mattyjay7

Stryder vs high-level abyss lich First loadout has 2 shock armor and 1 poison armor, and dagger and shield with -mag Second loadout has fire damage sword and shield with fire spells, 1 ring of cinders and 1 similar ring, necklace with fire enhancement and health, and gloves restoring stamina.

Elder scrolls blades arena match

Stole 1 round out of like 10 from Ofton

Consuming inferno in mortal mom at 2021

Winning one round vs H3PA3nb (currently 2nd player on elder scrolls leaderboard)

This guy was apparently a conscientious objector to violence.

Beat Ken with my frost loadout somehow.

Finally figured out that frost worked the best for me against these skeletons, who used to be a real pain in the abyss. Loadout is identical to the Fire Atronach loadout except I use tempered dragonplate armor and dragonplate frost shield.

Loadout is same as Fire Atronach

Loadout includes frost dagger, light daedric shield of depletion, health on gloves and jewelry.

Loadout is same as second loadout for Abyss Lich

Killing Troll from ‘Wrath of the Undying’ event with fire

This should defrost you a bit, my dear necromancer.

Taking out a skeleton w/fire spells

Intro: The Divergence Theorem may not be intuitively obvious just looking at it, but if you consider it, I imagine you’ll find it’s easier to understand intuitively than Green’s or Stokes’ Theorems. Intuition: basic example Let’s take a look at the triple integral on the right first and try to get an idea of what this is capturing. This triple integral is basically summing up the divergence of our vector field in all these little cubes inside our region E Let’s take a simple case where, say, the vector field represents the flow of water. And say we have a water source in the middle of this region. If that water is coming out from this point, we would say the vector field has a positive divergence there. The divergence is a ‘flux density’. It’s a sort of derivative of the vector Field F, and would be in terms in this case, of say, liters of water per cm^3 emanating from the source region. So if we integrate over the region E, we’d sort of ‘cancel out’ the cm^3 and end up with the Liter’s of water emanating from the region E. Now let’s look at the left side. This is a surface integral finding the FLUX of this vector field emanating through the the surface S, where we define S as the surface ENCLOSING our region E. Let’s imagine this is a sort of mesh surface the water will flow through. Say our water flow vector field has units of Liters/cm^2, so when we integrate over the surface we again sort of cancel out the ‘cm^2’ and are left with the FLUX, or the the total amount of water, leaving the surface enclosing the region E. So, hopefully you see, the left and right sides of the equation are both telling you the same thing – the amount of water leaving the region, which is the same thing as saying – the amount of water going through the surface enclosing the region. Intuition – more examples Now, say we added a drain within this region that is taking in the water. If we added these up we’d have a positive divergence source and a negative divergence drain, and if they were of the same magnitude they may cancel out so that the triple integral, or sum of the divergence over the region would be 0 And what would that look like on the surface integral side? Well, the vector field might show all the water from the source going out the drain, so there were no vector fields at the surface, or… it might show that the water was emanating from one side of the surface boundary but water was going in on the other side of the boundary. In either case, if the source and drain were of equal magnitude, we’d see the the flux exiting and entering the sides of the surface boundary would cancel out and you’d get a total flux of 0. So… hopefully that provides some intution behind the divergence theorem and perhaps, like me, you find it easier to understand than Green’s or Stokes’ Theorem, but it might be worth quickly noting some of the similarities betweeen the theorems. Similarity to Green’s/Stokes’ Theorem While Stoke’s Theorem has a closed line integral on the left side, the Divergence theorem has a closed surface integral including on the left side. The closed line or surface is a required criteria for both of the theorems. The closed integrals also correspond to the vector field F ⃗ in both cases Difference from Green’s/Stokes’ Theorem A major difference though is of course on the right side of the equation, Stokes’ Theorem uses the curl of F ⃗ while The Divergence theorem uses the divergence of F ⃗. Also, the divergence theorem operates on a sort of higher dimension on both sides, so it has a double instead of single integral on the left, and a triple instead of double integral on the right. Someimes it’s good to compare and contrast theories like this to help better scaffold our understanding of them. So I hope this helped provide a little bit better intuition on the Divergence Theorem, and until next time, take care!

Perhaps it's immediately obvious to you why a closed line integral over a vector field would equal the double integral of the curl of the vector field over the region D enclosed by the line integral. If so, feel free to move on. But if you're like me, this seems like a rather bizarre equality. So let's take a look at WHY it's true. Imagine we have a vector field in the xy plane. Now imagine we have a closed path traversing this vector field. And imagine we want to find this integral. What would this mean? Well, for example, if the vector field represented the wind velocity at each xy coordinate, the line integral would represent the work done by the wind on an object traveling along this path, in the given direction. So Green's Theorem tells us this line integral is equal to the double integral of the curl of the vector field, dotted with the k-hat, over the region D enclosed by the path. OK so here's what's going on. Imagine instead of finding the work done along that path, we were instead going to cut this up into a bunch of smaller regions, and find the work done along the counterclockwise path surrounding these little regions. If we added up the work done in all those regions, we'd find that much of it would cancel out. For example, these two regions are next to each other. Say the vector field was heading down here. This would make the work larger for that region on the right, since the vector field would be going along that counterclockwise direction. However, it would make the work on the left region negative, since it would be going against that counterclockwise direction. So if we added these together it would cancel out. And if we divided this entire D region into these little intervals, you could see everything within the D region would cancel out, but there wouldn't be anything to cancel out what's on the edges, so the work done 'on the edges' would be all that remained. So we could approximate this line integral is equal to the sum of all the line integrals within the region D Now here is where it gets interesting. We have been using the definition for curl as del cross F. This is really just a notation for curl though, because while this operation works in cartesian coordinates, it doesn't hold exactly for cylindrical and spherical. The more exact expression for curl is given as the curl (f) dot k-hat \approx line integral around f, divided by the area of the rectangle we are integrating around. So this shouldn’t seem to bizzare, because that line integral is expressing how much that vector is 'curling' around a given point, which is what we know curl represents, that amount of 'spin'. And of course we know our curl is positive in the z direction using the right hand rule, when the rotation is counter-clockwise. Solving this for that line integral, we get this:Now we can plug back this equation in for our line integral and we get this sum. And everything here will become exact now if we change this sum into an infinite sum, or double integral over this region D. And we are left here with Green's Theorem, which we can adjust a bit if we use our definition for the curl(F) as Q_x - P_y where F is our vector field in the xy plane, F=[P,Q,0]. So that's some intuition behind where we get Green's theorem from

Covers: Days before election Election day Overall results and anomalies unexpected patterns in state results glitches and the NYT-Edison data The Drop and Roll The One Key Vote Identified Irregularities and Impact The Likely Stolen Military ballots

Volume one: the immaculate deception Volume two: the art of the steal Volume three: yes, President Trump won (the receipts)

Why does the Discriminant let you determine max’s, min’s and saddle points? I was explaining to my students an exam problem on maxes and mins and why they got it wrong. I explained a little bit of WHY the 2nd derivative test works, they said they wished I explained this earlier, because it would have helped them understand and remember the rules for determining max’s, mins and saddle points. So that’s the point of this video: WHY does the discriminant tell you the max’s, min’s and saddle points. The part about the second derivative being greater or less than zero is perhaps the easiest part of this to understand. Recall back to finding critical points in differential calculus, once you found a critical point, you could tell if it was a max, min or inflection point by looking at it’s second derivative. If the second derivative was positive at the critical point, as it is at this point here, that means the curve is ‘concave up’ at that point. If concave up, then we know we’ll have a minimum at x=a. That’s basically what the first line of the second derivatives test is telling you. If the second derivative with respect to x is greater than 0, then that critical point is a local minimum. The second line states the opposite, if the second derivative is less than 0 , then it’s concave DOWN at this point and f(a,b) is a local maximum. This provides some intuition for lines 1 and 2, so let’s consider this example. Imagine the surface f(x,y)=x^2+y^2, or easier yet, look at it hear in CP3D. This will have a discriminant of D=4 which is greater than 0, it will be concave up in the x directions at the critical point (0,0) so we know it’s a local min. Throwing in some negative signs, the surface will still have a discriminant of D=4 is greater than 0, and it will be concave down in the x directions at the critical point (0,0) so we know that’s a local max now. But what if we changed it to f(x,y)=x^2-y^2? Now It’s still concave up in the x-direction as shown by the red trace, but it’s concave down in the y direction as shown by the blue trace, and this is no longer a local min, it’s a saddle point. The discriminant for this will be D=2(-2)-[0]^2 =-4-[0]^2 =-4 is less than 0 So the fact that the discriminant is less than 0 tells us it’s a saddle point. And why is the discriminant less than 0? Because the f_xx (a,b) f_yy (a,b) term is negative. And why is f_xx (a,b) f_yy (a,b) negative? Because f_xx (a,b) and f_yy (a,b) have different signs. So that’s the most basic way this discriminant gives you a negative number showing a saddle point. And consider if this f_xx (a,b) f_yy (a,b) term is negative and then you subtract the cross term, which is always positive since it’s squared, you still always end up with a negative D, regardless of the f_xy (a,b) term. But what about that term f_xy (a,b)? Why do we need to consider that? Well, the answer is things get a little more complicated in two dimensions, because now you have cross terms. So let me give you an example here. Consider the surface f(x,y)=x^2+y^2+axy If a=0, this is simply our earlier paraboloid that’s concave up in the x and y directions, and that critical point (0,0) is still a minimum, simple. But look what happens when we start increasing that a term. The critical point (0,0) transforms into a saddle point. So does that mean it’s concave up in the x direction but concave down in the y direction, or vice-versa? Nope, neither. We can see trace out the red line here along the surface in the x direction and you can see it is still concave up. We can see trace out the blue line along the surface in the y direction and you can see it is also still concave up. But even though they are both concave up, you can trace out this purple line along the diagonal which is obviously concave down. So that’s why we need to consider f_xy (a,b). In an equation like this, it basically accounts for how significant that cross term is: If a=0, then D is greater than 0 and (0,0) is a local min based on concavity. If a=2, then D=0 and we don’t have a max, min or saddle. If a=4, then D is less than 0 and (0,0) is a saddle point. That cross term is probably the least intuitive piece but hopefully this gives you a feel for how the discriminant and 2nd derivative test work to identify max’s, min’s and saddle points.

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Russ Ramsland explains 57000 votes changed in Texas and discusses Antrim County Audit

Col Phil Waldron gives update on Dominion vulnerabilities and China's election interference

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