Answer:
[tex]D(t)=100(0.4)^t[/tex]
Step-by-step explanation:
The temp is 100 at time t = 0
After 1 sec, the temp difference would be:
[tex]100-(\frac{100-60}{100})[/tex]
After 2 sec, the temp difference would be:
[tex]100-(\frac{100-60}{100})^2[/tex]
Similarly for 3 seconds, 4 seconds etc.
We notice that the parenthesis part is 40% of it, so we can also write:
100(40%)^t,
where
t is the time
40% can be written as 40/100 = 0.4
SO, the function is:
[tex]d(t)=100(0.4)^t[/tex]
Answer:
[tex]D(t)=100*0.4^t[/tex]
Step-by-step explanation:
Initially, the difference between the rod's and the water's temperatures is 100°
i.e D(t)=100 When t=0
After 1 seconds, the temp drops by 60%.
Therefore, the new value of D will be the old value multiplied by (100-60)%.
D(1)=100 X (100%-60%) = 100*0.4
After 2 seconds, the temp difference would be:
D(2)=100*0.4*0.4= [tex]100*0.4^2[/tex]
We notice that for any t, the percentage at which the difference is reduced is raised to the power of t.
Therefore, temperature difference in degrees Celsius, D(t), t seconds after the rod was plunged into the water is given as:
[tex]D(t)=100*0.4^t[/tex]
