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miskamm [114]
3 years ago
6

Q3 Help pls.. urgent!

Mathematics
1 answer:
Komok [63]3 years ago
7 0
Alright, so 3f-g=4 and f+2g=5. 
3f-g=4
f+2g=5
Multiplying the first equation by 2 and adding it to the second, we get 7f=13 and by dividing both sides by 7 we get f=13/7. Since f+2g=5, then we can plug 13/7 in for f to get 13/7+2g=5. Next, we subtract 13/7 from both sides to get 2g=3+1/7=22/7 (since 3*7=21 and 21+1=22). DIviding both sides by 2, we get 22/14=g. Plugging that into f/39g, we get (13/7)/(22*39/14)
= (13/7)/(858/14) 
= (13/7)*(14/858)
=182/6006
= 91/3003 (by dividing both numbers by 2)
= 13/429 (by dividing both numbers by 7)
 = 1/33 (by dividing both numbers by 13)
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Step-by-step explanation:

For this case we have the following differential equationÑ

\frac{du}{dt}= -k (u-T)

We can reorder the expression like this:

\frac{du}{u-T} = -k dt

We can use the substitution w = u-T and dw =du so then we have:

\frac{dw}{w} =-k dt

IF we integrate both sides we got:

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We can also express the solution in the following terms:

u(t) = (T_i -T_{amb}) e^{kt} +T_{amb}

For this case we know that k =-0.15 hr since w ehave a cooloing, T_{i}= 70 F, T_{amb}=11F, we have this model:

u(t) = (70-11) e^{-0.15t} +11

And if we want that the temperature would be 32F we can solve for t like this:

32 = 59 e^{-0.15 t} +11

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If we apply natural logs on both sides we got:

ln (\frac{21}{59}) =-0.15 t

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So it would takes approximately 6.9 hours to reach 32 F.

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