A prairie dog left its burrow 5 feet underground and started digging deeper into the ground. It digs straight down at a constant
rate of 2
feet per minute.
How long does it take the prairie dog to reach 13 feet underground?
Enter your answer in the box
min
Willing to give Brainliest No Links
feet per minute.
How long does it take the prairie dog to reach 13 feet underground?
Enter your answer in the box
min
Willing to give Brainliest No Links
1 answer:
Building and solving an equation to model the situation, it is found that it takes 4 min for the prairie dog to reach 13 feet underground.
- Initially, the dog is 5 feet underground.
- Each minute, the dog goes 2 more feet underground.
Hence, the underground height of the dog after t seconds is given by:
The time it takes for the dog to reach 13 feet underground is <u>t for which h(t) = 13</u>, hence:
It takes 4 min for the prairie dog to reach 13 feet underground.
A similar problem is given at brainly.com/question/25290003
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Answer:
305 °F
Step-by-step explanation:
The core temperature of the object after 4 hours can be found using an exponential decay formula to model the decay of the difference between core temperature and ambient.
<h3>Cooling Model</h3>The solution to the differential equation described by Newton's law of cooling is the exponential equation ...
y = ab^t +c
where 'a' is the initial core temperature difference from ambient, 'b' is the decay factor of that difference in 1 unit of time period t. 'c' is the ambient temperature.
For this problem, the ambient temperature is c=80, and the differences of interest are ...
a = 1200 -80 = 1120
b = (830 -80)/1120 = 75/112
Using these values in the model gives ...
y = 1120(75/112)^t +80 . . . . . . where y(t) is the core temperature at time t
Note that units of time are hours.
<h3>Application</h3>We want y when t=4.
y = 1120(75/112)^4 +80 ≈ 1120(0.20108) +80 ≈ 305.212
The core temperature after 4 hours is about 305 °F.
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<em>Additional comment</em>
The differential equation will have a solution of the form ...
where k = ln(75/112) ≈ -0.40101
In the above, we defined b = e^k = 75/112. Accuracy with this fraction can be better than using a truncated value of k.
