0
Link's Awakening DX is not available on the original DS however it is available for the 3DS on thee 3DS store.
Wiki User
∙ 10y ago
Add your answer:
Earn +20 pts
What are the differences between Zelda links awakening and Zelda links awakening DX?
Links awakening was released on the original gameboy. Links awakening DX was released as a gameboy color game. The only difference between the two is that DX has an extra bonus dungeon.
How do you wake the walrus in Links Awakening DX?
go back to mabe village and get marin to come with you (she is at the end of the beach)
What Zelda game is a dream?
Zelda: Link's Awakening (DX) is a dream.
Can you play a game for Nintendo DX on a Nintendo DS?
Dude,there's no such thing as a Nintendo DX
Should you get Oracle of seasons the minish cap or links awakening?
You should get Oracles of seasons the minish cap rather the links awakening.
Is the Legend of Zelda links awakening easy?
No way!
Where is the key for eagles tower in the legend of Zelda link's awakening dx?
In the clay pot on the left side o the tower
How do i save my game in The Legend of Zelda Link's Awakening DX?
press start,select,a,and b all at the same time.
How many Zelda games where there without Zelda?
Zelda does not appear in 2 games: Link's Awakening (DX) and Phantom Hourglass.In Phantom Hourglass, however Tetra is the alter ego of Princess Zelda, so you can still say that Princess Zelda did appear in this game, just in the form of Tetra. In Link's Awakening (DX) Zelda isn't even mentioned once.
What do you do when you defeat level 3 and get the angler key in legend links awakening?
find the doctor
What is johnathan coachman's entrance song?
dx dx dx dx dx dx dx dx dx dx dx dx dx dx dx dx dx d dx dx dx dx dx dx dx dx
How does one find the integral of secant cubed of x dx?
Find I = ∫ sec³ x dx. The answer is I = ½ [ log(sec x + tan x) + sec x tan x ]. * Here is how we may find it: Letting s = sec x, and t = tan x, we have, s² = 1 + t², dt = s² dx = (1 + t²) dx, and ds = st dx. Then, we obtain, dI = s³ dx = s dt. * Now, d(st) = s dt + t ds = dI + t ds = dI + st² dx = dI + s(s² - 1)dx = dI + s³ dx - s dx = 2dI - s dx; whence, 2dI = s dx + d(st). * Also, we have, s = (s² + st) / (s + t), whence s dx = (s² + st) dx / (s + t) = (dt + ds) / (s + t) = d(s + t) / (s + t) = d log(s + t). This gives us, 2dI = d log(s + t) + d(st). Integrating, we easily obtain, I = ½ [ log(s + t) + st ], which is the answer we sought. * Checking that we have arrived at the correct answer, we differentiate back: d(st) / dx = (st)'= st' + ts' = s³ + st² = 2s³ - s. d log(s + t) / dx = log'(s + t) = (s + t)' / (s + t) = (st + s²) / (s + t) = s. Thus, 2I' = [ st + log(s + t) ]' = 2s³; and I' = ½ [ st + log(s + t) ]' = s³, confirming that our answer is correct.
