18x 5
7 answers
The value of 18x times 5 is 90x.
A squirrel has stored its acorns in a hole that is 45 feet from the ground in a tall tree. The squirrel moved from a perch 100 feet above the ground to get 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)
Responses
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 −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 −19.05 feet/second.
Responses
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 −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 −19.05 feet/second.
The correct answer is: The quotient that describes the rate of change in the squirrel’s height above the ground is 19.05 feet/second.
Explain?
To find the rate of change in the squirrel's height above the ground, we divide the change in height by the time it took to make that change. In this case, the squirrel moved from a perch 100 feet above the ground to a hole 45 feet above the ground in 5.25 seconds.
The change in height is 100 feet - 45 feet = 55 feet.
To find the rate of change, we divide the change in height by the time it took: 55 feet / 5.25 seconds = 10.47619 feet/second.
Rounded to the nearest hundredth, the rate of change in the squirrel's height above the ground is 10.48 feet/second.
The change in height is 100 feet - 45 feet = 55 feet.
To find the rate of change, we divide the change in height by the time it took: 55 feet / 5.25 seconds = 10.47619 feet/second.
Rounded to the nearest hundredth, the rate of change in the squirrel's height above the ground is 10.48 feet/second.
Positive or negative
The rate of change in the squirrel's height above the ground is positive. Since the squirrel moved down from a higher position to a lower position, the height change and the rate of change are both positive.