All Answers 358091 Answers

A number is "factorific" if the sum of the factorials of each of its digits equals the number itself. Find the only three-digit factorific number.

Recall that a factorial is a product of all the integers from 1 to a specific number. For example 

1! = 001

4! = 024

5! = 120

1! + 4! + 5! = 145

-log(0.05) ?

$$\small{\text{$
\begin{array}{rcl}
-\log_{10}{(0.05)} &=& \log_{10}{(1)} - \log_{10}{(0.05)} \qquad | \qquad \log_{10}{(1)} =0 \\ \\
\log_{10}{(1)} - \log_{10}{(0.05)} &=& \log_{10}{
\left(\dfrac{1}{0.05} \right)
}\\\\
&=& \log_{10}{
\left(\dfrac{100}{5} \right)
}\\\\
&=& \log_{10}{(20)}\\\\
&=& \log_{10}{(2\cdot 10)}\\\\
&=& \log_{10}{(2)}+\log_{10}{10)} \qquad | \qquad \log_{10}{(10)} =1\\\\
&=& \log_{10}{(2)}+1\\\\
&=& 1+\log_{10}{(2)}\\\\
&=& 1+0.30102999566\\\\
-\log_{10}{(0.05)} &=& 1.30102999566\\\\
\end{array}
$}}$$

is the sequence arithmetic?if so, identify the common difference. 14,21,42,77...

This ist a ARITHMETIC SEQUENCE OF HIGHER ORDER.

The sequence  is arithmetic of order k if the differences of order k are equal.

We have the order k = 2. The second differences are equal = 14.

Let us see:

$$\small{\text{$
\begin{array}{lcccccccccc}
$Number $a &a_1=\textcolor[rgb]{1,0,0}{ 14}& & 21& & 42& &77 & &126 & \cdots \\
$First difference $D^1 & & D_0^1=\textcolor[rgb]{1,0,0}{7}& & 21 & & 35 & & 49 & \cdots \\
$Second difference $D^2 & & & D_0^2=\textcolor[rgb]{1,0,0}{14}& & 14& &14 & \cdots \\
\end{array}
$}}$$

If we have a arithmetic sequence of order k, we can find $$a_n$$ by :

$$\boxed{~~a_n = a_1 + \binom{n-1}{1}\cdot D_0^1 + \binom{n-1}{2}\cdot D_0^2+\cdots+\binom{n-1}{k}\cdot D_0^k
~~}$$

So the nth term is given by:

$$\small{\text{$
\begin{array}{rcl}
a_n &=& a_1 + \binom{n-1}{1}\cdot D_0^1 + \binom{n-1}{2}\cdot D_0^2\\\\
a_n &=& \textcolor[rgb]{1,0,0}{14} + \binom{n-1}{1}\cdot \textcolor[rgb]{1,0,0}{7} + \binom{n-1}{2}\cdot \textcolor[rgb]{1,0,0}{14} \qquad
| \qquad \binom{n-1}{1} = n-1 \qquad \binom{n-1}{2}=\dfrac{(n-2)(n-1)}{2}\\\\
a_n &=& \textcolor[rgb]{1,0,0}{14} + (n-1)\cdot \textcolor[rgb]{1,0,0}{7} + \dfrac{(n-2)(n-1)}{2}\cdot \textcolor[rgb]{1,0,0}{14} \\\\
a_n &=& \textcolor[rgb]{1,0,0}{14} + (n-1)\cdot \textcolor[rgb]{1,0,0}{7} + (n-2)(n-1)\cdot 7 \\\\
a_n &=& \textcolor[rgb]{1,0,0}{14} + 7(n-1)[1+(n-2)]\\\\
a_n &=& \textcolor[rgb]{1,0,0}{14} + 7(n-1)(n-1)\\\\
\mathbf{a_n} & \mathbf{=} & \mathbf{14 + 7(n-1)^2} \qquad | \qquad n \ge 1 \\
\end{array}
$}}$$

1 + 1  is not a  reasonable question !

9999999³  you are unable to tipp into the calculator?

I will do it for you :

$${{\mathtt{9\,999\,999}}}^{{\mathtt{3}}} = {\mathtt{999\,999\,700\,000\,029\,999\,999}}$$

I hope you are happy now !

- (5x+17a-6) - (-42a-5x)

First we can expand out the equations, since it's all addition and subtraction

-5x - 17a + 6 + 42a + 5x

Now seperate them according the the variables

[5x - 5x] + [42a - 17a] + 6

Simplify it:

25a + 6

We have 

$$e^{i\pi}=-1$$

This is one form of Euler's well known relationship.

$$e^{i\pi}\rightarrow \cos{\pi}+i\times\sin{\pi}\rightarrow -1+i\times0\rightarrow-1$$

.

It's probably simpler here just to cancel the √85 terms in the numerator and denominator, so that you are immediately left with -7/6.

.

Well, I am no stats expert but... this is a binomial distribution.

Each time the die is rolled there is a 1/6 = 0.16 repeater chance of rolling a 5.

You are asking if 23 fives is a probable outcome.

I am going to jump over to a binomial calculator 

Mila krbavcic's rogen 1942 in the village of rumenima after that is there in the village of rumenima after that is there in the village of krnavcici, ...

This does not seem to be a question :/

Ovo se ne čini da se pitanje :/

3 · {(300 - 80 ÷ 5) - [3 · 25 - (8 - 2 · 3)]} 

$$\\=3 \times \{(300 - 80 \div 5) - [3 \times 25 - (8 - 2 \times 3)]\}\\\\
$Do the stuff in the brackets, and use PEDMAS $\\
$Multiplication and division is done before addition and subtraction$\\\\
=3 \times \{(300 - 16) - [3 \times 25 - (8 - 6)]\}\\\\
=3 \times \{(284) - [3 \times 25 - (2)]\}\\\\
=3 \times \{284 - [3 \times 25 - 2]\}\\\\
=3 \times \{284 - [75 - 2]\}\\\\
=3 \times \{284 - [73]\}\\\\
=3 \times \{284 - 73\}\\\\
=3 \times \{211\}\\\\
=3 \times 211\\\\
=633$$ 

By rectangle triangle I assume that you mean a right angled triangle.

and that you have the 2 sides that are at right angles to one another.

The Hypoteunuse will be 10 cm 

Anyway here is your triangle.     atan is the same as inverse tan which is  $$tan^{-1}$$

$$\small{\text{$
\begin{array}{l}
\small{\text{here is an easier way, because: }}
0.28+0.72 = 1
\end{array} $}} \\\\
\small{\text{$
\begin{array}{rcl}
\mathbf{c }& \mathbf{=}& \mathbf{ -0.72 + 0.28a } \\
c &=& 0.28a-(1-0.28) \\
c &=& 0.28a - 1 + 0.28 \\
(1+c) &=& 0.28\cdot (1+a) \qquad | \qquad \text{ we set } C= 1+c \text{ and } A = 1+a\\
C &=& 0.28\cdot A \qquad | \qquad \cdot 25 \quad \text{ to get the lowest integer }\\
25C &=& 7A\\
25C-7A &=& 0 \qquad | \qquad \text{ we see } gcd{(25,7)}= 1 \\
&& \text{We have immediately the solution}\\
25 \cdot \textcolor[rgb]{1,0,0}{7} - 7\cdot \textcolor[rgb]{1,0,0}{25} &=& 0 \qquad | \qquad C = 7 \text{ and } A = 25\\
&& \text{and back}\\
C = 1+c &=& 7 \text{ so } \mathbf{c} = 7-1 \mathbf{=6} \\
A = 1+a &=& 25 \text{ so } \mathbf{a}= 25-1 = \mathbf{24} \\
\end{array} $}}$$

excursion:

We have also a function for a:

$$\small{
\begin{array}{rcl}
7a^4-6304a^3+379996a^2-5543568a-1008000&=&0\\
a_1 &=& -0.18 \\
a_2 &=& 24 \\
a_3 &=& 39.92 \\
a_4 &=& 836.83\\\\
\end{array}
}$$

$${\frac{{\mathtt{\,-\,}}\left({\mathtt{7}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{85}}}}\right)}{\left({\mathtt{6}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{85}}}}\right)}}$$

Put the co-efficiencts into the surds by squaring them...

$${\frac{{\mathtt{\,-\,}}{\sqrt{{\mathtt{85}}{\mathtt{\,\times\,}}{\mathtt{49}}}}}{{\sqrt{{\mathtt{85}}{\mathtt{\,\times\,}}{\mathtt{36}}}}}}$$

Simplify...

$${\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{4\,165}}}}}{{\sqrt{{\mathtt{3\,060}}}}}}$$

Which can be represented as:

$${\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\sqrt{{\mathtt{3\,060}}}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{4\,165}}}}\right)$$

So we can put the surds together, again by squaring it:

$${\mathtt{\,-\,}}{\sqrt{{\mathtt{4\,165}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{{\sqrt{{\mathtt{3\,060}}}}}^{\,{\mathtt{2}}}}}\right)}}$$

Simplify...

$${\mathtt{\,-\,}}{\sqrt{{\frac{{\mathtt{4\,165}}}{{\mathtt{3\,060}}}}}}$$

Now we can simplfy the fraction:

$${\mathtt{\,-\,}}{\sqrt{{\frac{{\mathtt{833}}}{{\mathtt{612}}}}}}$$

And further...

$${\mathtt{\,-\,}}{\sqrt{{\frac{{\mathtt{49}}}{{\mathtt{36}}}}}}$$

11666666666666666666666

Let's add some commas to make it easier to read... (Or periods in some countries)

11,666,666,666,666,666,666,666

Now, the level of magnitude of the 11 is 10^21, also known as "Sextillion".

So we have:

11 Sextillion, 666 Quintillion, 666 Quadrillion, 666 Trillion, 666 Million, 666 thousand, 6 hundred and 66

or: 

Eleven Sextillion, six hundred and sixty six Quintillion, six hundred and sixty six Quadrillion, six hundred and sixty six Trillion, six hundred and sixty six Million, six hundred and sixty six thousand, six hundred and sixty six.

$${\sqrt{{\mathtt{x}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}}}$$

Let's combine the surds. To do this, square one of them (so put it in terms of it's square root)

$${\sqrt{{{\sqrt{{\mathtt{x}}}}}^{\,{\mathtt{2}}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$

Now simplify since obviously the square of a square root is...

$${\sqrt{{\mathtt{x}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$

Expand out the equation inside the surd and you get:

$${\sqrt{{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}}}$$

Assuming it is not compound interest (Since if it was, the daily gain would change on a yearly basis):

1 year of interest = 10,000 * 5%

= 10,000 * ¹/₂₀

= ^10,000/₂₀

= 500

1 day of interest = ⁵⁰⁰/₃₆₅

= $1.37

200 * ¹/₄ * ¹/₂ * ¹/₅

= 100 * ¹/₄ * ¹/₅

= 25 * ¹/₅

= 5

Depends how you mean "add".

If you mean something like -7 + 4, then look at which number is larger. The larger number subtracted by the smaller number gives you the resulting value, and if the larger number is the negative one, then the resulting value is negative. so -7 + 4 = - (7  -4) = - (3) = -3

Consider positives as "removing" negative values, and negatives "removing" positive values. It's easier to work this way as long as you identify which value is bigger.

If you mean adding a negative number as though it was a positive number, this is represented as "|x|" meaning you get the value of x in it's positive form (So if x = -8 then |x| = 8, and if x = 4 then |x| = 4).

-3 + -5 + 1

Consider that "+ -" is just another way of saying "minus" or " - " (I actually do this all the time personally. Makes it easier to understand numbers as negatives and not negative as a function).

So we have:

- 3 - 5 + 1

Combine the negatives together:

- 8 + 1

Now combine these two together:

 -7

to type the inverse trigonomic function, enter "a" before it.

This gives "acos", "asin" and "atan" as the invesre functions. (In your case it would be "asin")

364 / 18   = divide numerator and denominator by 2  

182 / 9   =   20 + 2/9     ...and that's it !!!!

Yep....geometric.....the ratio is (2/3)

The series is given by:

(1/3)(2/3)^(n - 1)   for n ≥ 1

This series is not arithmetic, because there is no common difference ......the nth term - for n ≥ 2 - is given by:

14 + 7(n-1)^2