There are: \(2.3=6\left(numbers\right)\)

So the answer is A: \(6\)

There are \(3\) small triangles. (Only contains 1 triangle/each)

There are \(2\) medium triangles. (Contains 2 small triangles/each)

There are \(1\) big triangle.  (Contains 3 triangles)

So there are: \(3+2+1=6\left(triangles\right)\) in this figure

From the question, we have:

\(34.35+150\le5x\)

<=> \(1190+150\le5x\) 

<=> \(1340\le5x\)

The smallest value of \(x\) will be: \(x=1340:5=268\)

The possible answer of the question is \(268\) so there is no possible answer A,B,C,D to choose.

\(x\in\left\{-3;-2;-1;0;1;2;3;4;5;6;7;8;9\right\}\)

Tran Nhat Duong, do not copy my answer !!!

(CONTINUE) 

=> The value of \(x+y+z=7+2+5=14\)

So the answer is C: 14

We have: \(x.y=14\) and \(y.z=10\) and \(z.x=35\)

<=>\(x.y.y.z.z.x=14.10.35\)

<=>\(x^2.y^2.z^2=4900\)

<=>\(\left(x.y.z\right)^2=70^2\)

=> \(x.y.z=70\) <1>

From the context and <1>

=> \(x=\left(x.y.z\right):\left(y.z\right)=70:10=7\)

=> \(y=\left(x.y\right):x=14:7=2\)

=> \(z=\left(z.x\right):x=35:7=5\)

So \(x=7;y=2;z=5\)

Arrcoding to the question, we have the ratio of the salt and fresh water in Protaras is 7:193

=> The ratio of the salt and the sea water = 7:(193+7) = 7:200

First, we exchange: \(1000hg=100kg\) (sea water).

Because the ratio of the salt and sea water is 7:200

=> \(7kg\) salt are in \(200kg\) sea water.

\(200kg\) sea water are more than \(100kg\) sea water: \(200:100=2\left(times\right)\)

In \(100kg\) sea water there is: \(7:2=3,5\left(kg\right)\) salt.

(This question has no answer because there is no 3,5 kg salt.

If there is an answer, you must change \(1000hg\) sea water -> \(1000kg\) sea water or change the questions into: How many hectograms of salt are there in 1000hg of sea water)

What for?

The largest number of stories are: \(30:2=15\)

So the answer is A: \(15\)

The score of the game they draw must be \(0-0\) because if the score is 1-1 or 2-2 or more, then they will have no game to lose. (They only let 1 goal in).

The score of the game they lose must be \(0-1\) because they only let 1 goal in from the 3 matches, and if they score more goals then the score will be a draw or a win.

From the context, the score in the game they win must be \(3-0\) ( they scored 3 goals and 1 goal been let in was the lost match).

So the answer is B: \(3-0\)

There are: \(25-12=13\left(students\right)\) study only Biology.

There are: \(20-12=8\left(students\right)\) study only Chemistry.

There are: \(50-\left(13+8+12\right)=17\left(students\right)\) took neither Biology nor Chemistry.

ANSWER: 17 students

\(A=\dfrac{2x-5}{x+2}=\dfrac{2x+4-9}{x+2}=\dfrac{2x+4}{x+2}-\dfrac{9}{x+2}\)

Because \(\dfrac{2x+4}{x+2}\) is an integer

=> \(\dfrac{9}{x+2}\) must be an integer.

=> \(x+2\in GCD\left(9\right)\)

=> \(x+2\in\left\{\pm1,\pm3,\pm9\right\}\)

=> \(x\in\left\{-1,-3,1,-5,7,-11\right\}\)

=> \(x\in\left\{\pm1,-3,-5,7,-11\right\}\)

\(A=\dfrac{2x-5}{x+2}=\dfrac{2x+4-9}{x+2}=\dfrac{2x+4}{x+2}-\dfrac{9}{x+2}\)

Because \(\dfrac{2x+4}{x+2}\) is an integer

=> \(\dfrac{9}{x+2}\) must be an integer

Because multiplies of 6 are all even numbers.

The least possible number is \(6\) and the biggest possible number is \(198\).

We have: \(\left(198-6\right):6+1=33\left(numbers\right)\) are even numbers and multiplies of 6.

ANSWER: 33 numbers.

There is only 1 condition: \(x\in N\) \(\left(x\ge0\right)\)

\(\left|x+1\right|=\pm x+1\)

=> There are 2 conditions with \(\left|x+1\right|\)

Condition 1: \(x+1\ge0\left(x\in N\right)\)

=> \(x+1=x+1\)

Condition 2: \(x+1< 0\)

=> \(x+1\ne x+1\)

\(\dfrac{-5}{6}+\dfrac{-29}{6}+\dfrac{16}{6}\le x\le\dfrac{-1}{2}+\dfrac{4}{2}+\dfrac{5}{2}\)

\(-3\le x\le4\)

\(x=-3;-2;-1;0;1;2;3;4\)

\(x.\left[\left(1+100\right)\cdot100:2\right]=10100\)

\(x.5050=10100\)

\(x=10100:5050\)

\(x=2\)

\(=\left(\dfrac{1}{2}+\dfrac{-1}{2}\right)+\left(\dfrac{-1}{3}+\dfrac{1}{3}\right)+\left(\dfrac{1}{4}+\dfrac{-1}{4}\right)+\left(\dfrac{-1}{5}+\dfrac{1}{5}\right)+\left(\dfrac{1}{6}+\dfrac{-1}{6}\right)+\left(\dfrac{-1}{7}+\dfrac{1}{7}\right)+\dfrac{1}{8}\)

\(=0+0+0+0+0+0+\dfrac{1}{8}\)

\(=\dfrac{1}{8}\)

The answer is B: 1,2cm.

The sum of the perimeters of the 3 small triangles are: \(1,2\cdot3\cdot3=10,8\left(cm\right)\)

The length of the side of the green hexagon is: \(6-1,2\cdot2=3,6\left(cm\right)\)

Because there are 2 sides of the hexagon are not in the perimeter of the big triangle so that the perimeter of the big triangle is: \(10,8\cdot2-\left(10,8:6\cdot2\right)=18\left(cm\right)\) <1>

The perimeter of the big triangle is: \(6\cdot3=18\left(cm\right)\) <2>

Because <1> = <2> \(\left(18cm=18cm\right)\)

=> The side of the small triangle is B: 1,2cm.