A squirrel has stored its acorns in a hole that is 45 feet from the ground in a tall tree. The squirrel starts on a perch 100 feet above the ground. The squirrel moves from the perch down to its stored acorns in 5.25 seconds. Interpret the quotient to describe the rate of change in the squirrel's height above the ground. Give your answer to the nearest hundredth. (1 point)
The quotient is the change in height divided by the change in time. In this case, the change in height is 100 feet (from the perch to the ground) and the change in time is 5.25 seconds.
Therefore, the quotient is 100/5.25 = <<100/5.25=19.05>>19.05 feet per second.
This means that the squirrel's height above the ground is decreasing at a rate of approximately 19.05 feet per second.
A dolphin jumped above the surface of the water. It reached an altitude of 3.5 meters above the surface of the water and then dove 10 meters below the surface of the water. It went from its highest point above the water to its lowest depth in 12.5 seconds. Interpret the quotient to describe the average rate of change in the dolphin's position. Give your answer to the nearest hundredth. (1 point)
The quotient is the change in position divided by the change in time. In this case, the change in position is 3.5 meters above the water and 10 meters below the water, for a total change in position of 3.5 + 10 = 13.5 meters.
The change in time is 12.5 seconds.
Therefore, the quotient is 13.5 / 12.5 = <<13.5/12.5=1.08>>1.08 meters per second.
This means that on average, the dolphin's position is changing by approximately 1.08 meters per second.
A scuba diver was at a depth of 15 meters below the surface when she saw something interesting about 10 meters lower. She made the descent in 10.1 seconds. Interpret the quotient to describe the rate of change in the diver's depth. Give your answer to the nearest hundredth. (1 point)
The quotient is the change in depth divided by the change in time. In this case, the change in depth is 10 meters, and the change in time is 10.1 seconds.
Therefore, the quotient is 10/10.1 = <<10/10.1=0.99>>0.99 meters per second.
This means that the diver's depth is changing at a rate of approximately 0.99 meters per second.
Solve using division of rational numbers. If a car's value decreases by $4,000 in the first 12 months, how much does the value decrease per month? (Round to the nearest hundredth.) (1 point)
To find the value decrease per month, we need to divide the decrease in value by the number of months. The decrease in value is $4,000 and the number of months is 12.
Dividing $4,000 by 12 gives us $333.33.
Thus, the value of the car decreases by approximately $333.33 per month.
It is recommended that hikers drink 24 ounces of water an hour. If a canteen has 24 ounces of water, how much should the amount of water in the canteen change every 10 minutes? (1 point)
To find the amount of water the canteen should decrease every 10 minutes, we need to divide the recommended amount of water per hour by 6, since there are 6 sets of 10 minutes in an hour.
The recommended amount of water per hour is 24 ounces. Dividing 24 by 6 gives us 4 ounces.
Thus, the amount of water in the canteen should decrease by approximately 4 ounces every 10 minutes.
To find the rate of change in the squirrel's height above the ground, we need to calculate the change in height divided by the time elapsed.
The initial height of the squirrel is 100 feet, and it moves down to a height of 45 feet. So the change in height is 100 - 45 = 55 feet.
The time elapsed is given as 5.25 seconds.
To find the rate of change, we divide the change in height (55 feet) by the time elapsed (5.25 seconds):
Rate of change = Change in height / Time elapsed
Rate of change = 55 feet / 5.25 seconds
Evaluating this expression, we get:
Rate of change ≈ 10.48 feet/second
Rounding to the nearest hundredth, the rate of change in the squirrel's height above the ground is approximately 10.48 feet per second.