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tester [92]
3 years ago
15

WILL GIVE BRAINLIEST, PLEASE HELP ASAP!!! MY LIFE DEPENDS ON THIS!!!

Mathematics
1 answer:
ahrayia [7]3 years ago
7 0

h(x) = 3 * (2)^x

Section A is from x = 1 to x = 2

h(1) = 3 * (2)^1 = 3 * 2 = 6

h(2) = 3 * (2)^2 = 3 * 4 = 12

so

the average rate of change  = (12 - 6)/(2 - 1) = 6


Section B is from x = 3 to x = 4

h(3) = 3 * (2)^3 = 3 * 8 = 24

h(4) = 3 * (2)^4 = 3 * 16 = 48

so

the average rate of change  = (48 - 24)/(4 - 3) = 24


Part B: How many times greater is the average rate of change of Section B than Section A? Explain why one rate of change is greater than the other. (6 points)

the average rate of change of section B is 24 and the average rate of change of section A is 6

So 24/6 = 4

The average rate of change of Section B is 4 times greater than the average rate of change of Section A

It's exponential function, not a linear function; so the rate of change is increasing.

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Solution:

Use Newton's Law of Cooling.  


T = T_s + (T_0 - T_s)*e^(-kt)  


where  

T = temperature at any instant  

T_s = temperature of surroundings  

T_0 = original temperature  

t = elapsed time  

k = constant  


Now, we need to find this constant. We are given that after one hour, the temperature drops to 13° C in a 7°C Environment.  

T = 14, T_0 = 24, T_s = -15, t = 1, k = ?  

T = T_s + (T_0 - T_s)*e^(-kt)  

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T = T_s + (T_0 - T_s)*e^(-kt)  

==> T = -15 + (24 –(-15))*e^[ -(0.774)(2) ]  

==> T = -15 + 39*e^(-1.548)  

==> T ≈ 15.72° C  

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