Nelson Functions 11
1st Edition
ISBN: 9780176332037
Textbook solutions
All Solutions
Section 6-1: Periodic Functions and Their Properties
Exercise 1
Step 1
1 of 2
A and C are periodic because they are occur at regular alternating intervals. In other words the figure should display a repeating pattern. B and D are not periodic because they do NOT have a repeating pattern.
Result
2 of 2
(a) and (c) are periodic because the pattern repeats at regular intervals.
Exercise 2
Step 1
1 of 5
Refer to the periodic graph on the textbook. Notice that the minimum value of $f(x)$ is $2$ and the maximum value is $10$. Thus the range can be written as
$$
{ f(x) in bold{R};|;2leq f(x)leq 10}
$$
For a function $f(x)$, domain is the set of all possible values of $x$ while range is the set of all possible values of $f(x)$.
Step 2
2 of 5
A crest or maxima can be found at $x=1$ and $x=5$. The period is therefore $5-1=4$
Period is defined as the change in $x$ between corresponding points for one complete cycle. This can be seen as the distance between two crests (maxima) or the distance between two trough (minima).
Step 3
3 of 5
From the graph $f(x)_{max}=10$ and $f(x)_{min}=2$
Therefore, the equation of axis is
$y=dfrac{10+2}{2}$
$$
y=6
$$
To find the equation of axis of $f(x)$ can be found as
$$
y=dfrac{f(x)_{max}+f(x)_{min}}{2}
$$
Step 4
4 of 5
The amplitude is
$A=dfrac{10-2}{2}$
$$
A=4
$$
The amplitude A is the distance from the axis (equilibrium) to the maxima or minima. This can be calculated as
$$
A=dfrac{f(x)_{max}-f(x)_{min}}{2}
$$
Result
5 of 5
${ f(x) in bold{R};|;2leq f(x)leq 10}$ ; $period =4$ ; $y=6$ ; $amplitude=4$
Exercise 3
Step 1
1 of 8
a.) From the graph, $y$ is minimum $x=0.95$ and $x=1.95$. Therefore the period is
$period=1.95-0.95=1$ s
Period is the change in $x$ for one complete cycle. Generally, this is the distance between two maxima or minima. Since this function has several maxima within a cycle, we will consider the minima.
Step 2
2 of 8
b.) By inspection, the maximum distance is 1.5 cm.
The maximum distance can be seen from the maxima of the function.
Step 3
3 of 8
c.) By inspection, the graph ranges from $d=0$ to $d=1.5$, thus, the range can be written as
$$
{ d in bold{R};|0;leq d;leq 1.5}
$$
The range of $f(x)$ is the set of all possible values of $f(x)$.
Step 4
4 of 8
d.) $k=5$ cycles
period = $1$ s
$t = dfrac{5; cycles}{1; s ;per; cycle}=5$ s
The time for $k$ cycles is given as
$t=dfrac{k}{period}$.
Step 5
5 of 8
e.) $f(x)_{max}=1.5$
$f(x)_{min}=0$
$y=dfrac{1.5+0}{2}$
$y=0.75$ cm
The equation of the axis is given by
$$
y=dfrac{f(x)_{max}+f(x)_{min}}{2}
$$
Step 6
6 of 8
f.) The amplitude is
$A=dfrac{1.5-0}{2}$
$A=0.75$ cm
The amplitude is the distance between the equilibrium position to the maxima or minima which can be calculated as
$$
A=dfrac{f(x)_{max}-f(x)_{min}}{2}
$$
Step 7
7 of 8
g.) The plateau at $d=1.5$ means that the distance is not changing with time, hence the device is not moving. The segment with negative slope means the device is approaching to the appliance as the distance between them decreases. When the bolt is finally attached by the device to the appliance, it starts to move away which results in segment with positive slope.
We shall interpret the physical meaning of the graph.
Result
8 of 8
a.) $period = 1$ s
b.) $d_{max}=1.5$ cm
c.) ${ d in bold{R};|;0;leq d;leq 1.5}$
d.) $t=5$ s
e.) $y=0.75;cm$
f.) $A=0.75;cm$
Exercise 4
Step 1
1 of 7
The function is periodic since it produces a graph that has a regular repeating pattern over a
constant interval.
constant interval.
Part A
Step 2
2 of 7
The function is periodic since it produces a graph that has a regular repeating pattern over a
constant interval.
constant interval.
Part B
Step 3
3 of 7
The function is NOT periodic since it does NOT produce a graph that has a regular repeating pattern over a
constant interval.
constant interval.
Part C
Step 4
4 of 7
The function is NOT periodic since it does NOT produce a graph that has a regular repeating pattern over a
constant interval.
constant interval.
Part D
Step 5
5 of 7
The function is periodic since it produces a graph that has a regular repeating pattern over a
constant interval.
constant interval.
Part E
Step 6
6 of 7
The function is periodic since it produces a graph that has a regular repeating pattern over a
constant interval.
constant interval.
Part F
Result
7 of 7
See Solutions
Exercise 5
Step 1
1 of 7
The action is periodic since it produces a graph that has a regular repeating pattern over a
constant interval since the teeth are evenly spaced.
constant interval since the teeth are evenly spaced.
Part A
Step 2
2 of 7
The function is periodic since it produces a graph that has a regular repeating pattern over a
constant interval since we assume that he his jumping at a constant and consistent pattern
constant interval since we assume that he his jumping at a constant and consistent pattern
Part B
Step 3
3 of 7
The function is NOT periodic since it does NOT produce a graph that has a regular repeating pattern over a
constant interval since the cost vary from individual to individual.
constant interval since the cost vary from individual to individual.
Part C
Step 4
4 of 7
The function is NOT periodic since it does NOT produce a graph that has a regular repeating pattern over a
constant interval since the graph will constantly be increasing at a steady rate and doesn’t go back to an equilibrium position.
constant interval since the graph will constantly be increasing at a steady rate and doesn’t go back to an equilibrium position.
Part D
Step 5
5 of 7
The function is NOT periodic since it does NOT produce a graph that has a regular repeating pattern over a
constant interval since the balls bounce height will decrease with each subsequent bounce.
constant interval since the balls bounce height will decrease with each subsequent bounce.
Part E
Step 6
6 of 7
The action is periodic since it produces a graph that has a regular repeating pattern over a
constant interval since it rotates at a constant speed.
constant interval since it rotates at a constant speed.
Part F
Result
7 of 7
See Solutions
Exercise 6
Step 1
1 of 5
The function is periodic since it produces a graph that has a regular repeating pattern over a
constant interval due to the y-values repeating in a pattern.
constant interval due to the y-values repeating in a pattern.
Part A
Step 2
2 of 5
The function is periodic since it produces a graph that has a regular repeating pattern over a
constant interval due to the y-values repeating in a pattern.
constant interval due to the y-values repeating in a pattern.
Part B
Step 3
3 of 5
The function is NOT periodic since it does NOT produce a graph that has a regular repeating pattern over a
constant interval since the y-values to not repeat in a pattern.
constant interval since the y-values to not repeat in a pattern.
Part C
Step 4
4 of 5
The function is NOT periodic since it produces a graph that does NOT have a regular repeating pattern over a
constant interval due to the y-values repeating in a pattern.
constant interval due to the y-values repeating in a pattern.
Part D
Result
5 of 5
(a) Periodic
(b) Periodic
(c) Not Periodic
(d) Not Periodic
Exercise 7
Step 1
1 of 9
Refer to the graph.
Step 2
2 of 9
a.) The function repeats every 8 minutes, thus it is $bold{periodic}$
A function is said to be periodic if the repetition is at regular interval.
Step 3
3 of 9
b.) The pump is turn on when the depth reaches $bold{10 cm}$.
We can determine when the pump is turned on when the depth starts decreasing.
Step 4
4 of 9
c.) The depth decreased between 6 min and 8 min. Thus, the pump is turned on for $bold{2 minutes}$.
We know that the pump is working when the depth continues to decrease.
Step 5
5 of 9
d.) A complete cycle occurs every 8 minutes, thus, the period is $bold{8 minutes}$.
The period is defined as the time required for the function to complete one cycle.
Step 6
6 of 9
e.) The depth can only take values from 4 to 10. Thus, the range is
$$
{ d in bold{R};|;4leq dleq10 }
$$
The range is the set of all possible values of the dependent variable; in this case, the depth.
Step 7
7 of 9
f.) From the graph, at $t=3$ minutes, $d=7;cm$
We shall find the depth at 3 minutes from the graph
.
.
Step 8
8 of 9
g.) The graph reaches $d=10cm$ at $t=6 minutes$. Since this cycle will repeat every 8 minutes, the depth would be 10 cm at values of $t$ such that
$t=6+8n$ where $n = 0,1,2,3..$
We shall find the times at which the depth is $10 cm$
Step 9
9 of 9
At $62 minutes$, the cycle has gone though $7 cycles$ since $7cdot 8=56$ and $8cdot 8=64$. At 56 minutes, the cycle restarts, so the time elapsed from $56$ to $62$ is 6 minutes. From the figure, the depth is 10 m at $t=6$, thus, the depth at $t=62$ minutes is $bold{10 m}.$
We shall determine the depth at 62 minutes.
To do this, we need to find the number of complete cycles that has elapsed at 62 minutes, and restart the pattern from there.
To do this, we need to find the number of complete cycles that has elapsed at 62 minutes, and restart the pattern from there.
Exercise 8
Step 1
1 of 8
a.) From the graph, $y$ is minimum $x=0$ and $x=8$. Therefore the period is
$period=8-0=8$ s
Period is the change in $x$ for one complete cycle. Generally, this is the distance between two maxima or minima.
Step 2
2 of 8
b.) $f(x)_{max}=7$
$f(x)_{min}=1$
$h=dfrac{7+1}{2}$
$h=4$ m
The equation of the axis is given by
$$
y=dfrac{f(x)_{max}+f(x)_{min}}{2}
$$
Step 3
3 of 8
c.) The amplitude is
$A=dfrac{7-1}{2}$
$A=3$ m
The amplitude is the distance between the equilibrium position to the maxima or minima which can be calculated as
$$
A=dfrac{f(x)_{max}-f(x)_{min}}{2}
$$
Step 4
4 of 8
d.) By inspection, the graph ranges from $h=1$ to $h=7$, thus, the range can be written as
$$
{ h in bold{R};|1;leq h;leq 7}
$$
The range of $f(x)$ is the set of all possible values of $f(x)$.
Step 5
5 of 8
e.) At 24 seconds, the system has gone 3 cycles and is at minima at 24 seconds. Thus, the next minima is one period after this, which is $24+8=32$seconds
We shall find the time for the next minima after 24 s.
Step 6
6 of 8
f.) From the graph, the first maximum occurs at $t=4$ s. Since the period is $8$ s, the maxima shall occur at
$x = 4+8n$ where $n=0,1,2,3..$
We shall find the time at which Trevor is at the top of the wheel (maxima).
Step 7
7 of 8
g.) Notice that 2 seconds after the minima, the height becomes $4$ m, since the graph is at minima at $t=24;s$, then at 26 seconds, the height must be 4 m. Also, after 6 seconds from the minima, the height is also 4 m, thus, the height at $30$ (which is 6 seconds after 24 s) must be 4 m.
We shall find Trevor’s distance from the ground at 26 and 30 seconds.
Result
8 of 8
a.) period = $8$ s
b.) $h=4$ m
c.) A = 3m
d.) ${ h in bold{R};|1;leq h;leq 7}$
e.) 32 s
f.) $x = 4+8n$ where $n=0,1,2,3..$
g.) 4 m
Exercise 9
Step 1
1 of 3
[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} : T= left| {{x_2} – {x_1}} right| hfill \
end{gathered} ]
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} : T= left| {{x_2} – {x_1}} right| hfill \
end{gathered} ]
Step 2
2 of 3
In this case, period $T=20$, $A=6$ and the equation of the axis is $y=7$. The graph must be something like this
Result
3 of 3
See graph inside.
Exercise 10
Step 1
1 of 3
[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} : T= left| {{x_2} – {x_1}} right| hfill \
end{gathered} ]
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} : T= left| {{x_2} – {x_1}} right| hfill \
end{gathered} ]
Step 2
2 of 3
In this case, $T=4$ , $y_{min}=-2$ and $y_{max}=5$
The amplitude is $A=dfrac{5-(-2)}{2}=3.5$
The equation of axis is $y=dfrac{5+(-2)}{2}=dfrac{3}{2}$
Therefore, the graph must be something like this.
Result
3 of 3
See graph inside.
Exercise 11
Step 1
1 of 4
[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} : T= left| {{x_2} – {x_1}} right| hfill \
end{gathered} ]
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} : T= left| {{x_2} – {x_1}} right| hfill \
end{gathered} ]
Step 2
2 of 4
[begin{gathered}
{text{wheel diameter = 64 cm}} hfill \
{text{speed = 21}}{text{.6 km/h}} hfill \
hfill \
{text{To be able to graph the height above the ground}} hfill \
{text{with respect to time, we must know the period }}T, hfill \
{text{minimum height and maximum height}}{text{. }} hfill \
{text{The minimum height is zero (at the ground) while}} hfill \
{text{the maximum height is the diameter of the wheel (64 cm)}}{text{.}} hfill \
hfill \
{text{To determine the period, we must know how long it takes}} hfill \
{text{to complete one cycle}}{text{. One cycle corresponds to }} hfill \
{text{one complete revolution}}{text{. The distance travelled per}} hfill \
{text{revolution is just the circumference of the wheel }} hfill \
hfill \
C = pi d = pi cdot 64{text{ cm}} = 201.06{text{ cm per cycle}} hfill \
hfill \
{text{The period is the time for one complete cycle}}{text{.}} hfill \
{text{Remember that }} hfill \
{text{distance = speed}} times {text{time}} hfill \
Rightarrow {text{ period = time per cycle = }}frac{{{text{distance per cycle}}}}{{{text{speed}}}} hfill \
hfill \
{text{We shall convert the speed to cm/s to make}} hfill \
{text{the units consistent}} hfill \
{text{21}}{text{.6}}frac{{{text{km}}}}{{text{h}}} cdot left( {frac{{1000{text{ m}}}}{{1{text{ km}}}}} right) cdot left( {frac{{100{text{ cm}}}}{{1{text{ m}}}}} right)left( {frac{{1{text{ h}}}}{{3600{text{ s}}}}} right) = 600{text{ cm/s}} hfill \
hfill \
{text{We can know calculate the period as}} hfill \
{text{ time per cycle = }}frac{{{text{ }}201.06{text{ cm }}}}{{600{text{ cm/s}}}} = 0.335{text{1}};{text{s}} hfill \
{text{With}} hfill \
T = 0.3351 hfill \
{y_{min }} = 0 hfill \
{y_{max }} = 64{text{ cm}} hfill \
{text{The graph for 2 seconds interval can be sketched as follows}} hfill \
hfill \
hfill \
hfill \
hfill \
hfill \
hfill \
end{gathered} ]
{text{wheel diameter = 64 cm}} hfill \
{text{speed = 21}}{text{.6 km/h}} hfill \
hfill \
{text{To be able to graph the height above the ground}} hfill \
{text{with respect to time, we must know the period }}T, hfill \
{text{minimum height and maximum height}}{text{. }} hfill \
{text{The minimum height is zero (at the ground) while}} hfill \
{text{the maximum height is the diameter of the wheel (64 cm)}}{text{.}} hfill \
hfill \
{text{To determine the period, we must know how long it takes}} hfill \
{text{to complete one cycle}}{text{. One cycle corresponds to }} hfill \
{text{one complete revolution}}{text{. The distance travelled per}} hfill \
{text{revolution is just the circumference of the wheel }} hfill \
hfill \
C = pi d = pi cdot 64{text{ cm}} = 201.06{text{ cm per cycle}} hfill \
hfill \
{text{The period is the time for one complete cycle}}{text{.}} hfill \
{text{Remember that }} hfill \
{text{distance = speed}} times {text{time}} hfill \
Rightarrow {text{ period = time per cycle = }}frac{{{text{distance per cycle}}}}{{{text{speed}}}} hfill \
hfill \
{text{We shall convert the speed to cm/s to make}} hfill \
{text{the units consistent}} hfill \
{text{21}}{text{.6}}frac{{{text{km}}}}{{text{h}}} cdot left( {frac{{1000{text{ m}}}}{{1{text{ km}}}}} right) cdot left( {frac{{100{text{ cm}}}}{{1{text{ m}}}}} right)left( {frac{{1{text{ h}}}}{{3600{text{ s}}}}} right) = 600{text{ cm/s}} hfill \
hfill \
{text{We can know calculate the period as}} hfill \
{text{ time per cycle = }}frac{{{text{ }}201.06{text{ cm }}}}{{600{text{ cm/s}}}} = 0.335{text{1}};{text{s}} hfill \
{text{With}} hfill \
T = 0.3351 hfill \
{y_{min }} = 0 hfill \
{y_{max }} = 64{text{ cm}} hfill \
{text{The graph for 2 seconds interval can be sketched as follows}} hfill \
hfill \
hfill \
hfill \
hfill \
hfill \
hfill \
end{gathered} ]
Step 3
3 of 4
Result
4 of 4
See graph inside.
Exercise 12
Step 1
1 of 8
[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} : T= left| {{x_2} – {x_1}} right| hfill \
end{gathered} ]
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} : T= left| {{x_2} – {x_1}} right| hfill \
end{gathered} ]
Step 2
2 of 8
a.) We shall plot the given data as follows.
Step 3
3 of 8
b) Notice that the cycle is repeating in regular intervals. Thus, this is a periodic graph.
Step 4
4 of 8
c) We cab obtain the period from the time difference between two peaks. thus,
period: $T=60-12=48$
The period represents the time for one complete revolution of the spacecraft around the Earth.
Step 5
5 of 8
d) From the graph, we can estimate that at 8 minutes, the spacecraft is roughly 900 km from the earth.
Step 6
6 of 8
e) The spacecraft is farthest from the earth after 12 minutes and every 48 minutes after that. This can be written as
$t=12+48n,;;$ where $n=0,1,2,3…$
Step 7
7 of 8
f) Six orbits would take $6times 48=288$ minutes
Therefore, the domain is
${tin bf{R}$ $| ;;0leq tleq 288}$
Result
8 of 8
a) see graph
b) periodic graph
c) period = 48 min
d) roughly 900 km
e) $t=12+48n$ where $n=0,1,2,3…$
f) ${tin bf{R}$ $| ;;0leq tleq 288}$
Exercise 13
Step 1
1 of 8
[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} = left| {{x_2} – {x_1}} right| hfill \
end{gathered} ]
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} = left| {{x_2} – {x_1}} right| hfill \
end{gathered} ]
Step 2
2 of 8
a) We shall plot the given data as follows.
Step 3
3 of 8
b) The graph appears to be periodic since the pattern repeats at regular intervals.
Step 4
4 of 8
c) In this case
$y_{min}=10$ , $y_{max}=40$
For obtaining the period, since there are several peaks, we shall consider the trough. The distance between two trough (minimum point) is 7 minutes. Thus, the period is 7 minutes.
The equation of the axis is $y=dfrac{10+40}{2}implies y=25$
The amplitude is $A=dfrac{|40-10|}{2}=15$
Step 5
5 of 8
d) The container is filled when the depth goes up. The rate at which the water goes up is the slope of the line between $t=0$ and $t=3$
$$
r=dfrac{40-10}{3-0}=10 text{cm/min}
$$
Step 6
6 of 8
e) The container is drained when the depth goes down. The rate at which the water goes down is the slope of the line between $t=5$ and $t=7$
$r=dfrac{10-40}{7-5}=-15 text{cm/min}$
The negative sign indicates that the depth is decreasing, so the actual rate is simply $15$ cm/min.
Step 7
7 of 8
f) If the cycle continues, the container could NOT be empty since the minimum depth is 10 cm.
Result
8 of 8
a) see graph
b) periodic
c) $T=7$ , $y=25$ , $A=15$
d) 10 cm/min
e) 15 cm/min
f) No
Exercise 14
Step 1
1 of 5
A periodic function is a function that repeats at regular intervals.
Here are some examples of periodic functions.
Step 2
2 of 5
Step 3
3 of 5
Step 4
4 of 5
These are periodic functions because the cycle repeats at regular intervals.
Also, any vertical line touches the graph only once.
Result
5 of 5
Answers can vary. See examples inside.
Exercise 15
Step 1
1 of 4
It starts at 60 cm and stays there for 3 seconds, and then goes down to 30 within 0.5 s, and then stays at the bottom for 2 seconds, then goes back up within 0.5 s.
The plot can be sketched as follows.
Step 2
2 of 4
b) The period is 6 seconds based on the graph.
Step 3
3 of 4
c) The range is the possible values of distance between the minimum and maximum point. In this case,
${dinbf{R;}$ $|;;30leq dleq 60}$
Since Denis repeats this 3 times, the maximum $t$ is $6times 3=18$ s, Therefore
$$
{ tin bold{R};|;0leq tleq 18}
$$
Result
4 of 4
a) see graph
b) 6 seconds
c) range: ${dinbf{R;}$ $|;;30leq dleq 60}$
domain: ${ tin bold{R};|;0leq tleq 18}$
Exercise 16
Step 1
1 of 2
From the given graph, you can see that that initially, the CBR is 40 cm away from the paddle, and remains stationary for 1 second, after which, the paddle moves away by 30 cm within 0.5 s and then goes back to initial position within another 0.5 s. The cycle is repeated until $t=4$ where the paddle stays at $d=40$
Result
2 of 2
See explanation inside.
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