I agree with Quoc Tran Anh Le

Mr. Bee correctly

Number of students specialized in business is:

100 . 72% = 72 (students)

Number of students specialized math is:

100 . 58% = 58 ( students)

Number of students specialized both is:

( 72 + 58 ) - 100 = 30 (students)

The answer is : 30 students

Because each row is separated by 1 unit and each subsequent row adds a number

So we have : 

Row 5 : 10 12 14 16

I am sorry : 

4 + 4 : 4 = 2

4 : 4 + 4 : 4 = 2

4 + 4 - 4 = 2

...

Still can

We have : 

4 + 4 - 4 = 2

and 4 : 4 + 4 : 4 = 2

...

You should answer question  clearly

Question clearly

We have : Distance numbers are : 142,857

Should the next number be is : 714,285 + 142,857 = 857,142

Answer is : 857,142

This is 1009 and 2017

 Simplify the expression 

You should give the same denominator

Later Addition numerator

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P/S : Question spam

We have : 

\(\dfrac{1}{10.11}=\dfrac{1}{10}-\dfrac{1}{11},\dfrac{1}{11.12}=\dfrac{1}{11}-\dfrac{1}{12},...,\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)

So : 

\(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{10}-\dfrac{1}{100}=\dfrac{9}{100}\)

=> S = \(\dfrac{9}{100}\)

We have :

\(\left(x-2\right)^2\ge0\)

=> \(\left(x-2\right)^2+10+5+3\ge18\)

So \(B\) reach the smallest value equal 18 when the :

\(\left(x-2\right)^2=0< =>x-2=0=>x=2\)

We have : 

\(\left(x-2\right)^2\ge0\)

=> \(\left(x-2\right)^2+5+10+3\ge18\)

So \(x\) reach the smallest value equal 18 when the :

\(\left(x-2\right)^2=0< =>x-2=0=>x=2\)

So June has 30 days, so the number of water reduced is 30%

On the last day of June, the number of water reduced is : 30% - 1% = 29%

The answer is:

10000 : 100% . ( 100% - 29%) = 7100 (water)

10.000 : 100% . 30% = 3000

Call the number of questions that Johnny correctly answers as \(x\) ( \(x\) \(\in\) Z)

We have : \(x+2=60\%.30\)

<=> x = 16

So Johnny answered 16 questions correct

Consider prime numbers from 1 to 100 : (2; 3; 5; 7; ...; 97)

Consider : 100 - 2 = 98 (removed)

100 - 3 = 97 (to choose)

100 - 5 = 95 (removed)

...

So do we get 12 numbers : (97; 3) , (3; 97) , (89; 11) , (11; 89) , (83; 7) , (7; 83) , (71; 29) , (29; 71) , (59; 41) , (41; 59) , (53; 47) , (47; 53)

We have : BCNN (2;3;5) = 30

=> BC (2 , 3 ,5) =  \(\left\{30;60;90;120;...\right\}\)

These numbers are less than 1000

=> BC ( 2, 3 , 5) < 1000 =  \(\left\{30;60;90;120;...;990\right\}\) (1)

Rally (1) has it all : 

\(\dfrac{990-30}{30}\) +1 = 33 (numbers)

Other way , in rally (1) the numbers are B (8) = \(\left\{120;240;...;960\right\}\)  (2)

Rally (2) has it all : 

\(\dfrac{960-120}{120}\) = 8 (numbers)

So has all positive integers in the set of numbers from 1 to 1000 :

33 - 8 = 25 ( numbers) is a multiple of 2, 3 and 5 but not 8

We have: mn = p , n = mp, m = np

mp . np = p <=> mnp² = p

With p = 0, we have : m = n = p = 0

With p others 0, we have : mp . np = p <=> mnp = 1 <=> p² = 1

With p = 1, we have : mn = 1, m = n =>\(\left\{{}\begin{matrix}m=n=1\\m=n=-1\end{matrix}\right.\)

With P = -1, we have : mn = -1 , m= -n =>\(\left\{{}\begin{matrix}m=1,n=-1\\m=-1,n=1\end{matrix}\right.\)

So we have the sets of numbers (m, n ,p) satisfy is : (0;0;0), (1;1;1), (-1;-1;1), (1;-1;-1), (-1;1;-1)