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Determine the equation of the tangent to the curve where x=0. y=-1+x Oy=-x+1 Oy=1 O None of the above }= 2 H at the point on the curve

Determine the equation of the tangent to the curve where x=0. y=-1+x Oy=-x+1 Oy=1 O None of the above }= 2 H at the point on the curve

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Determine the equation of the tangent to the curve where x=0. Oy=-1+x Oy=-x+1 Oy=1 None of the above } = 2 1 at the point on the curve
Transcribed Image Text:Determine the equation of the tangent to the curve where x=0. Oy=-1+x Oy=-x+1 Oy=1 None of the above } = 2 1 at the point on the curve
The velocity of a car is given by v(t) = 120(1 - 0.85'). Describe the acceleration of the car. The values of v(t) slowly rise in the range of about t between 0 and 15. The slope for these values is positive and steep. Then as the graph nears t=20 the steepness of the slope decreases and seems to get very close to 0. We can reason that the car quickly accelerates for the first 20 units of time. Then, it seems to maintain a constant acceleration for the rest of the time. To verify this, we could differentiate and look at values where v'(t) is increasing. The values of v(t) quickly rise in the range of about t between 0 and 15. The slope for these values is positive and steep. Then as the graph nears t=40 the steepness of the slope decreases and seems to get very close to 0. We can reason that the car quickly accelerates for the first 40 units of time. Then, it seems to maintain a constant acceleration for the rest of the time. To verify this, we could differentiate and look at values where v'(t) is increasing. The values of v(t) quickly rise in the range of about t between 0 and 15.The slope for these values is positive and steep. Then as the graph nears t=20 the steepness of the slope decreases and seems to get very close to 0. We can reason that the car quickly accelerates for the first 20 units of time. Then, it seems to maintain a constant acceleration for the rest of the time. To verify this, we could differentiate and look at values where v'(t) is increasing None of the above
Transcribed Image Text:The velocity of a car is given by v(t) = 120(1 - 0.85'). Describe the acceleration of the car. The values of v(t) slowly rise in the range of about t between 0 and 15. The slope for these values is positive and steep. Then as the graph nears t=20 the steepness of the slope decreases and seems to get very close to 0. We can reason that the car quickly accelerates for the first 20 units of time. Then, it seems to maintain a constant acceleration for the rest of the time. To verify this, we could differentiate and look at values where v'(t) is increasing. The values of v(t) quickly rise in the range of about t between 0 and 15. The slope for these values is positive and steep. Then as the graph nears t=40 the steepness of the slope decreases and seems to get very close to 0. We can reason that the car quickly accelerates for the first 40 units of time. Then, it seems to maintain a constant acceleration for the rest of the time. To verify this, we could differentiate and look at values where v'(t) is increasing. The values of v(t) quickly rise in the range of about t between 0 and 15.The slope for these values is positive and steep. Then as the graph nears t=20 the steepness of the slope decreases and seems to get very close to 0. We can reason that the car quickly accelerates for the first 20 units of time. Then, it seems to maintain a constant acceleration for the rest of the time. To verify this, we could differentiate and look at values where v'(t) is increasing None of the above
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