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)
To find the rate of change in the squirrel's height above the ground, we need to divide the change in height by the time taken.
The change in height is 100 feet (initial height) - 45 feet (final height) = 55 feet.
The time taken is 5.25 seconds.
Therefore, the rate of change in the squirrel's height above the ground is 55 feet / 5.25 seconds ≈ 10.48 feet/second.
So, the quotient describes the rate of change in the squirrel's height above the ground as approximately 10.48 feet/second.
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)
To find the rate of change in the squirrel's height above the ground, we need to divide the change in height by the time taken.
The change in height is 100 feet (initial height) - 45 feet (final height) = 55 feet.
The time taken is 5.25 seconds.
Therefore, the rate of change in the squirrel's height above the ground is 55 feet / 5.25 seconds ≈ 10.48 feet/second.
So, the quotient describes the rate of change in the squirrel's height above the ground as approximately 10.48 feet/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)
To find the average rate of change in the dolphin's position, we need to divide the change in position by the time taken.
The change in position is 10 meters (highest point above the water) - 3.5 meters (lowest depth below the water) = 6.5 meters.
The time taken is 12.5 seconds.
Therefore, the average rate of change in the dolphin's position is 6.5 meters / 12.5 seconds ≈ 0.52 meters/second.
So, the quotient describes the average rate of change in the dolphin's position as approximately 0.52 meters/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)
To find the rate of change in the diver's depth, we need to divide the change in depth by the time taken.
The change in depth is 10 meters.
The time taken is 10.1 seconds.
Therefore, the rate of change in the diver's depth is 10 meters / 10.1 seconds ≈ 0.99 meters/second.
So, the quotient describes the rate of change in the diver's depth as approximately 0.99 meters/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 how much the car's value decreases per month, we need to divide the total decrease in value by the number of months.
The total decrease in value is $4,000.
The number of months is 12 months.
Therefore, the value decreases per month is $4,000 / 12 months ≈ $333.33.
So, the car's value decreases by approximately $333.33 per month.
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 that describes the rate of change in the squirrel’s height above the ground is −10.48 feet/second.
The quotient that describes the rate of change in the squirrel’s height above the ground is negative 10.48 feet/second.
The quotient that describes the rate of change in the squirrel’s height above the ground is 10.48 feet/second.
The quotient that describes the rate of change in the squirrel’s height above the ground is 10.48 feet/second.
The quotient that describes the rate of change in the squirrel’s height above the ground is 19.05 feet/second.
The quotient that describes the rate of change in the squirrel’s height above the ground is 19.05 feet/second.
The quotient that describes the rate of change in the squirrel’s height above the ground is −19.05 feet/second.