Year 11 2013 Chit Chat Thread (1 Viewer)

[HSC2014]

It is equivalent to solving:

I don't quite understand how you arrived to that D:

 

[kazemagic]

Anyone know how to solve this inequality algebraically?
1/(x-1)(x-3) <= -1

idk how to latex lol

 

[HSC2014]

idk how to latex lol
View attachment 27912

Ah, i understand now - thanks! Didn't know it would factor to that, i just expanded everything which was a mistake haha. You drew your graph wrong btw! (double zero at x = 2)

 

[kazemagic]

Ah, thanks very much. I understand now. You drew your graph wrong btw!

yea it is lol i forgot about the squared sign
(thx aysce for pointing it out to me as well)

 

[Kurosaki]

Solve [(2^x)-4][(2^x)-1] <= 0

How would one even graph that

Remember, when worse comes to worst, try a table of values . By inspection, when x=0, and x=2, the function values will be 0.

 

[SuchSmallHands]

Hey guys can someone please help me with this? I don't think it's worthy of a new thread so I'm asking here.

Basically there's a dot point in the chem syllabus which goes "Identify light, heat and electricity as the common forms of energy that may be released or absorbed during the decomposition or synthesis of substances and identify examples of these changes occurring in everyday life."


So say if you were the one writing notes, would you just write down,
"Light, heat and electricity as the common forms of energy that may be released or absorbed during the decomposition or synthesis of substances", then list down the examples?

That's exactly what I wrote

 

[BLIT2014]

At the moment I don't really regret any of my subject choices, however that may change after finishing all the assessments are over.

 

[Fawun]

And what was it that you did wrong? Was it an essay? Structure problem?

Probably came unprepared to the exam lol.

idk i didn't get feedback. All i got was an exam paper with 7/15 on it LOL

nah i was actually prepared. I memorised heaps of quotes and techniques. It was probably the listening part that threw me off but all I know was that i fucked up that assessment lol.

 

[HSC2014]

Show that when a>0, b<0, a+b<= |a+b|
I don't know how to prove something so conceptually simple :'(

Btw - anybody want to be my math buddy on skype? I ask a lot of random questions and i don't have a good source of knowledge unless it's here...

 

[Kurosaki]

Show that when a>0, b<0, a+b<= |a+b|
I don't know how to prove something so conceptually simple :'(

Btw - anybody want to be my math buddy on skype? I ask a lot of random questions and i don't have a good source of knowledge unless it's here...

damn I thought about squaring them but that wouldn't work.
Sure you can PM me with your Skype but i'm not that good haha

 

[Kurosaki]

Idk how to prove it haha.
I don't think this would be accepted as a proof lol.

If a+b is negative, then |a+b| is positive, a+b is greater than or equal to zero, |a+b| = a+b. therefore, a+b is never greater than |a+b|, so a+b <=|a+b|

 

[cricketfan1997]

Show that when a>0, b<0, a+b<= |a+b|
I don't know how to prove something so conceptually simple :'(

Btw - anybody want to be my math buddy on skype? I ask a lot of random questions and i don't have a good source of knowledge unless it's here...

it says show, not prove.
just sub in a point for a and b and you should get full marks

a = 1, b = -2
-1 <= 1
LHS = RHS so when a>0, b<0, a+b<= |a+b|

 

[cricketfan1997]

or if you use a = 1, b = -1, one way to show it works is:
-a + a < = |-a+a|

0<= 0 so i guess it works

 

[RealiseNothing]

it says show, not prove.
just sub in a point for a and b and you should get full marks

a = 1, b = -2
-1 <= 1
LHS = RHS so when a>0, b<0, a+b<= |a+b|

You can't just sub in points, you have to show it's true for all a>0 and b<0.

 

[HSC2014]

@cricketfan1997
I use the words proof/show loosely. I still consider a show a proof, but yes in a proof question i would not do a 'show' (if that made any sense LOL).

The full question says to consider the cases (a>0,b>0),(a>0,b<0),(a<0,b<0),(a<0,b>0)
and show that a+b <= |a+b|

So basically if all cases satisfy the inequality, than it is true for all real a,b. I'm stuck on the second one (a>0,b<0) as it is quite hard to show that all a>0 and b<0 satisfy the inequality. :{
Either the question is stupid (pointless) or my math language isn't as good as I thought it was as i can't put it into a "proof." Or am i allowed to substitute numbers here as it only says to "consider" the cases...

 

[RealiseNothing]

@cricketfan1997
I use the words proof/show loosely. I still consider a show a proof, but yes in a proof question i would not do a 'show' (if that made any sense LOL).

The full question says to consider the cases (a>0,b>0),(a>0,b<0),(a<0,b<0),(a<0,b>0)
and show that a+b <= |a+b|

So basically if all cases satisfy the inequality, than it is true for all real a,b. I'm stuck on the second one (a>0,b<0) as it is quite hard to show that all a>0 and b<0 satisfy the inequality. :{
Either the question is stupid (pointless) or my math language isn't as good as I thought it was as i can't put it into a "proof." Or am i allowed to substitute numbers here as it only says to "consider" the cases...

Nope

 

[HSC2014]

Nope

Are you hinting at the fact the cases don't consider when a,b = 0 LOL. Otherwise I'm lost :'( /cry

 

[RealiseNothing]

Are you hinting at the fact the cases don't consider when a,b = 0 LOL. Otherwise I'm lost :'( /cry

I am hinting at that haha.

 

[HSC2014]

Consider |a| = |b|, |a| < |b| and |a| > |b| where a>0, b<0

When |a| = |b|
a+b = 0, and |a+b| = a+b = 0
When |a| < |b|
a+b < 0 and thus |a+b| = -(a+b) which is > 0
When |a| > |b|
a+b > 0 and thus |a+b| = a+b

Clearly, a + b <= |a+b| for all a>0, b<0

POMOPASMDPOSAMDPOASMDPOSAMDSPODMSPO PLEASE TELL ME THIS IS MATHEMATICALLY CORRECT. SPENT LIKE ALL NIGHT ON IT *OFF TO CELEBRATE EVEN THOUGH IT MIGHT BE WRONG

 

[RealiseNothing]

That should be fine.