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kvv77 [185]
3 years ago
10

Welp okie please help

Mathematics
2 answers:
Readme [11.4K]3 years ago
7 0

Answer:

8:15 am

Step-by-step explanation:

just subtract 9 hours and 45 minutes from 18 hours (6:00 pm is equal to 18:00)

Nataly [62]3 years ago
4 0

Answer:

8:15

Step-by-step explanation:


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Help ASPP<br><br>-3/5 x = 18 <br>-30<br>-10 4/5<br>10 4/5<br>30​
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Answer:

-30

Step-by-step explanation:

move and divide -3/5 to the right side

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x=18÷(-3/5)

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7 0
3 years ago
Newton's law of cooling is:
Mnenie [13.5K]

Answer:

t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr

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

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:

ln |w| = -kt +C

If we apply exponential in both sides we got:

w = e^{-kt} *e^c

And if we replace w = u-T we got:

u(t)= T + C_1 e^{-kt}

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

21=59 e^{-0.15 t}

\frac{21}{59} = e^{-0.15 t}

If we apply natural logs on both sides we got:

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

t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr

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

7 0
3 years ago
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