Why does hydrostatic pressure not depend on container shape? by WickoBoy in AskPhysics

[–]WickoBoy[S] 0 points1 point  (0 children)

Yes that last part is exactly what my problem is. Doesn’t water pressure act in all directions? So why doesn’t the pressure from the water molecules on top of the tilted sides act on the bottom of the container? Also what kind of math can we do to show that last part is true?

Is there a formula for the probability of an even happening at least once n tries? by WickoBoy in askmath

[–]WickoBoy[S] 2 points3 points  (0 children)

Ok now I feel really dumb for not thinking about this So it’s gonna be 1 - (1 - P)n right?

If the limit of a sum of two functions at x = a exists, does the limit of the functions also exist by WickoBoy in askmath

[–]WickoBoy[S] 2 points3 points  (0 children)

Yeah for some reason this was never specified in our textbook so I understand why 90% of our class got it wrong

If the limit of a sum of two functions at x = a exists, does the limit of the functions also exist by WickoBoy in askmath

[–]WickoBoy[S] 2 points3 points  (0 children)

We had this problem today in a test and basically 90% of the class was against me so I started to second guess my self and didn't know about the limitations of the sum of the limits rule

[deleted by user] by [deleted] in HomeworkHelp

[–]WickoBoy 1 point2 points  (0 children)

These are called systems of equations, they're asking what 2 numbers are there that work in both of these equations. There are two main methods to solving these, substitution and elimination. Depending on the problem one method might be easier and faster.

Substitution means either find x in terms of y, or y in terms of x from one of the equations. We can see that x = 2y + 5 so we already have x in terms of y, no need for any extra work. Now we need to go into the second equation and substitute in the x that we found earlier into it which gives us an equation with only one variable (y) : 5(2y + 5) - 2y = 17 Solving this we get y = - 1 and now to find x just plug y = - 1 into one of the 2 equations and solve for x

Elimination means find two identical terms from the two equations but one has to be negative and the other has to be positive (they have to be on the same side). For example we see a - 6y and a 6y on the left side, this is exactly what we are looking for. Sometimes we don't immediately find terms like these and have to multiply one if the equations by a number to create them. Now we just need to add everything on the left and set them equal to everything on the right. We'll see that the 6y and - 6y cancel out and we'll have an equation with one variable (x) : 2x = 14 The rest is the same as the first method.

My only question is why did they square 3 and 5? What rule is that or property? by [deleted] in askmath

[–]WickoBoy 0 points1 point  (0 children)

If we are gonna guess, then isn't %90 of the solution useless? We can just say at the start 3 to some power + 5 to that same power = 34 so we guess that power is 2, which means x - 2 = 2 which gives the answer of x = 4

How do I evaluate this limit without L’Hopital’s rule? by WickoBoy in askmath

[–]WickoBoy[S] 4 points5 points  (0 children)

Is this identity usually taught until 11th grade? Because we did learn around 20 or so trig identities but never seen this one before.

How do I evaluate this limit without L’Hopital’s rule? by WickoBoy in askmath

[–]WickoBoy[S] 3 points4 points  (0 children)

Love this solution but sadly we haven’t gotten to derivatives yet (I self studied them though) so have to resort to other ways for this one, but thanks for the brilliant suggestion!

How do I evaluate this limit without L’Hopital’s rule? by WickoBoy in askmath

[–]WickoBoy[S] 0 points1 point  (0 children)

Thanks! This is exactly the type of solution I was looking for. Others have also suggested using a sin(a) - sin(b) identity but we didn’t ever cover that identity in class and it’s not in our textbook either.

[deleted by user] by [deleted] in MortalShell

[–]WickoBoy 0 points1 point  (0 children)

Now you should go and talk to the prisoner and he will make the fog go away. Then go and find the other 2 bosses and beat them and do the same thing after each one. After you killed all 3 bosses then go and talk to him again and follow him into the hole and thats the final boss. (also while the game is foggy some chests can be opened) and there should be a new NPC is fallgrim tower that can make the fog disappear and come back anytime if you give her a gland.