To be honest i dont even know if this is geometry (apologies if it is't). Well there is question 11 (everything, since it all goes together) which i dont quite understand. If you can just thoroughly explain it to me (don't quite need the answer), i'd be very grateful for your help.
year 10 advanced mathematics geometry (i think) explaination
- Thread starter spiary
- Start date
To be honest i dont even know if this is geometry (apologies if it is't). Well there is question 11 (everything, since it all goes together) which i dont quite understand. If you can just thoroughly explain it to me (don't quite need the answer), i'd be very grateful for your help.
- Joined
- Feb 4, 2004
- Messages
- 16,582
It is not reasonably feasible to attempt here to provide you with the weeks or months of classroom instruction which apparently are missing (since you are not familiar enough with the material to be sure even of the topic). How was it that you were assigned this homework?
This is to see our overall knowledge, and i've just never seen this type of question. If you want to be rude of me requesting for explanation, i'd rather have you to just go away and have others help me with this question. Its not like i'm asking for the answers, i just want it explained so i can learn it.
Hello spiary:
Nobody has treated you rudely.
You told us that the purpose of this handout is to test your existing knowledge.
If your existing knowledge on question 11 (understanding xy-coordinates, how to read a graph) is nothing, then that is what you should tell the person who gave you the handout.
It's best to be honest, with placement exams. Otherwise, you might end up in a class for which you are unprepared.
We do not type up lessons, on these boards, because plenty of lessons already exist on the Internet. We mainly help students who are able to begin an exercise, but get stuck. They show us their efforts so far, and we provide hints and guidance based on their work.
If you would like lessons, and you are not able to locate any using Google, please let us know. We can post some links for you.
Cheers :cool:
PS: Did you happen to see the Read Before Posting info? Please check it out.
Nobody has treated you rudely.
You told us that the purpose of this handout is to test your existing knowledge.
If your existing knowledge on question 11 (understanding xy-coordinates, how to read a graph) is nothing, then that is what you should tell the person who gave you the handout.
It's best to be honest, with placement exams. Otherwise, you might end up in a class for which you are unprepared.
We do not type up lessons, on these boards, because plenty of lessons already exist on the Internet. We mainly help students who are able to begin an exercise, but get stuck. They show us their efforts so far, and we provide hints and guidance based on their work.
If you would like lessons, and you are not able to locate any using Google, please let us know. We can post some links for you.
Cheers :cool:
PS: Did you happen to see the Read Before Posting info? Please check it out.
There was no rudeness in the reply. It is simply necessary that tutors understand exactly where a student is mathematically in order to give coherent responses. And it was also pointing out our limitations as tutors; Your problems cover a *very* wide variety of area: Algebra, Geometry, Physics, data analysis...
The problems are covering many different topics. No simple explanation is possible without our knowing what *you* know about each topic.
year 10 advanced mathematics book help.
So there is this diagram, and my question is to calculate the average speed, when i had a look at the back of the book where the answers were, the answer as 3ms^-1 (^= to the power of). I dont know how it turned out to be an answer with scientific notation.
A description of the diagram is: a particle is moving on a standard, horizontal number line. the diagram shows its distance x, in metres, from the origin after t seconds. So can anyone tell me how to calculate the average speed for the answer to be 3ms^-1? I'm trying to learn these stuff in preparation for my year.
So there is this diagram, and my question is to calculate the average speed, when i had a look at the back of the book where the answers were, the answer as 3ms^-1 (^= to the power of). I dont know how it turned out to be an answer with scientific notation. A description of the diagram is: a particle is moving on a standard, horizontal number line. the diagram shows its distance x, in metres, from the origin after t seconds. So can anyone tell me how to calculate the average speed for the answer to be 3ms^-1? I'm trying to learn these stuff in preparation for my year.
- Joined
- Feb 4, 2004
- Messages
- 16,582
It isn't "scientific" notation; it's exponential notation. The exercise is a simple application of the "slope" concept within the context of word problems.
It'll probably be a lot simpler, and more effective, to wait until your class reaches this material, in order and with all the necessary background concepts, rather than attempting it, piecemeal, from out-of-context example exercises.
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,962
Hello spiary:
You posted this exercise on the boards three times. (Your posts have been collected into the original thread.)
I sent you a private message, requesting you to not post duplicate threads. Did you see that message?
:cool:
You posted this exercise on the boards three times. (Your posts have been collected into the original thread.)
I sent you a private message, requesting you to not post duplicate threads. Did you see that message?
:cool:
Because it's a position graph, Denis. The graph is not a picture of the particle's actual route. (The particle is traveling on the Real number line -- from right to left -- approaching zero.)
The graph tells us how far away from zero the particle is at time t.
The distance from zero is measured on the vertical axis of the graph, and time is measured on the horizontal axis.
The particle's distance from zero is a function of time.
The particle's velocity is constant, as the position graph is linear.
No, that's backwards. The change in position is 12 meters in the negative direction, and that position change happens over 4 seconds.
Cheers :cool:
Last edited:
Average speed for an object traveling at constant velocity is the absolute value of (change in position)/(elapsed time)
The graph shows that the particle's position changed by -12 meters, and it took 4 seconds for that to happen.
The change in position is negative because the particle is moving in the negative direction. (If the particle was moving from left to right, instead, then the change in position would be +12.)
The elapsed time is 4 seconds.
|-12/4| = 3
3ms^-1 means "3 meters per second".
Units which are ratios are often written using the notation above.
The units "meters per second" are m/s -- that's a ratio.
Another way (besides writing a ratio) to show that the unit for "seconds" is in a denominator is s^(-1)
(Stapel already told you that this is "exponential" notation.)
ms^(-1) means m/s
Manipulating units is a part of "dimensional analysis". You may google that, for additional explanations, lessons, and examples.
Last edited:
D
Deleted member 4993
Guest
Seems to be something wrong with your diagram.
If the distance travelled is along the horizontal line, why is it shown along the vertical line?
Seems to me that what is meant is 4 metres travelled in 12 seconds.
So average speed is then 4/12 = 1/3, which is 3^(-1).
If I'm wrong, I'll go stand in the corner
The diagram can be interpreted as:
When t = 0, x = 12
when t = 4, x = 0
You are supposed to be in the corner already - who let you out??? Must be Jeff!
Last edited by a moderator:
Oh good.
I went back and re-read the OP. Perhaps, my explanations went too deep.
That course defines average speed as (distance traveled)/(elapsed time).
In other words, they seem to be using a "simplified" version of "change in position".
"Distance traveled" is always a positive number, so they've dropped the directional information. In other words, they're not bothering to explain vector quantities. They're saying that moving 12 meters in the negative direction is simply 12 meters.
I suppose this interpretation allows them to avoid the notion that speed is the absolute value of velocity.
I've seen other courses which "skip over" the concept of vector quantities versus scalar numbers. I guess that's okay, if you're trying to keep things simple, as long as students don't take physics. :cool: