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Shalnov [3]
3 years ago
7

The cardboard box below is used by a company to package new laptops of the company plans to increase the length of the Box by 2i

n what is the minimum amount of extra cardboard needed to construct a new box The dimensions of the box is 5 in high 16 in in width and 20 in in length
A. 84in^2

B. 116in^2

C.132in^2

D.160in^2
Mathematics
1 answer:
harina [27]3 years ago
3 0

Answer:

D

Step-by-step explanation:

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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
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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}

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

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