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3 years ago

I need help asap !!! please !!!

Mathematics

1 answer:

salantis [7]3 years ago

6 0

Answer:

2^-84

Step-by-step explanation:

First simplify inside the parentheses

2^-10 / 4^2

Rewriting 4 as 2^2

2^-10 / 4^2^2

We know that a^b^c = a^(b*c)

2^-10 / 2^(2*2) = 2^-10 / 2^4

We know that a^b / a^c = a^(b-c)

2^-10 / 2^4 = 2^(-10-4) = 2^-14

Replace the term in side the parentheses with 2^-14

2^-14 ^7

We know that a^b^c = a^(b*c)

2^(-14*7)

2^-84

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Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

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Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

I need help asap !!! please !!!​

I need help asap !!! please !!!