212°F
147°F
23.2 min
1) Initial temperature for t=0
T(0)=68+144e^(0)=68+144=212°F
2) After 15 min
T(15)=68+144e^(-0.04*15)=68+144e^(-0.6)=147°F
3) 125°F -> t=?
125=68+144e^(-0.04t)
144e^(-0.04t)=125-68=57
e^(-0.04t)=57/144=0.4 you can apply ln on both sides:
-0.04t=ln(0.4) solving you get t=23.3 min
I think that it gets smaller. (not entirely sure)
Answer:
Should be in the picture linked.
Step-by-step explanation:
If the chart is labeled by five, ten, fifteen, and twenty at the bottom X axis for gallons, then this should be correct. If not, could you take another picture of the problem? I could redo it then. Hopefully I helped, good luck
I think it’s b cos u minus 43 and your left with 48x and 47x so divide by 47 and you get 1 point something x = to x so yeah probably none because no number will make that work so it’s b
What is the question so I can slove
