A ladder 25 feet long is leaning against the wall of a building. Initially, the foot of the ladder is 7 feet from the wall. The foot of the ladder begins to slide at a rate of 2 ft/sec, causing the top of the ladder to slide down the wall. The location of the foot of the ladder, its x coordinate, at time t seconds is given by x(t)=7+2t. wall y(1) 25 ft. ladder x(1) ground (a) Find the formula for the location of the top of the ladder, the y coordinate, as a function of time t. The formula for y(t)= √ 25² - (7+2t)² (b) The domain of t values for y(t) ranges from 0 (c) Calculate the average velocity of the top of the ladder on each of these time intervals (correct to three decimal places): . (Put your cursor in the box, click and a palette will come up to help you enter your symbolic answer.) time interval ave velocity [0,2] -0.766 [6,8] -3.225 time interval ave velocity -1.224 -9.798 [2,4] [8,9] (d) Find a time interval [a,9] so that the average velocity of the top of the ladder on this time interval is -20 ft/sec i.e. a= (e) Using your work above and this picture of the graph of the function y(t) given below, answer these true/false questions Enter a number. word "True" or "False")
A ladder 25 feet long is leaning against the wall of a building. Initially, the foot of the ladder is 7 feet from the wall. The foot of the ladder begins to slide at a rate of 2 ft/sec, causing the top of the ladder to slide down the wall. The location of the foot of the ladder, its x coordinate, at time t seconds is given by x(t)=7+2t. wall y(1) 25 ft. ladder x(1) ground (a) Find the formula for the location of the top of the ladder, the y coordinate, as a function of time t. The formula for y(t)= √ 25² - (7+2t)² (b) The domain of t values for y(t) ranges from 0 (c) Calculate the average velocity of the top of the ladder on each of these time intervals (correct to three decimal places): . (Put your cursor in the box, click and a palette will come up to help you enter your symbolic answer.) time interval ave velocity [0,2] -0.766 [6,8] -3.225 time interval ave velocity -1.224 -9.798 [2,4] [8,9] (d) Find a time interval [a,9] so that the average velocity of the top of the ladder on this time interval is -20 ft/sec i.e. a= (e) Using your work above and this picture of the graph of the function y(t) given below, answer these true/false questions Enter a number. word "True" or "False")
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter9: Quadratic Functions And Equations
Section9.4: Solving Quadratic Equations By Factoring
Problem 2GP
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![A ladder 25 feet long is leaning against the wall of a building. Initially, the foot of the ladder is 7 feet from the wall. The foot of the ladder begins to slide at a rate of 2 ft/sec, causing the top of the ladder to slide down the wall. The location of the foot of the ladder, its x coordinate, at time t seconds is given by
x(t)=7+2t.
wall
y(1)
25 ft. ladder
x(1)
ground
(a) Find the formula for the location of the top of the ladder, the y coordinate, as a function of time t. The formula for y(t)= √ 25² - (7+2t)²
(b) The domain of t values for y(t) ranges from 0
(c) Calculate the average velocity of the top of the ladder on each of these time intervals (correct to three decimal places):
. (Put your cursor in the box, click and a palette will come up to help you enter your symbolic answer.)
time interval
ave velocity
[0,2]
-0.766
[6,8]
-3.225
time interval
ave velocity
-1.224
-9.798
[2,4]
[8,9]
(d) Find a time interval [a,9] so that the average velocity of the top of the ladder on this time interval is -20 ft/sec i.e. a=
(e) Using your work above and this picture of the graph of the function y(t) given below, answer these true/false questions Enter a number. word "True" or "False")](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6a3800f2-fc56-4294-9acb-319dcebd7fd4%2Fc4e91a3b-93d7-4115-8c5c-e0898d862d4e%2Fiwbo922_processed.png&w=3840&q=75)
Transcribed Image Text:A ladder 25 feet long is leaning against the wall of a building. Initially, the foot of the ladder is 7 feet from the wall. The foot of the ladder begins to slide at a rate of 2 ft/sec, causing the top of the ladder to slide down the wall. The location of the foot of the ladder, its x coordinate, at time t seconds is given by
x(t)=7+2t.
wall
y(1)
25 ft. ladder
x(1)
ground
(a) Find the formula for the location of the top of the ladder, the y coordinate, as a function of time t. The formula for y(t)= √ 25² - (7+2t)²
(b) The domain of t values for y(t) ranges from 0
(c) Calculate the average velocity of the top of the ladder on each of these time intervals (correct to three decimal places):
. (Put your cursor in the box, click and a palette will come up to help you enter your symbolic answer.)
time interval
ave velocity
[0,2]
-0.766
[6,8]
-3.225
time interval
ave velocity
-1.224
-9.798
[2,4]
[8,9]
(d) Find a time interval [a,9] so that the average velocity of the top of the ladder on this time interval is -20 ft/sec i.e. a=
(e) Using your work above and this picture of the graph of the function y(t) given below, answer these true/false questions Enter a number. word "True" or "False")
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