Atrisiniet 456+789 | Microsoft matemātikas risinātājs

Pāriet uz galveno saturu

456 + 789

= 1245

Izrēķināt

1245

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Sadalīt reizinātājos

3 \times 5 \times 83

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Viktorīna

456 + 789

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Adding 2 and 3 Digit Numbers

Saistītie jēdzieni

Saskaitīšana

Saskaitīšana jeb summēšana ir matemātiska darbība, kas atbilst divu objektu apvienošanai. Šo darbību apzīmē ar simbolu "+". Piemēram, izteiksme 3 + 2 nozīmē, ka tiek saskaitīti skaitļi 3 un 2. Skaitļus 3 un 2 šajā izteiksmē sauc par saskaitāmajiem, bet izteiksmes 3 + 2 vērtību sauc par summu. Saskaitīšanas rezultātu pieraksta šādi: 3 + 2 = 5 — apvienojot trīs ābolus un divus ābolus, iegūst piecus ābolus. Skaitīšana ir saskaitīšanas speciāls gadījums, kad atkārtoti tiek pieskaitīts skaitlis 1. Saskaitīšanai pretējā darbība ir atņemšana.

Reizināšana

Dalīšana

Līdzīgas problēmas no meklēšanas tīmeklī

Permutation as disjoint cycles

https://math.stackexchange.com/questions/2774402/permutation-as-disjoint-cycles

Given the permutation

(123) (45) (16789) (15)

, you can find the disjoint union of cycles by considering what this permutation does to each element. Observe that, from ...

finding overlapping permutations

https://math.stackexchange.com/questions/415610/finding-overlapping-permutations

We count the complement, the bad permutations where

7

8

9

. How many permutations have

7

9

6

places for the pair to go, and they can be in either ...

What is the probability of selecting two aces and two hearts when selecting

4

cards from a deck that does not include face cards?

https://math.stackexchange.com/q/2625242

The denominator is the number of ways we can select

4

40

cards in the deck. There are two cases to consider: Neither ace is a heart: You handled this case correctly. One ace is a heart: ...

Three length 3 “gapped” straights vs. one length 6 straight. Which is more likely and by how much?

https://math.stackexchange.com/questions/1870540/three-length-3-gapped-straights-vs-one-length-6-straight-which-is-more-likel

It seems A has the advantage, not B like I originally thought. I put extra checks into the program to make sure it is working correctly. I originally missed a few cases for A winning, thus the ...

1 - 9

https://math.stackexchange.com/questions/1799134/entry-of-1-9-in-magic-box

As you have already pointed out, no solution exists by using each digit only once. But without the constraint, solution can be achieved. Note that Row

5

in first column can only take

9

British Maths Olympiad (BMO) 2003 Round 1 Question 4 what is wrong with this approach, can it be fixed?

https://math.stackexchange.com/q/2777373

There are two good ways to do this problem. Let

a_{n}

be the number of wicked subsets of

{1, 2, \dots, n}

. You can show that

a_{n} = a_{n - 1} + a_{n - 2} + a_{n - 3}

n \geq 3

a_{n - 1}

Koplietot

Kopēts starpliktuvē

1245

Saskaitiet 456 un 789, lai iegūtu 1245.

Piemēri

Kvadrātiskais vienādojums

x^{2} - 4 x - 5 = 0

4 sin θ cos θ = 2 sin θ

Lineārs vienādojums

y = 3 x + 4

699 * 533

[2534] [2 - 1 0135]

Vienlaicīgs vienādojums

{8 x + 2 y = 46 7 x + 3 y = 47

Diferencēšana

\frac{d}{d x} \frac{( 3 x ^{2} - 2 )}{( x - 5 )}

\int_{0}^{1} x e^{- x^{2}} d x

x \to - 3 lim \frac{x ^{2} - 9}{x ^{2} + 2 x - 3}