Answer:
The temperature of the water after 96 minutes is 
Step-by-step explanation:
we know that
<u><em>Newton's Law of Cooling</em></u> states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature
so
![[T(t) - T_{a}] = [T(0) - T_{a}] e^{-kt}](https://tex.z-dn.net/?f=%5BT%28t%29%20-%20T_%7Ba%7D%5D%20%3D%20%5BT%280%29%20-%20T_%7Ba%7D%5D%20e%5E%7B-kt%7D)
where
T(t) ----> is the temperature of the water at time t
T_a ---> is the ambient temperature (temp of the freezer)
T(0) ---> is the initial temperature of the water
k ---> is the cooling constant
we have
----> Temperature of the water at time 24 minutes
----> Temperature of the water at time 96 minutes
step 1
Find the value of k
substitute the given values
![[50 - 0] = [100 - 0}] e^{-k(24)}](https://tex.z-dn.net/?f=%5B50%20-%200%5D%20%3D%20%5B100%20-%200%7D%5D%20e%5E%7B-k%2824%29%7D)
Applying ln both sides
step 2
Determine 
(Temperature of the water at time 96 minutes)
![[T_9_6 - T_{a}] = [T(0) - T_{a}] e^{-kt}](https://tex.z-dn.net/?f=%5BT_9_6%20-%20T_%7Ba%7D%5D%20%3D%20%5BT%280%29%20-%20T_%7Ba%7D%5D%20e%5E%7B-kt%7D)
substitute the given values
![[T_9_6 - 0] = [100 - 0] e^{-0.02888*96}](https://tex.z-dn.net/?f=%5BT_9_6%20-%200%5D%20%3D%20%5B100%20-%200%5D%20e%5E%7B-0.02888%2A96%7D)
![T_9_6= [100] e^{-0.02888*96}](https://tex.z-dn.net/?f=T_9_6%3D%20%5B100%5D%20e%5E%7B-0.02888%2A96%7D)
Answer:
D
Step-by-step explanation:
What we need to do is find the operation between f(x) and g(x) that results in h(x).
Notice how there's a pattern, the value of h(x) is always equal to the values of g(x) and f(x) added together.
This can be represented by the expression (g+f)(x), the expression in D.
0.21jbvghhjjjjhhhyyhhhhhhjjjj
Answer:
3. AGH= 150
4. CHF= 110
Step-by-step explanation:
3. AGH corresponds to CHF, so they equal the same. AGH = 150
4. BGE is vertical to AGH, so they equal the same AGH = 110. AGH corresponds to CHF, so they equal the same. CHF=110.
