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

Shadowdude

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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.
 
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powlmao

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<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
 
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bleakarcher

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lol i havent seen a question like this b4 which is good. i dont really know how to do it though
 

Alkanes

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<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

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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

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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.
 
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Shadowdude

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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

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

AAEldar

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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.
 

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