Risolvi {left(15/3.6right)}^2 | Microsoft Math Solver

Calcola

\frac{6 2 5}{3 6} \approx 17.361111111

Procedura della soluzione

(\frac{1 5}{3 . 6})^{2}

Espandi

\frac{1 5}{3 . 6}

moltiplicando numeratore e denominatore per

10

(\frac{1 5 0}{3 6})^{2}

Riduci la frazione

\frac{1 5 0}{3 6}

ai minimi termini estraendo e annullando

6

(\frac{2 5}{6})^{2}

Calcola

\frac{2 5}{6}

alla potenza di

2

e ottieni

\frac{6 2 5}{3 6}

\frac{6 2 5}{3 6}

Scomponi in fattori

\frac{5 ^{4}}{2 ^{2} \cdot 3 ^{2}} = 17 \frac{1 3}{3 6} \approx 17.361111111

Quiz

Arithmetic

5 problemi simili a:

(\frac{1 5}{3 . 6})^{2}

Problemi simili da ricerca Web

What is the remainder of

\frac{2 ^{2015}}{3 6}

https://math.stackexchange.com/questions/1564191/what-is-the-remainder-of-frac2201536

Since

g cd (2, 9) = 1

, we have from Euler's theorem, that

2^{ϕ (9)} \equiv 1 (m o d 9) ⟹ 2^{6} \equiv 1 (m o d 9)

This gives us that

2^{2010} \equiv 1 (m o d 9) ⟹ 2^{2015} \equiv 2^{5} (m o d 9) \equiv 5 (m o d 9)

...

Probability question (Birthday problem)

https://math.stackexchange.com/questions/140242/probability-question-birthday-problem

The basic idea was right, and a small modification is enough. Line up the people in some arbitrary order. There are, under the usual simplifying assumption that the year has

365

days,

36 5^{23}

...

Probability of Getting a Yahtzee of Fives Given Two Fives

https://math.stackexchange.com/questions/1826535/probability-of-getting-a-yahtzee-of-fives-given-two-fives

Notice that this expression is in the form of:

a^{3} + 3 b a^{2} + 3 b^{2} a + b^{3}

This is a form that you might see a lot in math problems like this, so you'll simply have to learn to recognize it, just like ...

n

-sect using bisection geometrically.

https://math.stackexchange.com/q/2737093

Minor mistake in

(b) (i)

To get

\frac{1}{4}

of the original length, you need two bisections as you have stated in part

(a)

. For part

(b) (i i)

, copy

A D

n

times. if

n = 3

, copy it

3

times. The ...

Largest integer less than

2013

obtained by repeatedly doubling an integer

x

https://math.stackexchange.com/questions/1885216/largest-integer-less-than-2013-obtained-by-repeatedly-doubling-an-integer-x

If you start with a number less than

100

, you can double it at least

4

times without exceeding

2012

, so the answer must be a multiple of

2^{4} = 16

. The largest multiple of

16

less than

2013

...

Copia

Copiato negli Appunti

\left(\frac{150}{36}\right)^{2}

Espandi \frac{15}{3.6} moltiplicando numeratore e denominatore per 10.

\left(\frac{25}{6}\right)^{2}

Riduci la frazione \frac{150}{36} ai minimi termini estraendo e annullando 6.

\frac{625}{36}

Calcola \frac{25}{6} alla potenza di 2 e ottieni \frac{625}{36}.

Esempi

Equazione quadratica

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

Trigonometria

4 sin θ cos θ = 2 sin θ

Equazione lineare

y = 3 x + 4

Aritmetica

699 * 533

Matrice

[2534] [2 - 1 0135]

Equazione simultanea

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

Differenziazione

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

Integrazione

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

Limiti

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

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