magento2 – Magento 2.3.5 Address Verification Suddenly Fails

For about month, my store has been making and completing sales. Now, suddenly it fails on address verification.

Vertex is not being used and it clears if we choose Shipping Option 2 (Overnight shipping at $34.95)

If a person chooses standard shipping, it fails Option 1 (Mageplaza Shipping methods module).

We even switched their amounts and it seems to fail only if 8.95 shipping is used not matter if it is option 1 or 2. Seems odd to be tied to amount.

8 – Email verification field for content types

If easy = doable in a small snippet on a SO post, then the answer is no.

To create a workflow like this with the Drupal API, you’ll need to do something like:

  1. Hook a custom submit handler into your content type form.
  2. Have your submission handler take action on unverified user content (e.g. email user via Drupal API, generate unique validation link)
  3. Create a custom route for validation of email validation links.

If you’re looking for a low code solution, look at different workflow modules and try to build a solution out of what they offer.

integration – Verification of an Cauchy’s contour Integral of Complementary Error function?

I tried to find an integral of the following,$DeclareMathOperator{erfc}{erfc}$

$intlimits_0^{2pi} erfc(a + bcos(theta))erfc(c + dsin(theta)),dtheta $

Where, $a,b,c,d in Bbb R$

Now, $cos(theta) = frac{e^{jtheta} + e^{-jtheta}}{2};quadsin(theta) = frac{e^{jtheta } – e^{-jtheta }}{2j}$

Let $z = {e^{jtheta }}$

Therefore, I can rewrite the integral as,
$ointlimits_{|z| = 1} {erfcleft( {a + bleft( {frac{{z + {z^*}}}{2}} right)} right)erfcleft( {c + dleft( {frac{{z – {z^*}}}{{2j}}} right)} right)frac{{dz}}{{iz}}} $

Again, let $f(z) = erfcleft(a + bleft(frac{z + z^*}{2}right)right)erfcleft(c + dleft(frac{z – z^*}{2j}right) right)$

Hence , the final integral can be written as,

$ointlimits_{|z| = 1}f(z)frac{dz}{iz} = 2pi f(0) = 2pierfc(a)erfc(c)$

As $0$ falls inside $|z| = 1$.

Can you tell me if I am correct or wrong or it needs more reasoning? Thank you.

solution verification – How to proof this statement. “There are not odd integers x,y,z according to $(x-z)^2$ + $(y-z)^2$ = $(x+y)^2$.”

I did try to proof this by contradiction(I guess it works). So I suppose this statement is false but I have confused about the meaning of it.

If this statment is false it means “There are odd integers x,y,z according to $(x-z)^2$ + $(y-z)^2$ = $(x+y)^2$.” or “There are odd integers x,y,z according to $(x-z)^2$ + $(y-z)^2$ $neq$ $(x+y)^2$.” or…..

Is proof by contradiction a good method? and Could you help me how to do next from the statement that suppose to be false?

Thank you very much.

abstract algebra – proof verification: algebraic extension (of F) containing all zeros of F[x] is algebraically closed.

This is Exercise 32.34 in Fraleigh’s book, ‘A first course in abstract algebra’.

Show that if $E$ is an algebraic extension of a field $F$ and contains all zeros of every $f(x) in F(x)$, then $E$ is algebraically closed field.

Below is my proof which I want to get verified. I’m quite certain with my proof, but since this is the first time for me to study algebra, it feels like I’m missing something.

Suppose that there exist polynomial $g(x) in E(x)$ which has no zero in $E$. Then by Kronecker’s Theorem, there exists field $K$ such that $E le K$ and $K$ contains zero of $g$, which I will denote by $alpha$. Then we have $F le E le E(alpha) le K$, since $E(alpha)$ is the smallest field which contains both $E$ and $alpha$. Since $E$ is an algebraic extension of $F$ (by assumption) and $E(alpha)$ is an algebraic extension of $E$ (because $g(alpha)=0$), $E(alpha)$ is an algebraic extension of $F$. Thus, $alpha$ is a zero of polynomial in $F(x)$. By assumption in question, we obtain $alpha in E$, a contradiction.

Thank you for reading my question. Any comments (including non-mathematical advices) will help me really lot.

I have a limitless google business verification code


Who wants google business verification code
Don't wait for your code to arrive by mail, verify as many businesses as you want.
Add your business to google maps as many times as you want

We deliver your code in 5 minutes just 50$

solution verification – Can this be accepted as a proof of the question.

$displaystyle begin{array}{{>{displaystyle}l}}
Let a and b in +mathbb{Z} such thatfrac{a²+b²}{ab+1} =k.\
Proove that k is square of an integer.

$displaystyle begin{array}{{>{displaystyle}l}}
Soln. For k to be perfect square\
frac{a²+b²}{ab+1} must be in form frac{mp^{n+2}}{mp^{n}}\
where pin +mathbb{Z} and n,min mathbb{Q}\
hence frac{a²+b²}{ab+1} =frac{mp^{n+2}}{mp^{n}} on comparison of numerator and \
denominator we have: a²+b²=mp^{n+2} …..( 1) \
and ab+1=mp^{n} …..( 2)\
Solving for a and b we get:\
b=sqrt{frac{mp^{n+2} pm sqrt{m²p^{2n+4} -4m²p^{2n} +8mp^{n} -4}}{2}} and a=left( mp^{n} -1right)sqrt{frac{2}{mp^{n+2} pm sqrt{m²p^{2n+4} -4m²p^{2n} +8mp^{n} -4}}}\
so k will be perfect square when a and b\
can be expressed in this\
form where pin +mathbb{Z} and m,nin mathbb{Q} .

Can this be a proof.

i need Code to disable the verification of filling in forms and PHP fields

i have a sidebar form, i want to remove email address field and also disable the verification of filling in forms and also the disable the massage: Please fill all the fields,
php and html code

[Guide] Free SMS Verification Sites

Receive an SMS:

SMS Receive free:

Online SMS:

Receive SMS online:

Get a free SMS number:

Receive SMS:

Receive SMS Online.NET:



Receive free SMS:

Receive free SMS.NET:

Receive SMS Online.IN:

Receive SMS online:

See SMS:



Send SMS now:

Receive SMS online.EU:


Anon SMS:

Hide my numbers:


Free online phone:


SkyCallbd free virtual number:

Capture SMS:

SMS Get:


Receive SMS:


Text anywhere:

Receive SMS online.ME:

Temporary emails:

Purchase virtual number:

Free Receive SMS online:


SMS Listen:

Free virtual SMS number:

SMS Tibo:

Receive SMS number:

Free SMS code:

Online SMS numbers:

SMS reception:

Trash Mobile


authentication – Is revealing phone number during OTP verification process consifered vulnerability?

One of the common way of implementing 2FA is using phone number Text message or Call with OTP. As I can see, usually web services show something like “OTP was sent to the number +*********34”. Is is done because revealing the number is considered a vulnerability?
If yes, then which one, is it described anywhere?
I guess it has something to do with not wanting to show too much info about the user. This info might be used be social engineering, but maybe there is something else?

Having a link to a trusted location with the description would be great as well.