how big a change is yr 11 maths to yr 10 maths (1 Viewer)

[Shadowdude]

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.

 

[bleakarcher]

for the second one

[sqrt(2)+log[e](1+sqrt(2))]/2

 

[Shadowdude]

for the second one

[sqrt(2)+log[e](1+sqrt(2))]/2

Yep.

Now the first?

Once you do the substitution, it should pop out quite easily.

 

[bleakarcher]

is t in termz of u

t=[u+sqrt(4+u^2)]/2

 

[powlmao]

In the solution I have (5-6 lines), there's no binomial theorem. Keep at it!

There's a trick involved.



To get that? Latex equation editor.

 

[4025808]

Here's two that could be interesting:

1.

2.

my friend (who you met during the outing) doing 1151 gave me those two questions, 1. seems like a tricky question but I got it out in the end; it's actually not hard tbh

 

[powlmao]

<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

 

[bleakarcher]

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

 

[bleakarcher]

no. i was talking about the integral. wtf is that lol?

 

[Alkanes]

<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?

 

[powlmao]

Wtf kind of condition is that on the 2nd line?

I believe i wrote it wrong. I am new to that latex stuff.

I got the question from a textbook

 

[AAEldar]

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]

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.

 

[Shadowdude]

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.

Spoiler

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.

 

[AAEldar]

If you're solving for t, you're taking the long way. Quite simple and elegant is that solution.

Spoiler

test

Those integrals are from MATH1131 work.

I think I've got the first one

 

[powlmao]

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

 

[AAEldar]

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.

 

[powlmao]

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.

Good

I just used the LaTex equation wrong

Thanks

 

[bleakarcher]

the spoiler doesnt work for shit

 

[AAEldar]

1.

However I checked with Wolfram Alpha and I appear to be wrong...