Put \(x-y=a\)

\(\Rightarrow\dfrac{x-y+y-z}{y-z}=\dfrac{a+\left(y-z\right)}{y-z}=\dfrac{a}{y-z}+\dfrac{y-z}{y-z}=\dfrac{x-y}{y-z}+1\)

But \(\dfrac{x-y}{z-y}+\dfrac{x-y}{y-z}=0\)

\(\Rightarrow\dfrac{x-y}{y-z}+1=2+1=3\)

We easy see the diagonal of the square is diameter of smaller circle.

=> The diagonal of the square is: 4.2 = 8 [in]

=> The edge of the square is: \(\dfrac{8}{\sqrt{2}}=4\sqrt{2}\) [in]

=> The area of square is: \(\left(4\sqrt{2}\right)^2=32\left(in^2\right)\)

The area of smaller square is: 4².3,14 = 50,24 [in²]

The area of bigger square is: 8².3,14 = 200,96 [in²]

=> The total area of the shaded regions is:

\(\left(200,96-50,24\right)+32=182,72\left(in^2\right)\)

\(182,72\approx58,16\pi\)

So the answer is 58,16\(\pi\)

Consider triangle ABC right at B

=> tan \(\widehat{ACB}\) = \(\dfrac{AB}{BC}\)

\(\Rightarrow BC=\dfrac{AB}{tan\widehat{ACB}}=\dfrac{CD}{tan30^o}=\dfrac{40}{\dfrac{\sqrt{3}}{3}}=40\sqrt{3}=69,28\)

\(\Rightarrow S_{ABCD}=BC\cdot CD=69,28\cdot40=2771,28\left(mm^2\right)\)

Let A,B respectively with first and second bike's profit 

So we have:

30%A + 50%B = 45%(A+B)

\(\Rightarrow\dfrac{30}{100}A+\dfrac{50}{100}B=\dfrac{45}{100}\text{(A+B)}\)

\(\Rightarrow\dfrac{30A}{100}+\dfrac{50B}{100}=\dfrac{45\text{(A+B)}}{100}\)

\(\Rightarrow30A+50B=45\text{(A+B)}\)

\(\Rightarrow30A+50B=45A+45B\)

\(\Rightarrow50B-45B=45A-30A\)

\(\Rightarrow5B=15A\)

\(\Rightarrow B=3A\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{1}{3}\)

So the ratio of his cost for the first bike to his cost for the second bike is \(\dfrac{1}{3}\)

a quitch + 2 gritches = 20 points

2 quitches + a gritch = 25 points

=> 3 quitches + 3 gritches = 20 + 25 = 45 points

=> a quitch + a gritch = 45/3 = 15 points

=> each gritch worth: 20 - 15 = 5 points

40% of 80 is: 40%.80 = 32

32% of 75 is: 32%.75 = 24

So the positive differnce between 40% of 80 and 32% of 75 is:

32 - 24 = 8

So 75% of it is:

8.75% = 6

Minute hand was runned 30'

=> Location of minute hand at number 6 and equal 1/2 circle

Now location of hour hand at number 3 plus 1/2 distance from number 3 to number 4

=> Location of hour hand equal \(\dfrac{1}{12}\cdot3+\dfrac{1}{2}\cdot\dfrac{1}{12}=\dfrac{7}{24}\left(circle\right)\)

So the difference of minute hand and hour hand is: \(\dfrac{1}{2}-\dfrac{7}{24}=\dfrac{5}{24}\left(circle-of-clock\right)\)

A circle of clock = 360^o 

=> \(\dfrac{5}{24}-circle-of-clock=360^o\cdot\dfrac{5}{24}=75^o\)

So the measure of the smaller angle formed by the minute hand and the hour hand of a clock at 3:30 is 75^o

I redraw this picture

If this two squares don't overlap, the sum their length is:

12 + 12 = 24 [units]

So the new length of rectange lost:

24 - 20 = 4 [units]

=> MN = PQ = 4

=> AM = BN = DP = CQ = \(\dfrac{20-4}{2}=8\)

So the area of \(\Delta BCQ\) is:

\(\dfrac{12\cdot8}{2}=48\left(units^2\right)\)

If this person start go:

- At first door, he/she can exit at 6 rest doors 

- At second door, he/she can exit at 5 rest doors 

- At third door, he/she can exit at 4 rest doors 

- At fourth door, he/she can exit at 3 rest doors 

- At fifth door, he/she can exit at 2 rest doors 

- At sixth door, he/she can exit at 1 rest doors 

- And at last door, he/she can exit at 0 rest doors 

So this person will have:

6 + 5 + 4 + 3 + 2 + 1 + 0 = 21 [ways]

C is the midpoint of AD

\(\Rightarrow AC=CD=\dfrac{AD}{2}=5\left(unit\right)\)

\(\Rightarrow S_{\Delta CDE}=\dfrac{1}{2}\cdot CE\cdot5=30\left(units^2\right)\)

\(\Rightarrow CE=30\div5\div\dfrac{1}{2}=12\left(units\right)\)

\(\Rightarrow BC=\dfrac{2}{3}\cdot12=8\left(units\right)\)

\(\Delta ACB\) has \(\widehat{ACB}=\widehat{ECD}=90^o\left(\text{two opposite top angles}\right)\) 

=> \(\Delta ACB\) right at C

=> \(AB=\sqrt{8^2+5^2}=\sqrt{89}\left(units\right)\)

So AB = \(\sqrt{89}\) units

Tell the number to find is \(\overline{ab0}\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=12\\\overline{ba0}-\overline{ab0}=540\end{matrix}\right.;a< b\)because \(\overline{ba0}>\overline{ab0}\)

=> \(\left(100b+10a\right)-\left(100a+10b\right)=540\)

=> 100b + 10a - 100a - 10b = 540

=> 90b - 90a = 540

\(\Rightarrow90\left(b-a\right)=540\)

\(\Rightarrow b-a=6\)

\(So-\left\{{}\begin{matrix}a=\dfrac{12-6}{2}=3\\b=\dfrac{12+6}{2}=9\end{matrix}\right.\)

So the original number is 390

Paint \(MP\perp AC\) at P.

Consider two right-triangles: \(\Delta HAM-and-\Delta PAM\), we have:

- AM is the common edge \(\left(\text{hypothesis}\right)\)

- \(\widehat{HAM}=\widehat{MAP}-\left(\text{hypothesis}\right)\)

=> \(\Delta HAM=\Delta PAM-\left(\text{hypotenuse - acute angle}\right)\)

=> HM = PM

Consider two right-triangles: \(\Delta ABH-and-\Delta AMH\), we have:

- AH is the common edge \(\left(\text{hypothesis}\right)\)

- \(\widehat{BAH}=\widehat{MAH}\left(\text{hypothesis}\right)\)

=> \(\Delta ABH=\Delta AMH\left(\text{hypotenuse - acute angle}\right)\)

=> BH = MH

\(\Rightarrow BM=MH=\dfrac{1}{2}BC=\dfrac{1}{2}CM\)

\(\Delta MPC\) right have: \(MI=\dfrac{1}{2}CM\)

\(\Rightarrow\widehat{C}=30^o\Rightarrow\widehat{HAC}=60^o\Rightarrow\widehat{BAC}=\dfrac{3}{2}\widehat{HAC}=90^o\)

So \(\left[{}\begin{matrix}\widehat{A}=90^o\\\widehat{B}=60^o\\\widehat{C}=30^o\end{matrix}\right.\)

Put S_n = 4ⁿ + 15n - 10

With n = 1, S₁ = 4¹ + 15.1 - 10 = 9 

Suppose n = k \(\ge1\), so S_k = 4^k + 15k - 10 ⋮ 9

So we must prove that: S_k+1 ⋮ 9

We have: S_k+1 = 4^k+1 + 15(k+1) - 10

                         = 4(4^k + 15k - 10) - 45k + 55 - 10

                         =  4S_k - 9(5k - 5)

But S_k ⋮ 9 => 4S_k ⋮ 9. The other side, 9(5k - 5) ⋮ 9

So S_k+1⋮ 9

So 4ⁿ + 15n - 10 ⋮ 9

Done

a,

\(a^2-a=a\left(a-1\right)\)

\(\left[{}\begin{matrix}a⋮2\Rightarrow a\left(a-1\right)⋮2\Rightarrow a^2-a⋮2\\a⋮̸2\Rightarrow a-1⋮2\Rightarrow a\left(a-1\right)⋮2\Rightarrow a^2-a⋮2\end{matrix}\right.\)

So a² - a⋮2

\(a^3-a=a\left(a^2-1\right)\)

\(\left[{}\begin{matrix}a⋮3\Rightarrow a\left(a^2-1\right)⋮3\Rightarrow a^3-a⋮3\\a⋮3\Rightarrow a^2\equiv1\left(mod3\right)\Rightarrow\left(a^2-1\right)⋮3\Rightarrow a\left(a^2-1\right)⋮3\Rightarrow a^3-a⋮3\end{matrix}\right.\)So a³ - a⋮3

\(a^4-a=a\left(a^3-1\right)=a\left(a-1\right)\left(a^2+a+1\right)=2k\cdot\left[a\left(a+1\right)+1\right]=2k\left(2h+1\right)=4kh+2k\)So \(a^4-a⋮4\) when a = 4k+1 means is don't always

b, 

\(a^3-a⋮3\left(proof-on\right)\)

\(a^3-a=a\left(a^2-1\right)\)

\(\left[{}\begin{matrix}a⋮2\Rightarrow a\left(a^2-1\right)⋮2\Leftrightarrow a^3-a⋮2\\a⋮̸2\Rightarrow a^2⋮̸2\Rightarrow\left(a^2-1\right)⋮2\Rightarrow a\left(a^2-1\right)⋮2\Leftrightarrow a^3-a⋮2\end{matrix}\right.\)

So a³ -a⋮2

But \(\left(2,3\right)=1\Rightarrow a^3-a⋮6\)

a³ - 7a = a³ - a - 6a = \(\left(a^3-a\right)-6a=6k-6a=6\left(k-a\right)⋮6\) - Because a³-a⋮6

So a³-7a ⋮6

\(a^3+11a=\left(a^3-a\right)+12a=6k+12a=6\left(k+2a\right)⋮6\)

So a³+11a ⋮ 6

\(A=1\cdot2+2\cdot3+3\cdot4+...+n\left(n+1\right)\)

\(3A=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot4\cdot3+...+3n\left(n+1\right)\)

\(=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)\(=1\cdot2\cdot3-0\cdot1\cdot2+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)\(=n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow A=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)

There are  numbers in this series

So their sum are: \(\dfrac{26}{2}\left(103+3\right)=1378\)

\(\Delta ABC\) has \(\widehat{A}=60^o\Rightarrow\widehat{B}+\widehat{C}=120^o\)

and \(\dfrac{\widehat{B}}{3}=\widehat{C}\Rightarrow\widehat{B}=3\cdot\widehat{C}\Rightarrow\widehat{B}+\widehat{C}=3\cdot\widehat{C}+\widehat{C}=4\cdot\widehat{C}=120^o\)

\(\Rightarrow\widehat{C}=\dfrac{120^o}{4}=30^o\Leftrightarrow\widehat{B}=120^o-30^o=90^o\)

So \(\left\{{}\begin{matrix}\widehat{B}=90^o\\\widehat{C}=30^o\end{matrix}\right.\)

The length of the insect viewed under this magnifying glass is:

4.3 = 12 [cm]

The volume of rectangular prism measuring is:

4.3.3 = 36 [cm³]

So there are \(\dfrac{36}{0,5^3}=288\) cubes are needed to completely fill this rectangular prism measuring

The circumference : \(r\cdot2\cdot3,14\left(unit\right)\)

The area: \(r^2\cdot3,14\left(square-unit\right)\)

So to the circumference is the same numerical as area, the radius must equal 2 unit

Done