The capsule remain sealed prolonged in:

2017 - 1940 = 77 [years]

Answer: 77 years

\(4:\dfrac{2}{3}-5\)

\(=6-5=1\)

3 = 8.0 + 3

11 = 8.1 + 3

.....

100th term is: 8.99 + 3 

=> Their sum is: \(8\cdot\left(0+1+2+...+99\right)+3\cdot100=8\cdot\dfrac{99\cdot100}{2}+300=39600+300=39900\)

First number: 3 = 8.0 + 3

Second number: 11 = 8.1 + 3

Thirth number: 19 = 8.2 + 3

.....

=> 100^th number is: 8.99 + 3 = 795

Oh, sorry

128,5m = 12850cm

The answer is: 12850.5000 = 64250000 [cm]

The answer: 128,5.5000 = 642500 [m]

To complete this schedule, this cruise ship must cross the road:

22.10 = 220 [mi]

The road this ship cross in first 4 hours is:

16.4 = 64 [mi]

The left road is:

220 - 64 = 156 [mi]

The left hours are:

10 - 4 = 6 [hours]

So its speed must is:

156: 6 = 26 [mi/h]

\(x+\dfrac{4}{x}=y+\dfrac{4}{y}\)

\(\Leftrightarrow\dfrac{x^2+4}{x}=\dfrac{y^2+4}{y}\)

\(\Leftrightarrow y\left(x^2+4\right)=x\left(y^2+4\right)\)

\(\Rightarrow x^2y+4y=xy^2+4x\)

\(\Rightarrow x^2y-xy^2=4x-4y\)

\(\Rightarrow xy\left(x-y\right)=4\left(x-y\right)\)

\(\Rightarrow xy\left(x-y\right)-4\left(x-y\right)=0\)

\(\Rightarrow\left(xy-4\right)\left(x-y\right)=0\)

But x ≠ y 

\(\Rightarrow xy-4=0\)

\(\Rightarrow xy=4\)

There are \(4\cdot\dfrac{11\cdot12}{2}=264\left(different-team\right)\)

\(\dfrac{5\cdot5\cdot10}{3\cdot1\cdot1}=83,\left(3\right)>83\)

So the answer is 83

The interest rate that Marte receives after one year is:

$100 . 4,8% = $4,8

The sum of money that Marte receives after one year is:

$100 + $4,8 = $104,8 = 10480 cents

\(\dfrac{8}{81}=0,\left(098765432\right)\)

So there are 9 digits in parentheses is 0,9,8,7,6,5,4,3,2

But 2012 : 9 = 223 residual 5

So the 2012th digit is 6

Barbara cost \(\dfrac{14}{7-5}\cdot7=49\)$ for his bill

The answer is \(\dfrac{9}{90-1}=\dfrac{9}{89}\) because eliminate number 10 because the units digit is 0, but the tens digit not is 0

\(5\sqrt{x}-30=2\sqrt{x}+54\)

\(\Rightarrow5\sqrt{x}-2\sqrt{x}=54+30=84\)

\(\Rightarrow3\sqrt{x}=84\Leftrightarrow\sqrt{x}=28\)

\(\Rightarrow x=28^2\Leftrightarrow x^2=614656\)

Tell the age of Madison and Harper sequence is a and b

\(\Rightarrow\left\{{}\begin{matrix}a+3a=47\\\dfrac{a+2}{b+2}=2\end{matrix}\right.\)

=> a + 2 = 2b + 4

=> a = 2b + 2

=> a + 3b = 2b + 2 + 3b = 47

=> 5b = 45 

=> b = 9

=> a = 47 - 3.9 = 20

So the age of Harper is 9

- There are \(\dfrac{363-3}{10}+1=37\) digit 3 do units

- There are \(\dfrac{330-30}{100}+1=4\) digit 3 do tenth

- There are \(\dfrac{399-300}{1}+1=100\) digit do hundredth

So there are 100 + 4 + 37 = 141 digit 3 appear as part of a page number of this book

In a hour, two person can dig:

3 : 6 = 1/2 [hole]

Speed of Tom is:

\(\dfrac{1}{2}\div\left(2+1\right)\cdot2=\dfrac{1}{3}\left(holes\text{/}hour\right)\)

So Tom need \(12:\dfrac{1}{3}=36\left(hours\right)\) to dig 12 holes

For \(\left(x^2-1\right)\left(x^2-4\right)\left(x^2-7\right)\left(x^2-1\right)< 0\)

=> In this four numbers, have:

\(\left[{}\begin{matrix}\text{a positive number and 3 negative numbers}\\\text{a negative number and 3 positive numbers}\end{matrix}\right.\)

* If have a positive number and 3 negative numbers:

=> x²-1 > 0 > x²-4 > x²-7 > x²-10

=> x² - 1 > 0 and x² < 4

=> 1 < x² < 4 => No number is satisfy

* If have a negative number and 3 positive numbers:

=> x²-1 > x²-4 > x²-7 > 0 > x²-10

=> x²-7 > 0 and x²-10 < 0

=> 7 < x² < 10

=> x² = 9 because x is a positive integer <=> x² is a positive integer too.

=> x = 3 because x is a positive integer

For \(\left(x^2-1\right)\left(x^2-4\right)\left(x^2-7\right)\left(x^2-10\right)< 0\)

=> \(x^2-1;x^2-4;x^2-7;x^2-10\) have a positive and 3 negative or a nagative and 3 positive

* If have a positive and 3 negative:

=> x² - 1 > 0 > x² - 4 > x² - 7 > x² - 10

=> x² - 1 > 0 and x² - 4 < 0

=> 1 < x² < 4 => x² can equal 2 or 3.

 But x is integer so x² can't equal 2 or 3

* If have a negative and 3 positive:

=> x² - 1 > x² - 4 > x² - 7 > 0 > x² - 10

=> x² - 7 > 0 and x² - 10 < 0

=> 7 < x² < 10

=> x² = 9 because x is integer

=> x = \(\pm3\)

But x is the positive integer => x = 3