Ratkaise 256+15 | Microsoftin matematiikan ratkaisija

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256 + 15

= 271

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271

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271

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256 + 15

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Subtracting: three digit numbers and basic regrouping | Arithmetic | Khan Academy

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khanacademy.org

Subtraction "Across the Zeros" 4th Grade Math Lightboard

Samanlaisia ongelmia verkkohausta

14 b = 56

http://www.tiger-algebra.com/drill/14b=56/

14b=56 One solution was found : b = 4 Rearrange: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : ...

https://brainly.com/question/1218073

16.48 (character limit)

At least how many numbers should be selected from the set {1, 5, 9, 13, …125} to be assured that two of the numbers selected have a sum of 146?

https://math.stackexchange.com/q/1421899

This is a textbook case of the Pigeonhole principle . You already correctly identified that are 13 containers, each containing two numbers, such that selecting both numbers will yield a sum of 146:

{21, 125}, {25, 121}, \dots, {69, 77}

Fixed points of iterates of a certain map

N \to N

https://math.stackexchange.com/q/1406078

ϕ : N \to N

denote the map that applies the operation that breaks a natural number into the sequence of its digit, forms their sum, and squares the result. We are looking precisely ...

Computing (the number) of paths in a directed graph starting from the initial state

https://math.stackexchange.com/questions/1727182/computing-the-number-of-paths-in-a-directed-graph-starting-from-the-initial-st

There might be exponentially many maximal primitive paths – just take

n

copies of graph given by

V = {0, 1, 2, 3}, E = {(0, 1), (0, 2), (1, 3), (2, 3)},

and connect each

3

0

4 digit number be arranged from 7,2,6,8,5,1 and divisible by 5

https://math.stackexchange.com/questions/2925096/4-digit-number-be-arranged-from-7-2-6-8-5-1-and-divisible-by-5

We have four digits to fill. In order for the final number to be divisible by 5, we have to have 5 as the last digit. All other possible numbers as final digit will result in a number not divisible by ...

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Kopioitu leikepöydälle

271

Selvitä 271 laskemalla yhteen 256 ja 15.

Esimerkkejä

Toisen asteen yhtälö

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

4 sin θ cos θ = 2 sin θ

Ensimmäisen asteen yhtälö

y = 3 x + 4

699 * 533

[2534] [2 - 1 0135]

Samanaikainen kaava

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

\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}