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usa – Cost living in Knoxville, TN for 5 people

I am going to start my work in the USA currently and I can’t decide what I should do with my family (bring them with me or stay them at home). The main dilemma is how to correctly correlate the awaited spendings in comparison with income which will be 4333 US/month (net). We are 5 people, 2 adults and 3 children.
I am awaiting that my spouse will be at home with the youngest child when the two others will be visiting a public school. So, spending on children’s care is not required.

My around calculations per month are listed below.

  1. house rent – 1300
  2. bills – 300
  3. food – 1000
  4. medical care – 500
  5. other goods – 500

Total: 3600. Seems that 4333 will be enough.

What Have I forgot/mistaken/not take into account? What can other spendings be?
Do you have any recommendations for me?

The cost of Elimination

Reference: The Introduction to Linear Algebra, Gilbert Strang, 5th edition, page 101.

In the book mentioned above, the author states "to find each entry below pivot requires one subtraction and multiplication. We will roughly count this n** 2(n is the order of matrix) multiplication and n **2 subtraction…."
But Does it require n subtraction and multiplication since we are multiplying and subtraction one at a time so to exemplify:

A = ((2, 1, 0),
(1, 2, 1),
(3, 1, 2))

So to make every element under the first pivot zero we require one multiplication and subtraction for 1, 3 elements, accordingly.

Any recommendation.

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dnd 5e – Do Pact of the Tome Warlocks need to pay a cost to scribe their ritual spells?

2 Hours and 50gp for each level of the spell.

The feature you are referring to is the Eldritch Invocation Book of Ancient Secrets (PHB, pg. 110), which states:

Prerequisite: Pact of the Tome feature

You can now inscribe magical rituals in your Book of Shadows. (…)

On your adventures, you can add other ritual spells to your Book of Shadows. When you find such a spell, you can add it to the book if the spell’s level is equal to or less than half your warlock level (rounded up) and if you can spare the time to transcribe the spell. For each level of the spell, the transcription process takes 2 hours and costs 50 gp for the rare inks needed to inscribe it.

You don’t automatically receive this feature by selecting the Pact of the Tome, it must be chosen as one of your Eldritch Invocation options:

At 2nd level, you gain two eldritch invocations of your choice. Your invocation options are detailed at the end of the class description. When you gain certain warlock levels, you gain additional invocations of your choice, as shown in the Invocations Known column of the Warlock table.

Additionally, when you gain a level in this class, you can choose one of the invocations you know and replace it with another invocation that you could learn at that level.

So when you select Pact of the Tome at 3rd level, you can replace one of the invocations you learned at 2nd level with Book of Ancient Secrets.

find if the column of the PostgreSQL table is empty with minimum cost

As far as I know (but not sure), if there are no data at all for the column, there is some property in the database saying that the chunk or the table has no data for this column, at least this happens when adding a new column to the existing table, so the database doesn’t update all existing rows if the default value of a new column is NULL

If the row data ends early, it assumes all column after that for that particular row are NULL, or in newer versions, are whatever DEFAULT was at the time the column was added. But this says nothing at all about what might be in other rows.

You can’t avoid scanning the table unless you a had a very strange constraint which forces the values to all be NULL. But you should be able to scan the table just once, not once for every column.

select count(col1), count(co2), count(col3)...count(col100) from the_table;

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multi factor – Anonymous SMS-based auth: How to get phone numbers to receive a single 2FA authentication SMS with minimal cost and effort, multiple times

What is the simplest, fastest, cheapest, easiest way to get access to a ‘non-VOIP’ phone number just long enough to receive a single text message?

Traditional approaches that are not ideal:

  1. Use existing mobile phone number. Compromises anonymity.
  2. Go to the store, buy a burner SIM, buy some minutes, set it up. Probably the best approach, but relatively costly and time-consuming… to receive a single text message! Cheapest I’ve seen is about $10USD for the SIM, plus about >$10-20USD for some ‘days’. Also, considerable hassle to set this up, from going to the store to calling the carrier to scratching off blah blah blah.
  3. Use one of the many burner SMS services, eg burnersms.com, many others, or use Twilio, Google Voice, Skype, or other online-services. This doesn’t work on the site. Need a ‘non-VOIP’ number, presumably registered with a standard carrier, see below.
  4. Ask someone else to receive the text. Creepy, and, well just not good.

Why / use case
In order to create more than one account on a website that requires SMS-based 2FA as a test to prove that you’re human and allow registration for the site. The site doesn’t store the phone number for later 2FA as would be typical for password retrieval, so the number isn’t needed again later. It is only ever needed to receive a single SMS. Caveats regarding the number:

  1. It has to be a phone number that has never been used on the site before.
  2. It cannot be a phone number that the site deems ‘VOIP-based’, see above, or it won’t work.
  3. Want at least reasonably decent anonymity, so not trivial to connect the phone number to the registrant, as it would be for a personal mobile number in routine use.

Want to do this multiple times, so want an approach that is inexpensive and non-time-consuming. Does not matter which country the phone number is associated with it appears.

Thanks!

$T(n)=Tleft(frac{n}{2}right)+n^2$ cost of each level of this tree?

I am confused about this. Can anyone do it with substitution method?

$T(n)=Tleft(frac{n}{2}right)+n^2$

$=Tleft(frac{n}{4}right)+left(frac{n}{2}right)^2+n^2$ (expand $T(left(frac{n}{2}right)$)

$=Tleft(frac{n}{8}right)+left(frac{n}{4}right)^2+left(frac{n}{2}right)^2+n^2$ (expand $Tleft(frac{n}{4}right)$)

$T(n)=n^2+left(frac{n}{2}right)^2+left(frac{n}{4}right)^2+left(frac{n}{8}right)^2+⋯+left(frac{n}{2^{log_2 n}}right)^2 $

$=n^2+frac{n^2}{4}+frac{n^2}{16}+frac{n^2}{64}+⋯$