how big a change is yr 11 maths to yr 10 maths (3 Viewers)

[SpiralFlex]

how? i dont even know how to even begin to attempt.

Actually, hold up. I think either I copied the question from the board or my working is wrong. Doesn't match with answer.

 

[bleakarcher]

lol i worked it out

 

[SpiralFlex]

lol i worked it out

Post the solution.

 

[bleakarcher]

naa i fuked up

 

[SpiralFlex]

Language young one. Now I shall have to give my tutor a visit in a weeks time. Prehaps a HSCer can answer it.

 

[bleakarcher]

i will be bak lol

 

[AAEldar]

Yes, the final HSC marks - but to generalise it to everything is wrong.

Like my school said that and implied that the 'no half marks' thing also applied to raw marks. I think I can recall teachers refusing to give me half a mark because of that reason - and especially in English, where I was trailing the first-placed person - every portion of a mark would've helped me narrow that oh-so-very-important-assessment-mark gap.

My Physics teacher, whose marked for the HSC (I swear every teacher has..) says that they don't award half marks. They discuss what will get each mark before they start marking and then if they're not sure whether something should get a mark they refer it to the Senior Marker.

Then again this could only apply to the one subject.. who knows.

 

[cutemouse]

I think I've read in the examiner's report that half marks were not awarded.

 

[Shadowdude]

I have raw marks to show that they are.

I looked through and they give them in Legal Studies and English for sure. So perhaps some subjects don't have that.

 

[AAEldar]

Ship A is sailing due north at 24 km/h and ship B is sailing due west at 16 km/h. At 2 pm, B is 104 km due east of A. When will they be nearest and by how near?

Is the answer 4 pm at 86.5 km apart?

 

[bleakarcher]

Spiralflex, i got it.

Let the horizontal distance between the ships after a time t after 2pm be (104-x)
Let the vertical distance between between the ships after a time t be y
Let the average distance between the ships after a time t be z
By the Pythagorean theorem:
z^2=(104-x)^2+y^2
z^2=10816-208x+x^2+y^2
As x=16t, y=24t,
z^2=10816-208(16t)+(16t)^2+(24t)^2
z^2=10816-3328t+832t^2
Differentiating w.r.t. time t:
2z(dz/dt)=1664t-3328
Minimum distance occurs when dz/dt=0,
1664t=3228
t=2
Now, z^2=10816-3328+832t^2
When t=2,
z^2=10816-3328(2)+832(2)^2=7488 km
z=86.53 km
Therefore, the minimum distance of 86.53 km occurs between the ships when t=2 (ie at 4pm)

 

[SpiralFlex]

Well done. How did you know that was your answer and the other working wasn't?

 

[bleakarcher]

wat lol

 

[SpiralFlex]

wat lol

You got another solution before? How did you know this one was correct?

 

[bleakarcher]

lol coz 7448 is not the minimum distance lol

 

[SpiralFlex]

lol coz 7448 is not the minimum distance lol

Not that. The other one you said you stuffed up.

 

[AAEldar]

You got another solution before? How did you know this one was correct?

Because I posted right before him with the answer? Haha.

 

[bleakarcher]

coz i got minimum distance occuring when t=0 which is wrong

 

[SpiralFlex]

Because I posted right before him with the answer? Haha.

lol

coz i got minimum distance occuring when t=0 which is wrong

Never mind...

 

[bleakarcher]

lol i didnt know. i went to have my dinner.