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# Year 13 Maths Help Thread Watch

1. (Original post by kiiten)
Now i got 3 + sqrt 17 instead of 3 + sqrt 13???
Did you make equal the DOUBLE of the path?
2. (Original post by RDKGames)
Did you make equal the DOUBLE of the path?
No...

so would it be

4x^2 = (2x+2)(x+2) - 2x^2
3. (Original post by kiiten)
No...

so would it be

4x^2 = (2x+2)(x+2) - 2x^2
Almost. I see you took a different approach, which is still technically right.

Try:

Im not sure where you pulled that from.
4. (Original post by RDKGames)
Almost. I see you took a different approach, which is still technically right.

Try:

Im not sure where you pulled that from.
i multiplied the 2x^2 by 2 because it says the area of the lawn (2x^2) is twice that of the path hence 4x^2 ?

EDIT: i misunderstood .... again. Thanks
5. (Original post by RDKGames)
Almost. I see you took a different approach, which is still technically right.

Try:

Im not sure where you pulled that from.
its asking for the length of the lawn.
6. (Original post by kiiten)
its asking for the length of the lawn.
Yep.
7. (Original post by RDKGames)
Yep.
So its 6 +2 sqrt 13 (not 3 + sqrt 13)
8. (Original post by kiiten)
So its 6 +2 sqrt 13 (not 3 + sqrt 13)
Yep.
9. (Original post by RDKGames)
Yep.
yay finally i got something right

Question asks for a complementary function and a particular integral of and I've gotten the CF to be but I'm unsure where to go from there because subbing it into the original differential doesn't exactly favour maths as so there's no value for that will make this work, is there? What does this mean for the PI?
11. (Original post by RDKGames)

Question asks for a complementary function and a particular integral of and I've gotten the CF to be but I'm unsure where to go from there because subbing it into the original differential doesn't exactly favour maths as
Try a PI of the form where is just a real constant. You don't need to sub your complementary function into the ODE once you have found it. The arbitrary constant has to be there for the general solution and you can only find the value of if you're given boundary conditions. your general solution will just be .
12. (Original post by B_9710)
Try a PI of the form where is just a real constant. You don't need to sub your complementary function into the ODE once you have found it. The arbitrary constant has to be there for the general solution and you can only find the value of if you're given boundary conditions. your general solution will just be .
Ah yes I remember that now,
So and since it cancels, it only leaves
and therefore ?
13. (Original post by RDKGames)
Ah yes I remember that now,
So and since it cancels, it only leaves
and therefore ?
Yes that's right.
14. (Original post by B_9710)
Yes that's right.
Can you explain why the CF of is and not, as I found it, ? Or are they wrong?
Because when I used mine I get PI to be which is correct on their answers.
15. (Original post by RDKGames)
Can you explain why the CF of is and not, as I found it, ? Or are they wrong?
Because when I used mine I get PI to be which is correct on their answers.
The CF is .
16. (Original post by B_9710)
The CF is .
This book smh...
17. Am i doing this wrong?

Solve: 5+2sin(2x+1)=6 for 0<=x<=720

rearranged to sin(2x+1)= 1/2

let u = 2x+1
sinu=1/2
u=1/6 pi

change the bounds to 1 <=u<=4pi +1
using a cast circle i got the first answer to be 2.05 but it should be 0.81 ???

*sorry if this doesnt make sense - let me know and ill post the full working
18. Not quite sure where I'm going wrong.

but that doesn't seem right as I wouldn't get the left side from
19. (Original post by kiiten)
Am i doing this wrong?

Solve: 5+2sin(2x+1)=6 for 0<=x<=720

rearranged to sin(2x+1)= 1/2

let u = 2x+1
sinu=1/2
u=1/6 pi

change the bounds to 1 <=u<=4pi +1
using a cast circle i got the first answer to be 2.05 but it should be 0.81 ???

*sorry if this doesnt make sense - let me know and ill post the full working

Then:

and due to symmetry of sine:

For any integer n (try a few around 0 until your solutions go outside the range)

Try that.
20. (Original post by RDKGames)

Then:

and due to symmetry of sine:

For any integer n (try a few around 0 until your solutions go outside the range)

Try that.
??? I have no idea what you're talking about.
What is 2n and is arcsin inverse of sin?

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