Hopp til hovedinnhold
Microsoft
|
Math Solver
Løse
Praksis
Skuespill
Emner
Pre-Algebra
Bety
Modus
Største felles faktor
Minst vanlige multiplum
Rekkefølge av operasjoner
Fraksjoner
Blandede brøker
Førsteklasses faktorisering
Eksponenter
Radikaler
Algebra
Kombiner som termer
Løse for en variabel
Faktor
Utvide
Vurdere brøker
Lineære formler
Kvadratiske ligninger
Ulikheter
Ligningssystemer
Matriser
Trigonometri
Forenkle
Vurdere
Grafer
Løs formler
Beregning
Derivater
Integraler
Grenser
Algebra innganger
Trigonometri-innganger
Kalkulus innganger
Matrise innganger
Løse
Praksis
Skuespill
Emner
Pre-Algebra
Bety
Modus
Største felles faktor
Minst vanlige multiplum
Rekkefølge av operasjoner
Fraksjoner
Blandede brøker
Førsteklasses faktorisering
Eksponenter
Radikaler
Algebra
Kombiner som termer
Løse for en variabel
Faktor
Utvide
Vurdere brøker
Lineære formler
Kvadratiske ligninger
Ulikheter
Ligningssystemer
Matriser
Trigonometri
Forenkle
Vurdere
Grafer
Løs formler
Beregning
Derivater
Integraler
Grenser
Algebra innganger
Trigonometri-innganger
Kalkulus innganger
Matrise innganger
Grunnleggende
algebra
trigonometri
beregning
statistikk
Matriser
Tegn
mode(2,4,5,3,2,4,5,6,4,3,2)
Evaluer
2,4
Spørrelek
5 problemer som ligner på:
mode(2,4,5,3,2,4,5,6,4,3,2)
Lignende problemer fra nettsøk
mn+1 \equiv 0 \pmod{24} then : m+n \equiv 0 \pmod{24} using group theory
https://math.stackexchange.com/questions/2350421/mn1-equiv-0-pmod24-then-mn-equiv-0-pmod24-using-group-theory
You're trying to prove that if mn \equiv -1 \pmod{24} then m \equiv -n \pmod{24}. Let k = -n. Then you're trying to show that if -mk \equiv -1 \pmod{24} then m \equiv k \pmod{24}. Of ...
Can we ever have \Gamma \models \perp
https://math.stackexchange.com/questions/2639449/can-we-ever-have-gamma-models-perp
That's exactly right: "\Gamma\models\perp" is equivalent to "\Gamma has no model" (or "\Gamma is unsatisfiable").
Is this proof about Mersenne numbers acceptable?
https://math.stackexchange.com/questions/86429/is-this-proof-about-mersenne-numbers-acceptable
There is nothing incorrect, but there are a few things that could be changed. We only need p>2. From 2^p \equiv 2 \pmod {p} one should conclude M_p=2^p -1\equiv 1 \pmod{p} immediately, without ...
Solving system of linear congruence equations
https://math.stackexchange.com/questions/473711/solving-system-of-linear-congruence-equations
The way you express your congruences is rather unconventional. Given that 23d\equiv1\pmod{40}, 73d\equiv1\pmod{102}, and that 40=2^3\times5 and 102=2\times3\times17, it follows that 23d\equiv1\pmod5, ...
How to prove an element of a given structure is not definable?
https://math.stackexchange.com/questions/927915/how-to-prove-an-element-of-a-given-structure-is-not-definable
HINT: If x is a definable element in a structure \mathcal M, then any automorphism of \cal M must satisfy f(x)=x. To show that 2 is not definable, find an automorphism of \cal A such that ...
The deduction theorem according to AIMA
https://math.stackexchange.com/questions/13251/the-deduction-theorem-according-to-aima
In order for \alpha\Rightarrow\beta to be valid, it must hold in all models; for \alpha\Rightarrow\beta to not be valid, there must be a model where it is false. If there is a model where it is ...
Flere Elementer
Aksje
Kopi
Kopiert til utklippstavle
Lignende problemer
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)
Tilbake til toppen