Q 0.00 10 0.02 8.187 0.04 6.703 0.06 5.488 0.08 4.493 0.10 3.678 Solution In the following figure we plot the data and use it to sketch a curve that approximates the graph of the function Q(coulombs) 10 0.02 0.04 0.06 0.08 0.10 Given the points P(0.04, 6.703) and R(0.00, 10) on the graph, we find that the slope of the secant line PR, rounded to two decimal places, is as follows. 10-(x moo 0.00 -0.04 The following table shows the results of similar calculations for the slopes of other secant lines. R (0.00, 10) (0.02, 8.187) -74,20 (0.06, 5.488)-60.75 (0.08, 4.493) -55.25 (0.10, 3.678)-50.42 mp Q(coulombs) 104 8 x. From this table we would expect the slope of the tangent line at t-0.04 to lie somewhere between -74.20 and -60.75. In fact, the average of the slopes of the two closest secant lines is as follows. (-74.20-60.75) - (seconds) So, by this method, we estimate the slope of the tangent line, rounded to the nearest integer, to be about x Another method is to draw an approximation to the tangent line at P and measure the sides of the triangle ABC, as in the figure below. JAB J 0.02 0.04 0.06 0.08 0.10 (seconds) This gives an estimate of the slope of the tangent line as 5.362-8.044 5.362-8.094-67.05. 0.06 -0.02 。
Q 0.00 10 0.02 8.187 0.04 6.703 0.06 5.488 0.08 4.493 0.10 3.678 Solution In the following figure we plot the data and use it to sketch a curve that approximates the graph of the function Q(coulombs) 10 0.02 0.04 0.06 0.08 0.10 Given the points P(0.04, 6.703) and R(0.00, 10) on the graph, we find that the slope of the secant line PR, rounded to two decimal places, is as follows. 10-(x moo 0.00 -0.04 The following table shows the results of similar calculations for the slopes of other secant lines. R (0.00, 10) (0.02, 8.187) -74,20 (0.06, 5.488)-60.75 (0.08, 4.493) -55.25 (0.10, 3.678)-50.42 mp Q(coulombs) 104 8 x. From this table we would expect the slope of the tangent line at t-0.04 to lie somewhere between -74.20 and -60.75. In fact, the average of the slopes of the two closest secant lines is as follows. (-74.20-60.75) - (seconds) So, by this method, we estimate the slope of the tangent line, rounded to the nearest integer, to be about x Another method is to draw an approximation to the tangent line at P and measure the sides of the triangle ABC, as in the figure below. JAB J 0.02 0.04 0.06 0.08 0.10 (seconds) This gives an estimate of the slope of the tangent line as 5.362-8.044 5.362-8.094-67.05. 0.06 -0.02 。
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter4: Exponential Functions
Section4.4: Modeling Nearly Exponential Data
Problem 13E: Special Rounding Instructions For this exercise set, round all regression parameters to three...
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Please solve this homework
This math equation needs to be solved
![0.00 10
0.02 8.187
0.04 6.703
0.06 5.488
0.08 4.493
0.10 3.678
Solution
In the following figure we plot the data and use it to sketch a curve that approximates the graph of the function.
Q(coulombs)
10
8
Q
6
4
2
O
Given the points P(0.04, 6.703) and R(0.00, 10) on the graph, we find that the slope of the secant line PR, rounded to two decimal places, is as follows.
10-(x )
8
0.00 -0.04
The following table shows the results of similar calculations for the slopes of other secant lines.
R
6
0.02 0.04 0.05 0.08 0.10
4
(0.00, 10)
-82.43
(0.02, 8.187)
-74.20
(0.06, 5.488)
-60.75
(0.08, 4.493)
-55.25
(0.10, 3.678) -50.42
From this table we would expect the slope of the tangent line at t-0.04 to lie somewhere between -74.20 and -60.75. In fact, the average of the slopes of the two closest secant lines is as follows.
(-74.20-60.75) -
x
t (seconds)
mpst
2
So, by this method, we estimate the slope of the tangent line, rounded to the nearest integer, to be about
x
Another method is to draw an approximation to the tangent line at P and measure the sides of the triangle ABC, as in the figure below.
Q(coulombs)
104
x
A
P
……………………………… (seconds)
t
0.02 0.04 0.05 0.08 0.10
This gives an estimate of the slope of the tangent line as
A
MB]
5.362-8.044-67.05.
|BC|
0.06 -0.02](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F66fc88c0-8650-45c5-a829-43207bea1bea%2F4c350d9c-3d18-4886-8b40-6a6f0a5e9d80%2Fmov0uws_processed.png&w=3840&q=75)
Transcribed Image Text:0.00 10
0.02 8.187
0.04 6.703
0.06 5.488
0.08 4.493
0.10 3.678
Solution
In the following figure we plot the data and use it to sketch a curve that approximates the graph of the function.
Q(coulombs)
10
8
Q
6
4
2
O
Given the points P(0.04, 6.703) and R(0.00, 10) on the graph, we find that the slope of the secant line PR, rounded to two decimal places, is as follows.
10-(x )
8
0.00 -0.04
The following table shows the results of similar calculations for the slopes of other secant lines.
R
6
0.02 0.04 0.05 0.08 0.10
4
(0.00, 10)
-82.43
(0.02, 8.187)
-74.20
(0.06, 5.488)
-60.75
(0.08, 4.493)
-55.25
(0.10, 3.678) -50.42
From this table we would expect the slope of the tangent line at t-0.04 to lie somewhere between -74.20 and -60.75. In fact, the average of the slopes of the two closest secant lines is as follows.
(-74.20-60.75) -
x
t (seconds)
mpst
2
So, by this method, we estimate the slope of the tangent line, rounded to the nearest integer, to be about
x
Another method is to draw an approximation to the tangent line at P and measure the sides of the triangle ABC, as in the figure below.
Q(coulombs)
104
x
A
P
……………………………… (seconds)
t
0.02 0.04 0.05 0.08 0.10
This gives an estimate of the slope of the tangent line as
A
MB]
5.362-8.044-67.05.
|BC|
0.06 -0.02
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