Langkau ke kandungan utama
Microsoft
|
Math Solver
Selesaikan
Bermain
Latihan
Muat turun
Selesaikan
Latihan
Bermain
Pusat Permainan
Menyeronokkan + meningkatkan kemahiran = Menang!
Topik
Pra-algebra
Min
Mod
Faktor Sepunya Terbesar
Gandaan Sepunya Terkecil
Aturan Operasi
Pecahan
Pecahan Campuran
Pemfaktoran Perdana
Eksponen
Radikal
Algebra
Gabungkan Istilah Serupa
Selesaikan untuk mendapat pemboleh ubah
Faktor
Kembangkan
Menilai Pecahan
Persamaan Linear
Persamaan Kuadratik
Ketaksamaan
Sistem Persamaan
Matriks
Trigonometri
Permudahkan
Menilai
Graf
Selesaikan Persamaan
Kalkulus
Derivatif
Kamiran
Had
Kalkulator Algebra
Kalkulator Trigonometri
Kalkulator Kalkulus
Kalkulator Matriks
Muat turun
Pusat Permainan
Menyeronokkan + meningkatkan kemahiran = Menang!
Topik
Pra-algebra
Min
Mod
Faktor Sepunya Terbesar
Gandaan Sepunya Terkecil
Aturan Operasi
Pecahan
Pecahan Campuran
Pemfaktoran Perdana
Eksponen
Radikal
Algebra
Gabungkan Istilah Serupa
Selesaikan untuk mendapat pemboleh ubah
Faktor
Kembangkan
Menilai Pecahan
Persamaan Linear
Persamaan Kuadratik
Ketaksamaan
Sistem Persamaan
Matriks
Trigonometri
Permudahkan
Menilai
Graf
Selesaikan Persamaan
Kalkulus
Derivatif
Kamiran
Had
Kalkulator Algebra
Kalkulator Trigonometri
Kalkulator Kalkulus
Kalkulator Matriks
Selesaikan
algebra
trigonometri
statistik
kalkulus
matriks
pemboleh ubah
senarai
mode(2%2C4%2C5%2C3%2C2%2C4%2C5%2C6%2C4%2C3%2C2)
Nilaikan
2
Kuiz
mode(2%2C4%2C5%2C3%2C2%2C4%2C5%2C6%2C4%2C3%2C2)
Masalah Sama dari Carian Web
Compositeness of number k\cdot 2^n+1?
https://math.stackexchange.com/q/89871
There are zillions of such relations. For example: 2^6+1 is a multiple of 13, so k2^n+1 is composite if n\equiv6\pmod{12} and k\equiv1\pmod{13}. You can make as many of these as you want. ...
Semantic deduction theorem in first order logic for sentences
https://math.stackexchange.com/q/2721332
Your argument is correct: the issue is with the \vDash relation that, in some cases, is defined for sentences . For open \psi we have that M \vDash \psi is defined as follows : M \vDash \psi \text { iff } M \vDash \text{Cl}(\psi) ...
First orderer logic completeness and independence: the proof that disappear?
https://math.stackexchange.com/questions/3114517/first-orderer-logic-completeness-and-independence-the-proof-that-disappear
The complexity of T is indeed the issue, but on a much grander scale than you're considering. When you ask whether T is "computationally simple" (e.g. effectively axiomatizable) you're ...
AIME 2013 Solutions (divisiblity)
https://math.stackexchange.com/questions/1173167/aime-2013-solutions-divisiblity
Big hint: Note that the digits b and c can be chosen freely, (100 choices total); and then, whatever the choices for b and c, there are 2 choices for d. For instance if b and c are ...
Question about the proof of Hensel's Lemma
https://math.stackexchange.com/questions/1125270/question-about-the-proof-of-hensels-lemma
1.1 We need that the solutions continue to be congruent to 0\mod p^n\Bbb Z_p because we are using the fact that the values converge to \Bbb 0 in \Bbb Z_p. Recall that |x|_p=0\iff x=0, so ...
Proof involving Chinese Remainder Theorem.
https://math.stackexchange.com/questions/470030/proof-involving-chinese-remainder-theorem
Since d\mid a_1-a_2, there is an integer x with xd=a_1-a_2. Since (n_1,n_2)=d, we have ({n_1\over d}, {n_2\over d})=1, so by the chinese remainder theorem, there is an integer k withk\equiv 0\;(\mbox{mod}\;{n_1\over d}) ...
Lagi Item
Kongsi
Salin
Disalin ke papan klip
Masalah yang serupa
mode(1,2,3,2,1,2,3)
mode(1,2,3)
mode(20,34,32,35,45,32,45,32,32)
mode(2,4,5,3,2,4,5,6,4,3,2)
mode(10,11,10,12)
mode(1,1,2,2,3,3)
Kembali ke atas