Løs 26+27 | Microsoft Math Solver

Hopp til hovedinnhold

26 + 27

= 53

Evaluer

53

Tick mark Image

Faktoriser

53

Tick mark Image

Spørrelek

26 + 27

Videoer

Oops. Something went wrong. Please try again. | Khan Academy

khanacademy.org

Khan Academy Goes Way Beyond Math: New Courses Launch for 2024 School Year in STEM, Humanities, Computer Science, and More

Adding 2-Digit Numbers With Regrouping | Double-Digit Addition | Elementary Math with Mr. J

Lignende problemer fra nettsøk

4 t 2 + 27

http://www.tiger-algebra.com/drill/4t2_27/

4t2+27 Final result : 4t2 + 27 Reformatting the input : Changes made to your input should not affect the solution: (1): "t2" was replaced by "t^2". Step by step solution : Step 1 :Equation ...

Determining the probability of throwing

8

https://math.stackexchange.com/questions/387477/determining-the-probability-of-throwing-8-with-two-dice

If first dice gives

r, 1 \leq r \leq 6;

the second needs to give

8 - r

1 \leq 8 - r \leq 6 ⟹ 7 \geq r \geq 2

2 \leq r \leq 6

So, the required probability

\sum_{2 \leq r \leq 6} P (r) P (8 - r) = 5 \cdot (\frac{1}{6})^{2}

Solving equations that contain summations

https://math.stackexchange.com/questions/291351/solving-equations-that-contain-summations

In your case, here are some results for you to either derive or prove through induction:

\sum_{i = 1}^{n} 1 = n

\sum_{i = 1}^{n} i = \frac{1}{2} n (n + 1)

\sum_{i = 1}^{n} i^{2} = \frac{1}{6} n (n + 1) (2 n + 1)

\sum_{i = 1}^{n} i^{3} = \frac{1}{4} n^{2} (n + 1)^{2}

Formula to exponentially increase an amount over a set of points to get a desired sum

https://math.stackexchange.com/q/2536204

m

: Starting buy amount

r

: Increasing ratio

M

: Total amount Then the previously indicated formula is written as:

M = \sum_{i = 0}^{N - 1} m r^{i} = m \frac{1 - r ^{N}}{1 - r}

The starting buy amount ...

Solving a recurrence inequality

https://math.stackexchange.com/questions/92729/solving-a-recurrence-inequality

You can show easily by induction that

R (x, y) \leq (y x + y) n .

x = 0

y = 0

this is clearly true. Suppose this holds whenever

0 \leq x + y \leq n

x + y = n + 1

x, y > 0

. Then you have

R (x, y) \leq R (x - 1, y) + R (x, y - 1) \leq (y x + y - 1) n + (y - 1 x + y - 1) n = (y x + y) n .

Why does the Cauchy Theorem for triangle fail in this case?

https://math.stackexchange.com/questions/1542930/why-does-the-cauchy-theorem-for-triangle-fail-in-this-case

The Cauchy Integral theorem does not apply here because

f (x + i y) = x^{2} + y i

is not an analytic (holomorphic) function. The partial derivatives of all order exist, but the the Cauchy-Riemann equations ...

Aksje

Kopiert til utklippstavle

53

Legg sammen 26 og 27 for å få 53.

Eksempler

Kvadratisk ligning

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

4 sin θ cos θ = 2 sin θ

Lineær ligning

y = 3 x + 4

699 * 533

[2534] [2 - 1 0135]

Samtidig formel

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

Differensiering

\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}