Strecke
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
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Exercise:
In dieser Abbildung mit dem Radius r overlineAM r ist die Strecke overlineAS x gesucht. center tikzpicturescale. coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; draw thick A -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw thick M coordinate C -- ++resula:r coordinate D; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center
Solution:
Die gesuchte länge overlineAS lässt sich durch den Satz des Pytagoras berechnen. a sqrtb^ + c^ overlineAS sqrtoverlineAB^ +overlineBS^ Berechnung von overlineAB: * overlineAB overlineAM r em resulf * Berechnung von overlineBS: Die Strecke overlineBS ist zusammen gesetzt aus der Strecke overlineBy und der Strecke overlineyS. center tikzpicturescale. coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; coordinate labelright:yy at rresulg; draw thickred resuleresuld -- y; draw thickred r -- y; draw thick A -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw thick M coordinate C -- ++resula:r coordinate D; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; drawdashed thick -resull -- D nodemidway leftoverlineBy; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center Um die Strecke overlineBy auszurechnen verwen wir den Sinus von a^circ. * overlineBy sina^circ rem resuld * Um die Strecke overlineyS zu berechnen brauchen wir die Länge der roten Linie und den Winkel alpha. center tikzpicturescale . coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; coordinate labelright:yy at rresulg; draw thickred resuleresuld -- y; draw thickred r -- y; draw thick A -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw thick M coordinate C -- ++resula:r coordinate D; filldrawcolorred fillred!!white D--D+. arc :resulj:.--cycle; nodered at D+. alpha; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center Diese Winkel sind gleich weil sie Scheitelwinkel zueinander sind. Die Tangente steht rechtwinklig zu dem Radius r daher ist alpha resulj^circ * alpha ^circ - a^circem resulj^circ * Die Länge der roten Linie l_R ist zusammengesetzt aus dem Radius r r und aus der grünen Linie. center tikzpicturescale. coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; coordinate labelright:yy at rresulg; draw thickred resuleresuld -- y; draw thickred r -- y; draw thick A -- -resull; draw thick M -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw fadedgreen thick M -- -resull nodeleftbelow l_G; drawfadedgreen dashed thick -resull -- D; draw thick M coordinate C -- ++resula:r coordinate D; filldrawcolorred fillred!!white D--D+. arc :resulj:.--cycle; nodered at D+. alpha; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center Die grüne Linie l_G wird mit dem Kosinus von a^circ berechnet. * l_G cosa^circ rem cosa^circ rem resull * Für die Berechnung von der Strecke l_R rechnen wir die Linie l_G plus den Radius r r * l_R r + L_Gem r + cosa^circ rem resuln * Für die Berechnung von der Strecke overlineyS verwen wir den Tangens von alpha resulj^circ * overlineyS tanalpha l_Rem tanresulj^circ resulnem resulm * Da wir nun overlineBy und overlineyS kennen können wir overlineBS berechnen. * overlineBS overlineBy + overlineySem resuld + resulmem resulo * Nun können wir alles in die Gleichung vom Anfang einsetzten. * overlineAS sqrtoverlineAB^ +overlineBS^em sqrtresulf^ +resulo^em resulqem * Also erhalten wir als Ergebniss x approx numround-modeplaces round-precisionresulq
In dieser Abbildung mit dem Radius r overlineAM r ist die Strecke overlineAS x gesucht. center tikzpicturescale. coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; draw thick A -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw thick M coordinate C -- ++resula:r coordinate D; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center
Solution:
Die gesuchte länge overlineAS lässt sich durch den Satz des Pytagoras berechnen. a sqrtb^ + c^ overlineAS sqrtoverlineAB^ +overlineBS^ Berechnung von overlineAB: * overlineAB overlineAM r em resulf * Berechnung von overlineBS: Die Strecke overlineBS ist zusammen gesetzt aus der Strecke overlineBy und der Strecke overlineyS. center tikzpicturescale. coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; coordinate labelright:yy at rresulg; draw thickred resuleresuld -- y; draw thickred r -- y; draw thick A -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw thick M coordinate C -- ++resula:r coordinate D; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; drawdashed thick -resull -- D nodemidway leftoverlineBy; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center Um die Strecke overlineBy auszurechnen verwen wir den Sinus von a^circ. * overlineBy sina^circ rem resuld * Um die Strecke overlineyS zu berechnen brauchen wir die Länge der roten Linie und den Winkel alpha. center tikzpicturescale . coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; coordinate labelright:yy at rresulg; draw thickred resuleresuld -- y; draw thickred r -- y; draw thick A -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw thick M coordinate C -- ++resula:r coordinate D; filldrawcolorred fillred!!white D--D+. arc :resulj:.--cycle; nodered at D+. alpha; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center Diese Winkel sind gleich weil sie Scheitelwinkel zueinander sind. Die Tangente steht rechtwinklig zu dem Radius r daher ist alpha resulj^circ * alpha ^circ - a^circem resulj^circ * Die Länge der roten Linie l_R ist zusammengesetzt aus dem Radius r r und aus der grünen Linie. center tikzpicturescale. coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; coordinate labelright:yy at rresulg; draw thickred resuleresuld -- y; draw thickred r -- y; draw thick A -- -resull; draw thick M -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw fadedgreen thick M -- -resull nodeleftbelow l_G; drawfadedgreen dashed thick -resull -- D; draw thick M coordinate C -- ++resula:r coordinate D; filldrawcolorred fillred!!white D--D+. arc :resulj:.--cycle; nodered at D+. alpha; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center Die grüne Linie l_G wird mit dem Kosinus von a^circ berechnet. * l_G cosa^circ rem cosa^circ rem resull * Für die Berechnung von der Strecke l_R rechnen wir die Linie l_G plus den Radius r r * l_R r + L_Gem r + cosa^circ rem resuln * Für die Berechnung von der Strecke overlineyS verwen wir den Tangens von alpha resulj^circ * overlineyS tanalpha l_Rem tanresulj^circ resulnem resulm * Da wir nun overlineBy und overlineyS kennen können wir overlineBS berechnen. * overlineBS overlineBy + overlineySem resuld + resulmem resulo * Nun können wir alles in die Gleichung vom Anfang einsetzten. * overlineAS sqrtoverlineAB^ +overlineBS^em sqrtresulf^ +resulo^em resulqem * Also erhalten wir als Ergebniss x approx numround-modeplaces round-precisionresulq
Meta Information
Exercise:
In dieser Abbildung mit dem Radius r overlineAM r ist die Strecke overlineAS x gesucht. center tikzpicturescale. coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; draw thick A -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw thick M coordinate C -- ++resula:r coordinate D; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center
Solution:
Die gesuchte länge overlineAS lässt sich durch den Satz des Pytagoras berechnen. a sqrtb^ + c^ overlineAS sqrtoverlineAB^ +overlineBS^ Berechnung von overlineAB: * overlineAB overlineAM r em resulf * Berechnung von overlineBS: Die Strecke overlineBS ist zusammen gesetzt aus der Strecke overlineBy und der Strecke overlineyS. center tikzpicturescale. coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; coordinate labelright:yy at rresulg; draw thickred resuleresuld -- y; draw thickred r -- y; draw thick A -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw thick M coordinate C -- ++resula:r coordinate D; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; drawdashed thick -resull -- D nodemidway leftoverlineBy; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center Um die Strecke overlineBy auszurechnen verwen wir den Sinus von a^circ. * overlineBy sina^circ rem resuld * Um die Strecke overlineyS zu berechnen brauchen wir die Länge der roten Linie und den Winkel alpha. center tikzpicturescale . coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; coordinate labelright:yy at rresulg; draw thickred resuleresuld -- y; draw thickred r -- y; draw thick A -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw thick M coordinate C -- ++resula:r coordinate D; filldrawcolorred fillred!!white D--D+. arc :resulj:.--cycle; nodered at D+. alpha; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center Diese Winkel sind gleich weil sie Scheitelwinkel zueinander sind. Die Tangente steht rechtwinklig zu dem Radius r daher ist alpha resulj^circ * alpha ^circ - a^circem resulj^circ * Die Länge der roten Linie l_R ist zusammengesetzt aus dem Radius r r und aus der grünen Linie. center tikzpicturescale. coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; coordinate labelright:yy at rresulg; draw thickred resuleresuld -- y; draw thickred r -- y; draw thick A -- -resull; draw thick M -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw fadedgreen thick M -- -resull nodeleftbelow l_G; drawfadedgreen dashed thick -resull -- D; draw thick M coordinate C -- ++resula:r coordinate D; filldrawcolorred fillred!!white D--D+. arc :resulj:.--cycle; nodered at D+. alpha; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center Die grüne Linie l_G wird mit dem Kosinus von a^circ berechnet. * l_G cosa^circ rem cosa^circ rem resull * Für die Berechnung von der Strecke l_R rechnen wir die Linie l_G plus den Radius r r * l_R r + L_Gem r + cosa^circ rem resuln * Für die Berechnung von der Strecke overlineyS verwen wir den Tangens von alpha resulj^circ * overlineyS tanalpha l_Rem tanresulj^circ resulnem resulm * Da wir nun overlineBy und overlineyS kennen können wir overlineBS berechnen. * overlineBS overlineBy + overlineySem resuld + resulmem resulo * Nun können wir alles in die Gleichung vom Anfang einsetzten. * overlineAS sqrtoverlineAB^ +overlineBS^em sqrtresulf^ +resulo^em resulqem * Also erhalten wir als Ergebniss x approx numround-modeplaces round-precisionresulq
In dieser Abbildung mit dem Radius r overlineAM r ist die Strecke overlineAS x gesucht. center tikzpicturescale. coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; draw thick A -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw thick M coordinate C -- ++resula:r coordinate D; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center
Solution:
Die gesuchte länge overlineAS lässt sich durch den Satz des Pytagoras berechnen. a sqrtb^ + c^ overlineAS sqrtoverlineAB^ +overlineBS^ Berechnung von overlineAB: * overlineAB overlineAM r em resulf * Berechnung von overlineBS: Die Strecke overlineBS ist zusammen gesetzt aus der Strecke overlineBy und der Strecke overlineyS. center tikzpicturescale. coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; coordinate labelright:yy at rresulg; draw thickred resuleresuld -- y; draw thickred r -- y; draw thick A -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw thick M coordinate C -- ++resula:r coordinate D; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; drawdashed thick -resull -- D nodemidway leftoverlineBy; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center Um die Strecke overlineBy auszurechnen verwen wir den Sinus von a^circ. * overlineBy sina^circ rem resuld * Um die Strecke overlineyS zu berechnen brauchen wir die Länge der roten Linie und den Winkel alpha. center tikzpicturescale . coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; coordinate labelright:yy at rresulg; draw thickred resuleresuld -- y; draw thickred r -- y; draw thick A -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw thick M coordinate C -- ++resula:r coordinate D; filldrawcolorred fillred!!white D--D+. arc :resulj:.--cycle; nodered at D+. alpha; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center Diese Winkel sind gleich weil sie Scheitelwinkel zueinander sind. Die Tangente steht rechtwinklig zu dem Radius r daher ist alpha resulj^circ * alpha ^circ - a^circem resulj^circ * Die Länge der roten Linie l_R ist zusammengesetzt aus dem Radius r r und aus der grünen Linie. center tikzpicturescale. coordinate labelbelow left:AA at -r; coordinate labelbelow right:BB at r; coordinate labelbelow:MM at ; coordinate labelright:SS at rresulc; coordinate labelright:yy at rresulg; draw thickred resuleresuld -- y; draw thickred r -- y; draw thick A -- -resull; draw thick M -- B; draw thick B -- S; draw thickA -- S nodemidway belowx; draw A arcstart angle angle radiusr; draw fadedgreen thick M -- -resull nodeleftbelow l_G; drawfadedgreen dashed thick -resull -- D; draw thick M coordinate C -- ++resula:r coordinate D; filldrawcolorred fillred!!white D--D+. arc :resulj:.--cycle; nodered at D+. alpha; drawthick D -- ++resulb:; drawthick D -- ++resulb:-resulc; drawthick - arcstart angle angleresula radiuscm; node at -.. a^circ; drawthick resulh arcstart angle angle radiuscm; filldraw black resuli. circle pt; tikzpicture center Die grüne Linie l_G wird mit dem Kosinus von a^circ berechnet. * l_G cosa^circ rem cosa^circ rem resull * Für die Berechnung von der Strecke l_R rechnen wir die Linie l_G plus den Radius r r * l_R r + L_Gem r + cosa^circ rem resuln * Für die Berechnung von der Strecke overlineyS verwen wir den Tangens von alpha resulj^circ * overlineyS tanalpha l_Rem tanresulj^circ resulnem resulm * Da wir nun overlineBy und overlineyS kennen können wir overlineBS berechnen. * overlineBS overlineBy + overlineySem resuld + resulmem resulo * Nun können wir alles in die Gleichung vom Anfang einsetzten. * overlineAS sqrtoverlineAB^ +overlineBS^em sqrtresulf^ +resulo^em resulqem * Also erhalten wir als Ergebniss x approx numround-modeplaces round-precisionresulq
Contained in these collections
| Title | Creator | Matched on |
|---|---|---|
| Winkel in geometrischer Figur | uz | tags |
| Entfernung von Punkten | uz | tags |
| Strecke in Figur | uz | tags |
| Strecke in Figur berechnen | uz | tags |
| Winkel in Dreieck | uz | tags |
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| Title | Creator | Matched on |
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| Winkel in geometrischer Figur | uz | tags |
| Entfernung von Punkten | uz | tags |
| Strecke in Figur | uz | tags |
| Strecke in Figur berechnen | uz | tags |
| Winkel in Dreieck | uz | tags |
| Gleichschenkliges Dreieck | uz | tags |
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