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A student throws a ball upward from a height of 48 feet, initially at 32 ft/sec. The function that models the height of the ball t seconds after it was thrown is given by h(t) = -16t2 + 32t + 48. What is the maximum height (in ft) attained by the ball? Answer:

A student throws a ball upward from a height of 48 feet, initially at 32 ft/sec. The function that models the height of the ball t seconds after it was thrown is given by h(t) = -16t2 + 32t + 48. What is the maximum height (in ft) attained by the ball? Answer:

College Algebra
College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter5: Polynomial And Rational Functions
Section: Chapter Questions
Problem 6PT: Solve the following application problem. A rectangular field is to be enclosed by fencing. In...
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Answer the 3 questions (7,8,9) with complete solutions
A student throws a ball upward from a height of 48 feet, initially at 32 ft/sec. The function that models the height of the ball t seconds after it was thrown is given by h(t) = -16t2 + 32t + 48. What is the maximum height (in ft) attained by the ball? Answer: A rectangular lot having an area of 5000 sq. meters is to be fenced on three sides. Find the minimum amount of fencing needed. Answer: A box is to be constructed from a piece of square cardboard whose perimeter is 36 inches by cutting equal squares out of the corners and then turning up the sides. Find the maximum volume of the box that can be made this way. Answer:
Transcribed Image Text:A student throws a ball upward from a height of 48 feet, initially at 32 ft/sec. The function that models the height of the ball t seconds after it was thrown is given by h(t) = -16t2 + 32t + 48. What is the maximum height (in ft) attained by the ball? Answer: A rectangular lot having an area of 5000 sq. meters is to be fenced on three sides. Find the minimum amount of fencing needed. Answer: A box is to be constructed from a piece of square cardboard whose perimeter is 36 inches by cutting equal squares out of the corners and then turning up the sides. Find the maximum volume of the box that can be made this way. Answer:
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