Equazione cartesiana 10
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
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
footnoteN. Rusca Considera la circonferenza mathcalC di equazione x^-x+y^-y+. abclist abc Qual è la misura della corda avente come punto medio? abc Scrivi le equazioni delle rette tangenti alla circonferenza e parallele al diametro avente un estremo nel punto . abclist
Solution:
Riscriviamo la quadrica come x-^+y-^ dunque il centro è e il raggio è sqrt. abclist abc La distanza tra il punto medio della corda e il centro della circonferenza è sqrt dunque per Pitagora la lunghezza della corda è sqrtsqrt^-sqrt^sqrt. abc La retta che forma il diametro passante per è data dall'equazione x-y-. Dunque le rette parallele tangenti avranno le equazioni frac|- +d|sqrtsqrtiff din- dunque x-y e x-y- abclist
footnoteN. Rusca Considera la circonferenza mathcalC di equazione x^-x+y^-y+. abclist abc Qual è la misura della corda avente come punto medio? abc Scrivi le equazioni delle rette tangenti alla circonferenza e parallele al diametro avente un estremo nel punto . abclist
Solution:
Riscriviamo la quadrica come x-^+y-^ dunque il centro è e il raggio è sqrt. abclist abc La distanza tra il punto medio della corda e il centro della circonferenza è sqrt dunque per Pitagora la lunghezza della corda è sqrtsqrt^-sqrt^sqrt. abc La retta che forma il diametro passante per è data dall'equazione x-y-. Dunque le rette parallele tangenti avranno le equazioni frac|- +d|sqrtsqrtiff din- dunque x-y e x-y- abclist
Meta Information
Exercise:
footnoteN. Rusca Considera la circonferenza mathcalC di equazione x^-x+y^-y+. abclist abc Qual è la misura della corda avente come punto medio? abc Scrivi le equazioni delle rette tangenti alla circonferenza e parallele al diametro avente un estremo nel punto . abclist
Solution:
Riscriviamo la quadrica come x-^+y-^ dunque il centro è e il raggio è sqrt. abclist abc La distanza tra il punto medio della corda e il centro della circonferenza è sqrt dunque per Pitagora la lunghezza della corda è sqrtsqrt^-sqrt^sqrt. abc La retta che forma il diametro passante per è data dall'equazione x-y-. Dunque le rette parallele tangenti avranno le equazioni frac|- +d|sqrtsqrtiff din- dunque x-y e x-y- abclist
footnoteN. Rusca Considera la circonferenza mathcalC di equazione x^-x+y^-y+. abclist abc Qual è la misura della corda avente come punto medio? abc Scrivi le equazioni delle rette tangenti alla circonferenza e parallele al diametro avente un estremo nel punto . abclist
Solution:
Riscriviamo la quadrica come x-^+y-^ dunque il centro è e il raggio è sqrt. abclist abc La distanza tra il punto medio della corda e il centro della circonferenza è sqrt dunque per Pitagora la lunghezza della corda è sqrtsqrt^-sqrt^sqrt. abc La retta che forma il diametro passante per è data dall'equazione x-y-. Dunque le rette parallele tangenti avranno le equazioni frac|- +d|sqrtsqrtiff din- dunque x-y e x-y- abclist
Contained in these collections
| Title | Creator | Matched on |
|---|---|---|
| Equazione cartesiana 1 | gl | tagstitle |
| Equazione cartesiana 2 | gl | tagstitle |
| Equazione cartesiana 3 | gl | tagstitle |
| Equazione cartesiana 4 | gl | tagstitle |
| Equazione cartesiana 5 | gl | tagstitle |
Similar exercises (17)
| Title | Creator | Matched on |
|---|---|---|
| Equazione cartesiana 1 | gl | tagstitle |
| Equazione cartesiana 2 | gl | tagstitle |
| Equazione cartesiana 3 | gl | tagstitle |
| Equazione cartesiana 4 | gl | tagstitle |
| Equazione cartesiana 5 | gl | tagstitle |
| Equazione cartesiana 6 | gl | tagstitle |
| Equazione cartesiana 7 | gl | tagstitle |
| Equazione cartesiana 8 | gl | tagstitle |
| Equazione cartesiana 9 | gl | tagstitle |
| Equazione cartesiana 11 | gl | tagstitle |
| Equazione cartesiana 12 | gl | tagstitle |
| Equazione cartesiana 13 | gl | tagstitle |
| Equazione cartesiana 14 | gl | tagstitle |
| Equazione cartesiana 15 | gl | tagstitle |
| Equazione cartesiana 16 | gl | tagstitle |
| Equazione cartesiana 17 | gl | tagstitle |
| Equazione cartesiana 18 | gl | tagstitle |

