• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

how big a change is yr 11 maths to yr 10 maths (2 Viewers)

Shadowdude

Cult of Personality
Joined
Sep 19, 2009
Messages
12,145
Gender
Male
HSC
2010
In the solution I have (5-6 lines), there's no binomial theorem. Keep at it!

There's a trick involved.

WHichwebsite do u use, i cannot find a good one.
To get that? Latex equation editor.
 
Last edited:

powlmao

Banned
Joined
Feb 17, 2011
Messages
3,970
Location
Hogwarts
Gender
Male
HSC
2012
<a href="http://www.codecogs.com/eqnedit.php?latex=f(x)=\left\{\begin{matrix} 3x-2^x & for & x>-1 & & \\ 7\pi - x^2 & for & 0<x>10 & & \\ \pi^x -3x^4 & for & x>0 & & \end{matrix}\right.\displaystyle" target="_blank"><img src="http://latex.codecogs.com/gif.latex?f(x)=\left\{\begin{matrix} 3x-2^x & for & x>-1 & & \\ 7\pi - x^2 & for & 0<x>10 & & \\ \pi^x -3x^4 & for & x>0 & & \end{matrix}\right.\displaystyle" title="f(x)=\left\{\begin{matrix} 3x-2^x & for & x>-1 & & \\ 7\pi - x^2 & for & 0<x>10 & & \\ \pi^x -3x^4 & for & x>0 & & \end{matrix}\right.\displaystyle" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{f(1) @plus; (f4)}{f(6)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{f(1) + (f4)}{f(6)}" title="\frac{f(1) + (f4)}{f(6)}" /></a>

Answer
 
Last edited:

bleakarcher

Active Member
Joined
Jul 8, 2011
Messages
1,509
Gender
Male
HSC
2013
lol i havent seen a question like this b4 which is good. i dont really know how to do it though
 

Alkanes

Active Member
Joined
May 20, 2010
Messages
1,417
Gender
Male
HSC
2012
<a href="http://www.codecogs.com/eqnedit.php?latex=f(x)=\left\{\begin{matrix} 3x-2^x & for & x>-1 & & \\ 7\pi - x^2 & for & 0<x>10 & & \\ \pi^x -3x^4 & for & x>0 & & \end{matrix}\right.\displaystyle" target="_blank"><img src="http://latex.codecogs.com/gif.latex?f(x)=\left\{\begin{matrix} 3x-2^x & for & x>-1 & & \\ 7\pi - x^2 & for & 0<x>10 & & \\ \pi^x -3x^4 & for & x>0 & & \end{matrix}\right.\displaystyle" title="f(x)=\left\{\begin{matrix} 3x-2^x & for & x>-1 & & \\ 7\pi - x^2 & for & 0<x>10 & & \\ \pi^x -3x^4 & for & x>0 & & \end{matrix}\right.\displaystyle" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{f(1) @plus; (f4)}{f(6)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{f(1) + (f4)}{f(6)}" title="\frac{f(1) + (f4)}{f(6)}" /></a>


They are functions.
Wtf kind of condition is that on the 2nd line?
 

AAEldar

Premium Member
Joined
Apr 5, 2010
Messages
2,246
Gender
Male
HSC
2011
Can someone double check my working, first time on the equation thing and i may of written something wrong.
Your function conditions make absolutely no sense. How can there be two restrictions for the same value? Greater than -1 and greater than 0 yield the first and third functions, and the middle restriction makes no sense either. Really need to clear that up :\

The integrals Shadowdude posted - I shall also have a crack at them, fair sure I've done before though.
 

powlmao

Banned
Joined
Feb 17, 2011
Messages
3,970
Location
Hogwarts
Gender
Male
HSC
2012
Your function conditions make absolutely no sense. How can there be two restrictions for the same value? Greater than -1 and greater than 0 yield the first and third functions, and the middle restriction makes no sense either. Really need to clear that up :\

The integrals Shadowdude posted - I shall also have a crack at them, fair sure I've done before though.
I see what i have done wrong, i typed it wrong in the first place and went from there.
 
Last edited:

Shadowdude

Cult of Personality
Joined
Sep 19, 2009
Messages
12,145
Gender
Male
HSC
2010
is t in termz of u

t=[u+sqrt(4+u^2)]/2
If you're solving for t, you're taking the long way. Quite simple and elegant is that solution.

test

Your function conditions make absolutely no sense. How can there be two restrictions for the same value? Greater than -1 and greater than 0 yield the first and third functions, and the middle restriction makes no sense either. Really need to clear that up :\

The integrals Shadowdude posted - I shall also have a crack at them, fair sure I've done before though.
Those integrals are from MATH1131 work.
 

powlmao

Banned
Joined
Feb 17, 2011
Messages
3,970
Location
Hogwarts
Gender
Male
HSC
2012

Quick question, the way i answered was right? (even though the question was typed in wrong)
 

AAEldar

Premium Member
Joined
Apr 5, 2010
Messages
2,246
Gender
Male
HSC
2011
Quick question, the way i answered was right? (even though the question was typed in wrong)
Yes. There were 3 different conditions, so you find the one that satisfies the number that you put in and solve it just the way you did.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top