15? Please explain your answer

FrozenT [24]

0.125 is bigger than 0.15.

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3 years ago

(1 point) A tank contains 23402340 L of pure water. Solution that contains 0.050.05 kg of sugar per liter enters the tank at the

aksik [14]

Let $S(t)$ be the amount of sugar in the tank at time $t$ .

a. The tank contains only pure water at the start, so $\boxed{S(0)=0}$ .

b. The inflow rate of sugar is

${S_{\rm in}}'=\left(0.05\dfrac{\rm kg}{\rm L}\right)\left(5\dfrac{\rm L}{\rm min}\right)=\dfrac1{40}\dfrac{\rm kg}{\rm min}$

and the outflow rate is

${S_{\rm out}}'=\left(\dfrac S{2340}\dfrac{\rm kg}{\rm L}\right)\left(5\dfrac{\rm L}{\rm min}\right)=\dfrac S{468}\dfrac{\rm kg}{\rm min}$

so the net rate at which $S(t)$ changes over time is governed by

$S'=\dfrac1{40}-\dfrac S{468}\implies S'+\dfrac S{468}=\dfrac1{40}$

Multiply both sides by $e^{t/468}$ ,

$e^{t/468}S'+\dfrac{e^{t/468}}{468}S=\dfrac{e^{t/468}}{40}$

and condense the left side as the derivative of a product,

$\left(e^{t/468}S\right)'=\dfrac{e^{t/468}}{40}$

Integrate both sides to get

$e^{t/468}S=\dfrac{117e^{t/468}}{10}+C$

and solve for $S$ :

$S=\dfrac{117}{10}+Ce^{-t/468}$

With $S(0)=0$ , we find $C=-\dfrac{117}{10}=-11.7$ , so that

$\boxed{S(t)=11.7-11.7e^{-t/468}}$

c. As $t\to\infty$ , the exponential term will converge to 0, leaving a fixed amount of 11.7 kg of sugar in the solution.

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3 years ago