Total height of 2 boys are:

                131.6 x 2 = 263.2(cm)
Total height of 4 girls are:

                      128.3 x 4 = 513.2(cm)
Total height of all children are:  

                263.2 + 513.2 = 776.4(cm)
The average height of all children are:

               776.4 : 6 = 129.4(cm)

Fairy Tail

Mai has the stamp number is : 

40 + 7 = 47 ( stamps )

Answer : 47 stamps

Fairy Tail

 When mai brushing any 2 number and write their sum on the board , the end number is the sum of 10 number

The end number : 

1 + 2 + 3 + ..... + 9 + 10  = 55

Summer Clouds

 The edge of the square is : 

$\sqrt{9}=3$$\sqrt{\mathrm{<i>9</i>}}<i>=</i>\mathrm{<i>3</i>}$ (cm)

Hence,the diagonal OH of the square or the radius of the semicircle is $3\sqrt{2}$$\mathrm{<i>3</i>}\sqrt{\mathrm{<i>2</i>}}$cm.

The area of the semicircle is : $\frac{{\left(3\sqrt{2}\right)}^2\pi }{2}=9\pi$${\displaystyle \frac{{<i>(</i>\mathrm{<i>3</i>}\sqrt{\mathrm{<i>2</i>}}<i>)</i>}^{\mathrm{<i>2</i>}}\mathrm{<i>\pi </i>}}{\mathrm{<i>2</i>}}}<i>=</i>\mathrm{<i>9</i>}\mathrm{<i>\pi </i>}$(cm2)

Summer Clouds

 I solved the first way for you.This is the second way :

We have :$x^2-5x+4=0\Rightarrow a=1;b=-5;c=4$${\mathrm{<i>x</i>}}^{\mathrm{<i>2</i>}}<i>-</i>\mathrm{<i>5</i>}\mathrm{<i>x</i>}<i>+</i>\mathrm{<i>4</i>}<i>=</i>\mathrm{<i>0</i>}<i>\Rightarrow </i>\mathrm{<i>a</i>}<i>=</i>\mathrm{<i>1</i>}<i>;</i>\mathrm{<i>b</i>}<i>=</i><i>-</i>\mathrm{<i>5</i>}<i>;</i>\mathrm{<i>c</i>}<i>=</i>\mathrm{<i>4</i>}$

$\Rightarrow b^2-4ac=25-16=9>0$$<i>\Rightarrow </i>{\mathrm{<i>b</i>}}^{\mathrm{<i>2</i>}}<i>-</i>\mathrm{<i>4</i>}\mathrm{<i>a</i>}\mathrm{<i>c</i>}<i>=</i>\mathrm{<i>25</i>}<i>-</i>\mathrm{<i>16</i>}<i>=</i>\mathrm{<i>9</i>}<i>></i>\mathrm{<i>0</i>}$$\Rightarrow \sqrt{b^2-4ac}=3$$<i>\Rightarrow </i>\sqrt{{\mathrm{<i>b</i>}}^{\mathrm{<i>2</i>}}<i>-</i>\mathrm{<i>4</i>}\mathrm{<i>a</i>}\mathrm{<i>c</i>}}<i>=</i>\mathrm{<i>3</i>}$

$\Rightarrow [\begin{array}{c}x=\frac{5+3}{2}=4\end{array}$$\begin{array}{c}x=\frac{5-3}{2}=1\end{array}$

Summer Clouds 

 $x^2-5x+4=0$${\mathrm{<i>x</i>}}^{\mathrm{<i>2</i>}}<i>-</i>\mathrm{<i>5</i>}\mathrm{<i>x</i>}<i>+</i>\mathrm{<i>4</i>}<i>=</i>\mathrm{<i>0</i>}$

$\iff \left(x^2-4x\right)-\left(x-4\right)=0$$<i>\iff </i><i>(</i>{\mathrm{<i>x</i>}}^{\mathrm{<i>2</i>}}<i>-</i>\mathrm{<i>4</i>}\mathrm{<i>x</i>}<i>)</i><i>-</i><i>(</i>\mathrm{<i>x</i>}<i>-</i>\mathrm{<i>4</i>}<i>)</i><i>=</i>\mathrm{<i>0</i>}$

$\iff \left(x-1\right)\left(x-4\right)=0\Rightarrow [\begin{array}{c}x=1\end{array}$$\begin{array}{c}x=4\end{array}$

Summer Clouds 

$2^{32}={\left(2^4\right)}^8={16}^8$${\mathrm{<i>2</i>}}^{\mathrm{<i>32</i>}}<i>=</i>{<i>(</i>{\mathrm{<i>2</i>}}^{\mathrm{<i>4</i>}}<i>)</i>}^{\mathrm{<i>8</i>}}<i>=</i>{\mathrm{<i>16</i>}}^{\mathrm{<i>8</i>}}$ has 6 as the last digit

So the answer is 6

Summer Clouds

In a triangle,the sum of 2 arbitrary sides' lengths is greater than the length of the other side.We have :

2 + 3 > 4 ; 5 + 6 = 11 ; 4 + 4 = 8 ; 1 + 2 = 3

Hence,the answer is (A)

Maths

1dm = 10 cm

The area of shaded triangle :

10 x 36 : 2 = 180 (cm²)

Carter

Each time, a person takes 1, 2, 3, or 4 chips. The strategy for a player to win the game is taking the number of chips so that the number of chips left is the form of 5k + 1 .

If It is my turn, and there are 2014 chips in the pile, I shoud take 3 chips and the left is 2011 (=5.k +1). And if the opponent takes x chips (x = 1, 2, 3, 4) , i win take 5 - x chips to guarantee the left is always in the form 5k + 1. And finally, the opponent will take the last chip.

$\left(1\right)$ The first thing I did to solve this problem was write out every step, so it  would look something this:

$6+5.4+4.4+3.4+2.4+1.4$

$=6+15.4=6+60=66$

$\left(2\right)$  I was looking at the patterns that I had already found and then realized I needed to find a pattern between $n$ (the number of cubes high) and the total number of cubes in the tower so I made this chart:

I found that the number being multiplied to get the total number of cubes was going up by odd numbers.  I then noticed that it would be equal to $2n-1$.  I then found the equation for all n values to be:

Total number of cubes $=n\left(2n-1\right)$

Total number of cubes $=2n^2-n$

So total number of cubes $=2n^2-n$

Applying the converse of Pythagoras theorem,we have :

(1) $\{\begin{array}{c}{\left(\sqrt{1}\right)}^2+{\left(\sqrt{2}\right)}^2=3\\ {\left(\sqrt{3}\right)}^2=3\end{array}$$\Rightarrow {\left(\sqrt{1}\right)}^2+{\left(\sqrt{2}\right)}^2={\left(\sqrt{3}\right)}^2$

(2) $\{\begin{array}{c}{3}^2+4^2=25\\ 5^2=25\end{array}$=> 32 + 42 = 52

(3)$\{\begin{array}{c}{11}^2+{12}^2=263\\ {13}^2=169\end{array}$$\Rightarrow {11}^2+{12}^2\ne {13}^2$

(4)$\{\begin{array}{c}{1665}^2+{2220}^2={555}^2{.3}^2+{555}^2{.4}^2={555}^2.25\\ {2775}^2={555}^2{.5}^{}\end{array}$

We have : AE = AD = 10 cm (because of the folding)

Applying the Pythagoras theorem to the right $\Delta ABE$,we have :

$BE=\sqrt{A{E}^2-A{B}^2}=\sqrt{100-64}=\sqrt{36}=6$(cm)

=> EC = BC - BE = 10 - 6 = 4 (cm) 

The edge of the small square is :

                               8 - 6 = 2 (cm)

The area of the small square is :

                      22 = 4 (cm2)

The area of each right triangle is :

               $\frac{6\times 8}{2}=24$(cm2)

The area of the given square is :

                    4 + 24 x 4 = 100 (cm2)

Nguyễn Ngọc Mai

100 + 500 : 5 - 40

= 100 + 100 - 40

= 200 - 40

= 160

Bùi Vân                      

Price is : 

2400:80x100=3000$

Nguyễn Phương Linh

Time upon the road as:

 9h42'-8h30'=1h12'

  1h12'=1,2h

Motor of motor are : 

60:1,2=50(km/h)

The amount of money she spends in all is:

18.03 + 19.94 = 37.97 (dollars)

Solve : 

For (A) with nn people at the party, what are the possible number of hands each person could shake?

Show that it is impossible for both someone to have shaken hands with everyone and someone to have shaken hands with noone simultaneously.

Is there some way to use the pigeon-hole principle here?

For (B) count the number of handshakes that occur by adding up how many handshakes people participated in individually and recognize that this overcounted somehow (because each handshake we counted twice: once from the shorter person of the two and again from the taller person). This gives us a useful result called "the handshaking lemma."

What happens then if there are an odd number of people who shook an odd number of hands? Remember that the number of handshakes must be a whole number.

Maths

                              6 x 45 = 270 ( legs ) 

 MathYou