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deff fn [24]
3 years ago
9

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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1.45 rounded to the nearest tenth would be 1.5 and 2.38 rounded to the nearest tenth would be 2.40 (or 2/4)

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At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel
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Use Law of Cooling:
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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:
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This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
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