So as to have the function be increasing, 2016-m² must be positive. Therefore, 2016 must be greater than m². Since 2016 is a positive number, m has to range from \(-\sqrt{2016}\) to \(\sqrt{2016}\). In other words, m has to be between -44.899 and 44.899. Since m is a whole number, the minimal value of m is -44 and the maximum is 44. To calculate the number of values, we use this formula:
\(N=\frac{44-(-44)}{1}+1=89\)

To conclude, there are 89 values of m such that the function is increasing.

The sum of the three numbers is 120.
The sum of the two remaining numbers is 101.
The lowest possible value for the second number is 20.
So the highest number would be 81.

Analyzing the numbers first, we can see that each number increases by 1, 2, 3, 4, 5 ... progressively arithmetically increasing.
So the number that comes next is definitely 22
Analyzing the letters, it will be the order of the letter in the Latin alphabet.
So the letter is V
So it's V22

Mininum divisors? That is the number 1, obviously

Or did you mean the minimum NUMBER of divisors?
If so than it's 9, how you ask?
To create less permutations with the divisors, the factors should overlap the most, hence, the 6 divisor number is a⁵ and the 4 divisor number is a³, so the product would be a⁸, having 9 divisors.

M=2
A=1
T=7
H=8
MATH=2178
How did I solve this? Continuous trial and error.

\(x+4567108=\left(-4566109\right)\)
\(x=\left(-4566109\right)-4567108=-9133217\)
\(x\times\left(-8\right)=\left(-9133217\right)\times\left(-8\right)=73065736\)
I have no idea how one is supposed to do this without a calculator

We have the domain of f(x), time to replace all of these values into the variables...
After a while of repetition, we will get the range of f(x) is {-3;-2;1;6;13}
So the intersection of these two sets is {-3;-2;1}

*written that

7 numbers: 0,1,2,3,4,5,6 written that in both bases

b=2772 and a=757

I don't think this needs much explanation, G is obviously 1.
L=5 and S=8, hence M=9

I'm not sure, probably 2 because...
2=1+1
5=4+1
8=7+1
11=10+1
14=13+1
17... IDK
I'm not sure, probably 2

So I've said, it's 2x²-x+7

Simplifying? Okay
\(\dfrac{2}{x-1}+\dfrac{2x-1}{x^2+x+1}-\dfrac{x^2+6x+2}{1-x}=\dfrac{2}{x-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{x^2+6x+2}{x-1}\)

Condition for x: \(x\ne1\)
\(\dfrac{2\left(x^2+x+1\right)}{x^3-1}+\dfrac{\left(2x-1\right)\left(x-1\right)}{x^3-1}+\dfrac{\left(x^2+6x+2\right)\left(x-1\right)}{x^3-1}\)
\(\Rightarrow2x^2+2x+2+2x^2-2x-x+1+x^3+6x^2+2x-x^2-6x-2\)
\(=x^3+3x^2-5x+1\)
(I can't factorize that polynomial )

a. \(11^2=121;20^2=400\) so the answer is 400
b. \(3^3=27;4^3=64\) so the answer is 64

1) The smallest 3 digit prime number is 101. The biggest 4 digit prime number is 9973. And the sum of those numbers is 10074.
2) Anna walked \(8762-80^2=8762-6400=2362\left(m\right)\) more than Becky. (Those mad girls walking such long distances for no reason)
3) The biggest number among them is C)8977

\(v_{tb}=\dfrac{s+s}{t_1+t_2}=\dfrac{2s}{\dfrac{s}{v_1}+\dfrac{s}{v_2}}=\dfrac{2}{\dfrac{v_1+v_2}{v_1v_2}}=\dfrac{2v_1v_2}{v_1+v_2}=\dfrac{2\cdot4\cdot12}{4+12}=\dfrac{96}{16}=6\left(mph\right)\)

After each round, the number of people will divide by half.
So 32 => 16 => 8 => 4 => 2 => Champion
The number of matches each round will be determined by the value of HALF the number of participants:
So there will be 16+8+4+2+1=31(matches)

The 500th number will be 5+6(500-1)=2999

The sum of each numbers of each set is 15
So the last digit in set g is 4

1 second before crash, each of them will need to travel 1 second until they trash
So #1 has to travel 15 mers, and #2 has to travel 25 mers, so the distance will be 40 mers

16,637 because the sum of the digits isn't equal to 20

Think of these numbers as points on a number line.
Between b and e, there would be exactly 3 equal segments.
So each of those segment would be \(\dfrac{10-5.5}{3}=1.5\)
Therefore a would be \(5.5-1.5=4\)

How are we supposed to calculate without a given value of x?

Sorry but... I have no idea how to calculate that when you haven't given us any values of x to work with...

Wait what? Calculate? But you didn't even give me what x is... Well, I guess what you meant is... SOLVE 

\(A=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}-\dfrac{\left(1-x\right)}{9-x^2}=\dfrac{x+1}{x-3}+\dfrac{x-1}{x+3}+\dfrac{1-x}{x^2-9}\)

The condition for x is \(x\ne\pm3\)
\(A=\dfrac{\left(x+1\right)\left(x+3\right)}{x^2-9}+\dfrac{\left(x-1\right)\left(x-3\right)}{x^2-9}+\dfrac{1-x}{x^2-9}\)
\(\Leftrightarrow x^2+4x+3+x^2-4x+3+1-x=2x^2-x+7\)
Wait what? Are you sure you typed that question correctly?

Okaym let';s assume it's (1-x)²
Then everything will go normally, until we get:
\(\Leftrightarrow x^2+4x+3+x^2-4x+3+x^2-2x+1=3x^2-2x+7\)
Still... can't solve that...
I think you might have typed the question wrong...

\(\dfrac{4x-5}{2x-1}=\dfrac{2\left(2x-1\right)-3}{2x-1}=2-\dfrac{3}{2x-1}\)
Thế thì 2x-1 phải thuộc ước của 3, là các số \(\pm1;\pm3\)
Lần lượt tìm x với giá trị của đơn thức là các ước của 3, ta tìm được các nghiệm nguyên là -1;0;1;2

Case 1: Following what was given in the question, the original price of the second shirt is \(\dfrac{130\%}{70\%}=185,\left(714285\right)\%\)of the first shirt's original price.
So the total loss is \(\left(185,\left(714285\right)\%+100\%\right)-2\cdot130\%=25,\left(714285\right)\%\)of the first shirt's original price.

Case 2: Following what was given in the question, the original price of the first shirt is \(\dfrac{70\%}{130\%}=53,\left(846153\right)\%\)of the second shirt's original price
So the total loss is \(\left(53,\left(846153\right)+100\%\right)-2\cdot70\%=15,\left(846153\right)\%\)of the second shirt's original price.

WHAT'S UP WITH THESE QUESTIONS WITH SUCH AMBIGUOUS PERCENTAGES?
Anyways, the numbers are quite... odd and the solution for them seems really inappropriate for any math question ever so I think I might be wrong, but I don't know how to understand the question in any other way so... UWU

https://imgur.com/w5IOA1h
Ugh, please MathYou staff, try and fix the uploading images problem. For the time, I'm just going to use imgur. And for those of you who don't know what imgur is, don't worry, it's not any viruses, just simple images posted online via a link.

I don't understand this question that well... but I'll try my hardest.

I would think of it this...
So there would be \(20-1=19\)(knots) assuming that Michael isn't stupid enough to tie a ROPE into a freaking CIRCLE. And those knots would be \(19\cdot5=95\)(cm) in total
For the 18 lengths in the middle, each of them will have sacrificed 5cm on each end, therefore sacrificing 10cm in total, leaving only \(50-10=40\)(cm) if we are not counting the knots, so all of them would be \(40\cdot18=720\)(cm)
The two end lengths are different, because each of them will sacrifice only 5cm, making each of them \(50-5=45\)(cm) and making \(45\cdot2=90\)(cm) when we count all of the end lengths.
So in the end, we have a \(40+720+90=850\)(cm) rope. Not bad, Michael.

FOUR cubes with the edge length of 1m will have the total volume of \(4\cdot1^3=4\)(m³)\(=4000000\)(cm³)
Each smaller cube would have the volume of \(4^3=64\)(cm³)
So if we cut the four bigger cubes up, we will get \(\dfrac{4000000}{64}=62500\)(smaller cubes)
Since each of these cubes has the edge length of 4, the length of the line would be \(62500\cdot4=250000\)(cm)\(=2500\)(m)
Right, I would be ashamed if I converted the units wrong or something, but I'm pretty sure the line would be 2500m
I could imagine a snake as long as that roaming around in my nightmares...

Oh, forgot to mention, this is basing of the speculation that they make pies at a very constant speed :)

880 PIES IN 8 HOURS??? I'm very impressed. But who's gonna eat those pies... Are they going to sell it as ... a super popular product in the super market? Maybe Alex+Benito=Pie_Factory... I wonder if they will have to trademark "made by Alex" and "made by Benito" when they export it... But that would mean more diabetes for us all.

Okay, now going to solving the question.
In 1 hour, the both of them made \(\dfrac{880}{8}=110\)(pies)
So the sum is 110, the absolute difference is 10.
Alex made more pies.
So Alex made \(\dfrac{110+10}{2}=60\)(pies) in 1 hour
Alex made 60 pies in 1 hour.
 

Call the number of students in group 1 x (students) satisfying 0<x<108 & x\(\in\)N*
So the number of students in group 2 is 4x (students)
The number of students in group 3 is 2x+2 (students)
The number of students in group 4 is 2x-2(students)
The sum of the number of students in all the groups equals 108
\(\Rightarrow x+4x+2x+2+2x-2=108\)
\(\Leftrightarrow9x=108\)
\(\Leftrightarrow x=12\)(students)
Huh, 12 satisfies all conditions... So group 1 has 12 students.

BONUS ROUND: Fixing Huỳnh Anh Phương's answer
Clearly there's one thing you missed when you looked at the question: that is the words "TWO TIMES"
:) But I reading the way you tried to solve it.

The city hall has 4 regions to choose from
The school has 4 regions to choose from
The movie theater has 3 regions to choose from (Because it can't be the same region as the school)
So there are \(4\cdot4\cdot3=48\) ways these buildings can be built...

Wait what? FOUR BUILDINGS? I wonder if that's a typo.
Okay, so IF there are 4 buildings and only 1 condition above, the fourth building will aso have 4 regions to choose from.
That would mean \(4\cdot4\cdot3\cdot4=192\) ways these FOUR buildings can be built.

ARGH, this hurts my head. So... we know the maximum weight, but we don't know the minimum weight so it is quite challenging...
I think we need at least 4 trucks to carry these, but it's just my guess.
Since you can just chuck as much as you can on the first three trucks, and the rest can be chucked onto the fourth one, because any problems you may encounter can be solved easily by shuffling the boxes around from one truck to another.

Starting with 1, the first number formed is definitely going to be 51 being divisible by 17.
Then it's 85 divisible by 17.
Then it's 68 divisible by 17.
Now it's 46 divisible by 23.
And it's 34 divisible by 17.
And it's 23 divisible by 23.
And now it's 92 divisible by 23.
Here, it's 69 divisible by 23
And it's 46 divisible by 23
And it repeats...
SO the patter is always ...(92346)851 (The things in the parenthesis are repeating)
Since it's 2017 digits, the loop would end on the (2017-3) mod 5 = 4th number being 2
THE FIRST DIGIT IS 2

Oh, sorry, I mixed up the divisor and the dividend 
Okay, so that number is divisible by both 18 and 28.
\(18=2\cdot3^2\)
\(28=2^2\cdot7\)
The least common multiple of these two numbers is... that's 3 digits...

18k is divisible by 28, is that what you mean?
That means 9k is divisible by 14, which means k is at least 14 and...
That's still 3 digits...

So what if it's 18k+10 being divisible by 28, adding 10 to both the divisor and the dividend.
So.... k can only be 1, which means the number Anna chose is 18?

Please, next time, have proper grammar in your sentences for us to understand  The question should've been:
"Anna chose a 2 digit number. That number is divisible by 28. If you add 10 to the divisor and the dividend, the new dividend is still divisible by the new divisor. What number did Anna choose?"