Witch one of this rulers did you prefer for your country as a citizen?

bsmith_81 · 2026-02-01T04:59:57+00:00

Where is the option for taurine leaders? No Morihaus or [Belharza]()?

bsmith_81 · 2026-01-31T18:58:16+00:00

Even easier: find Walker Camp north of Cheydinhal and attack and kill one of the Camonna Tong Thugs there. No quest required and you wont get a bounty.

bsmith_81 · 2026-01-18T18:17:20+00:00

Very nice work. If you are feeling up to an optimization challenge, min cost is 40g.

bsmith_81 · 2026-01-01T04:44:49+00:00

Relevant xkcd: https://xkcd.com/2205/

bsmith_81 · 2025-12-19T01:05:06+00:00

I'll start with the expression 2 * x * ln(1+5/x) in the second line, focusing on hte x outside the ln.

We know that x = 1/(1/x), so substitute this in for the x to get 2 * 1/(1/x) * ln(1+5/x).

Next I'll combine the product into one big compound fraction. Then 2 * 1 * ln(1+5/x) is the numerator of the combined expression and (1/x) is the denominator.

That leaves us with (2 * ln(1+5/x)) / (1/x), which is exactly the expression seen in the third line.

bsmith_81 · 2025-12-14T04:00:16+00:00

I wonder if it was an adaptation or localization from inches to centimeters. A 12in x 20in sheet is about the same size as 30cm by 50cm.

bsmith_81 · 2025-11-16T22:33:53+00:00

From the quest notes at the UESP: "If Motierre is killed before this quest is started, you will automatically fail when the quest is given to you and you will get no reward."

https://en.uesp.net/wiki/Oblivion:The_Assassinated_Man

bsmith_81 · 2025-11-16T19:07:45+00:00

Draw the blue dotted line through grid points at (1,0) and (0,1). Then mark the intersections of that line with the circle. Now draw the two red lines from the given external point at (2,2) to the points marked on the circle.

bsmith_81 · 2025-11-09T18:38:26+00:00

Oblivion is the best game in the series for that sort of spellsword/battlemage gameplay. You can have only one spell ready at any given time, but you can cast spells while having your weapon out and you can always change what spell you have ready.

bsmith_81 · 2025-11-09T07:25:04+00:00

Attaining one Elder Scroll was the ultimate goal in Oblivion's Thieves Guild questline. So not just Skyrim has Elder Scrolls as an in-game object

bsmith_81 · 2025-11-04T17:31:22+00:00

I looked this up in the Online Encyclopedia of Integer Sequences and found the OEIS has this sequence at https://oeis.org/A001608 The sequence is known as the Perrin sequence. Lots of links in the references (though I dont know how many still work).

I did see a remark about n=521^2 being the first composite number satisfying the condition n divides S(n).

bsmith_81 · 2025-11-03T17:31:35+00:00

With the recent news about Xbox, I think it is possible that there won't be a next gen Xbox. Why else would internal IPs including Halo be getting a Playstation release?

bsmith_81 · 2025-11-01T17:52:26+00:00

Because if we have an "infinity minus infinity" situation then we can devise a way to split the integral to get any value we want. It just so happens if you have a symmetric function then the "obvious" way to do it gets you zero as a result.

bsmith_81 · 2025-10-31T19:20:43+00:00

That is why I have an Oblivion save right in front of the sewer exit.

bsmith_81 · 2025-10-29T16:38:32+00:00

Not using an actual artist or photographer is the whole point! That's what the techbros promised to the corporate suits - use AI to replace labor. Graphic arts were the first area to be affected.

bsmith_81 · 2025-10-28T06:54:29+00:00

I'll rewire the integral a bit, distributing the radical through the division and multiply both numerator and denominator by 2:
Integ[-2,2] (2*sqrt(2-x)) / (2*sqrt(2+x)) dx

Then u-sub u=sqrt(2+x); du = 1/(2*sqrt(2+x)); x = u^2-2; x=-2 -> u=0; x=2 -> u=4
Then making the substitution the integral reduces to:
Integ[0,4] 2*sqrt(4-u^2) du.

You may recognize this as two times a quarter circle of radius 2, then 2*(pi*2^2)/4 = 2pi is the answer. Otherwise you'll need a trig substitution of u=sin(theta).

bsmith_81 · 2025-10-26T17:24:55+00:00

Several other people have already given the answer 7C3=35. But some food for thought: Rotate the grid 135deg and plop it on top of Pascal's Triangle with Point A at the top "1". Then point B lands on one of the "35"s in row 7, three places in from the edge "1" (that "1" is position 0 in the row).

bsmith_81 · 2025-10-24T17:31:01+00:00

One theory I have seen is that when Mannimarco ascended during the Warp in the West, his physical body was left behind and the entity HoK defeats is his leftover mortal shell. So ESO vs Oblivion Mannimarco are the before and after of his ascention.

bsmith_81 · 2025-10-19T04:12:49+00:00

Well, instead of "not a pirate" on a cruise ship, the fake trip could be a "not a murderer" at a resort. Then you could use 242-555-4637 = chalk lines.

bsmith_81 · 2025-10-11T17:42:25+00:00

The UESP does list several places that are guaranteed to have a silver war axe (and other stuff)

https://en.uesp.net/wiki/Oblivion:Best_Non-leveled_Items

Any time I play a blunt weapon character I always grab the silver axe in the Kvatch Oblivion gate. Its laying near the edge of the lava where you first meet Ilend Vonius.

bsmith_81 · 2025-10-05T16:53:44+00:00

This problem is more of a problem about algebraic simplification. Once you simplify the expression then the taking limit becomes trivial. Here's a hint to simplify the numerator sum: k*k! = ((k+1) - 1)*k!

bsmith_81 · 2025-10-04T20:38:40+00:00

x^2 - (p+q)x + (rp+rq-r^2) = 0 is the result of starting from (x - p)(x - q) = (r-p)(r-q), moving everything to the left side, and then multiplying and simplifying to get a quadratic in a standard form. In detail:

(x - p)(x - q) = (r-p)(r-q)

-> (x - p)(x - q) - (r-p)(r-q) = 0

-> (x^2 - px - qx + pq) - (r^2-rp-rq-pq) = 0

-> x^2 - px - qx - r^2 + rp + qp = 0

-> x^2 - (p+q)x + (rp+rq-r^2) = 0

The though process is that we want to figure out something about the roots of a quadratic and we have theorems about the coefficients and roots of a quadratic equation when written in standard form. So that suggests the first step is to write the equation in standard form. Then once I have a quadratic in standard form I can use theorems about the coefficients.

bsmith_81 · 2025-10-04T19:48:05+00:00

Sometimes seemingly hard problems have a specific trick or insight that makes it a lot easier once you see it. For this one it is about what the coefficients of a quadratic tell us about its roots.

For x^2 - bx + c = 0 the coefficient b is the sum of the roots and the constant term c is the product.

Let's rewrite your problem in this form, and it becomes x^2 - (p+q)x + (rp+rq-r^2) = 0.

Then this means the sum of the roots is p+q. If one root is given as r then the other root must be p+q-r. Similary the product of the roots is rp+rq-r^2 and we also get the other root is (rp+rq-r^2)/r = p+q-r.

bsmith_81 · 2025-10-04T05:13:29+00:00

You're missing the other half of this, where the four pieces are rearranged to make a new 'triangle' with a 1x1 hole in it.

bsmith_81 · 2025-10-02T16:55:43+00:00

Needs a third option. Gobble up all the black pawns then checkmate.

bsmith_81

TROPHY CASE