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3 years ago

According to your online content assignment, a good schedule keeps a student from:

Mathematics

1 answer:

seraphim [82]3 years ago

8 0

Answer:

Step-by-step explanation:

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Evaluate by distributing and find the expression

Elden [556K]

Answer:

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3 years ago

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

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3 years ago

Which of the following data could be graphed using a dot plot?Select all that apply.

ExtremeBDS [4]

Answer:

what are the options?

Step-by-step explanation:

7 0

3 years ago

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Translate the figure 7 units left and 2 units down.

pshichka [43]

Answer:

2nd quad 5,7

hope this helps (:

Step-by-step explanation:

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3 years ago

A lot of points. help please asap

IgorC [24]

The answer is the first option. 7 left is not - 7 in this case. With these questions, you switch the signs. The vertical movement signs stay the same. Therefore, the answer is the first option.

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3 years ago

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