We see that, if x satisfies the equation above => 

\(-8< 3x+4< -32\) and \(8< 3x+4< 32\)

So we have \(28\cdot2=56\) possible values of x.

Solve each equations we will have the exact values of x.

\(\sqrt{2+4+6+x}\) (Condition \(-12\le x< 8\)).

\(\Leftrightarrow\sqrt{12+x}=n\left(n\in N\right)\)

So x will satisfy \(x=\left\{-12;-11;-8;-3;4\right\}\)

The sum of \(x=-12-11-8-3+4=-30\)

Answer : - 30.

Call the percent increase is x (%)

Because the number of participants increase each year study.

We have equation : 

\(\left\{{}\begin{matrix}P=40+40\cdot x\%\\P+P\cdot x\%=90\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}P=40+40\cdot x\%\\40+40\cdot x\%+\left(40+40\cdot x\%\right)\cdot x\%=90\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}P=40+40\cdot x\%\\x\%=50\%\end{matrix}\right.\)

So \(P=40+40\cdot50\%=60\)

Answer P = 60.

This is "Math Q&A" not "English Q&A"

Please put this question on different page.

Thanks for your cooperation.

Call the number is a.

We have \(a\cdot x=a:0,125\)

\(\Leftrightarrow a\cdot x=a:\dfrac{1}{8}\)

\(\Leftrightarrow a\cdot x=a\cdot8\Rightarrow x=8\)

So the value of x is 8.

Answer : 8.

The shape must be placed in the shaded space is triangle :

Answer : Triangle \(\Delta\)

When Samir completes a ten-day job with the second ways, we have the total amounts he can earn:

\(1+2+4+8+16+32+64+128+256+512=1023\)

So the positive difference of the total amounts Samir can be paid :

\(1023-1000=23\)

Answer : 23.

The degree  measure of an interior angle of a regular pentagon is :

\(\left(5-2\right)\cdot\dfrac{180}{5}=3\cdot36=108^o\)

Answer : \(108^o\)

We have : y = x + a.

See : 5 = 2 + a => a = 3.

6 = 3 + a => a = 3.

7 = 4 + a => a = 3.

So the value of a is 3.

Answer : 3.

The first odd number is 21.

The last odd number is 157.

So there will be \(\dfrac{157-21}{2}+1=69\) (odd numbers) between 20 and 158.

Answer : 69 odd numbers.

We know that all the sides of a square are equal.

So we have : \(x-2=3\)

\(\Rightarrow x=5\)

So the value of x is 5.

Answer : 5.

The number of red grapes : \(50\cdot20\%=10\left(grapes\right)\)

So there will be \(50-10=40\left(grapes\right)\) are not red.

Answer : 40 grapes.

Area of the 120 mm2 triangle: 
Area = 1/2 x height x base 
120mm² = 1/2 x 10 x base 
Find out the base: 
120mm² = 5 x base 
base = 120/5 
=> base = 24mm. 

We know that two triangles that are similar will have same increase in height and base dimensions. 
Since height is doubled, then base is doubled. 

So :

height is 20mm 
base is 48mm 
The area of a similar triangle with a height of 20 mm is :

\(\dfrac{1}{2}\cdot20\cdot48=480\left(mm^2\right)\)

Answer : 480 mm2.

We have : \(Side=2\cdot Radius\)

So one side of the square is \(2\cdot4=8\left(ft\right)\)

The area of the square : \(8\cdot8=64\left(ft^2\right)\)

Answer : 64.

The sum of books : 6 + 3 + 7 = 16 (books)

The number of math books is : \(3+7=10\left(books\right)\)

The probability is : \(\dfrac{10}{16}=\dfrac{5}{8}\)

Answer : \(\dfrac{5}{8}\)

There are \(5\cdot3\cdot2\cdot4=120\) different uniforms can he wear to school.

Answer : 120 ways.

We have :

1 year : 12 month.

35 years : x month

So \(x=12\cdot35=420\left(months\right)\)

Answer : 420 months.

We have the formula :

Volume = Length x Width x Height.

So \(120=5\cdot4\cdot height\)

So the value of the height is \(\dfrac{120}{5\cdot4}=6\left(in\right)\)

Answer : 6 in.

\(FACT\left(96\right)=2^5\cdot3\)

The greatest prime factor of 96 is 3.

Answer : 3.

Cause \(2\cdot\left(3+4\right)>2\cdot3+4\)

So the positive difference between the value of 2 × (3 + 4) and the value of 2 × 3 + 4 is :

\(2\cdot\left(3+4\right)-\left(2\cdot3+4\right)=14-10=4\)

Answer : 4.

The whole angles of the square is \(360^0\)

The pizza is cut into 12 congruent slices => Each slice has the angle of \(\dfrac{360}{12}=30^0\)

So the sum of the central angles of the slices that were not eaten after ate 2 slices is :

\(360-30\cdot2=360-60=300^0\)

Answer : \(300^0\)

Call  the width is x.

We have \(78=2x+6\)

So \(x=36\)  <=> the width is 36 ft.

Answer : 36 ft.

We have : The ratio : Charity A : Charity B = 2 : 3.

The money Sam contributes to Charity B is : \(\dfrac{2500}{2+3}\cdot3=1500\)$.

Answer : 1500$.

A day the man floats \(3-2=1\) (mi north)

So until 47 days the man can go 47 mis north ( Cause 1 mi / day).

In the last day (day 48) the man floats 3 mi north so the distance he went is 47 + 3 = 50 (mi north).

So it will take him 48 days to reach the location.

Let's take the frist 20 positive intergers 1 to 20.

The number of odd prime numbers : 3;5;7;11;13;17;19.

The probability is : \(\dfrac{7}{20}\cdot100\%=35\%\)

\(5\cdot\left(11+4:4\right)\)

\(=5\cdot\left(11+1\right)\)

\(=5\cdot12=60\)

Cut in half so 72/2=36 is the diameters of two circles.

Sum of the areas : \(2S=\dfrac{d^2}{4}\cdot3,14\cdot2=2034,72\left(cm^2\right)\)

Answer : 2034,72 \(cm^2\).

We have \(\overline{a0b}⋮9\Rightarrow a+0+b\in\left\{9;18;27;36;...\right\}\)

But \(a+0+b\le18\)

So \(a+b=\left\{9;18\right\}\)

We have pair (a;b) = (1;8) , (8;1) , (2;7) , (7;2) , (6;3) , (3;6) , (5;4) , (4;5) , (9;0).

But there is only the pair : (4;5) match the condition \(\overline{a0b}=\overline{ab}\cdot9\).

So the answer of the two digit number \(\overline{ab}\) is 45.

We have \(7^{2017}=\left(7^4\right)^{504}\cdot7\)

We have \(7^4\equiv1\left(mod12\right)\)

\(\Rightarrow\left(7^4\right)^{504}\equiv7^{2016}\equiv1\left(mod12\right)\)

\(\Rightarrow7^{2017}\equiv1\cdot7\equiv7\left(mod12\right)\)

So the remainder is 7.

Call : \(\dfrac{\left(n+1\right)^2}{n+23}=Q_{\left(n\right)}+\dfrac{R}{n+23}\)

n + 23 = 0 => n = -23.

\(R=\left(-23+1\right)^2=\left(-22\right)^2=484\)

\(\Rightarrow\dfrac{\left(n+1\right)^2}{n+23}=Q_{\left(n\right)}+\dfrac{484}{n+3}\in Z\)

\(\Rightarrow n+23\inƯ_{\left(484\right)}=\left\{\pm1;\pm2;\pm4;\pm11;\pm22;\mp44;\pm121;\pm242;\pm484\right\}\)

\(\Rightarrow n+23_{MAX}=484\Rightarrow n=484-23=461\)

Answer : 461.

Hey guy (LeQuocTranAnh), the question asked the sum of the digits not numbers.

I think you've wrong in reading thread :<

The sum of the digits 0-9 : 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9, which equals 45.

Next, the sum of the digits of the numbers 10-19 : \(\left(10\cdot1\right)+\left(0+1+...+9\right)\)

Then, the sum of the digits of the numbers 20-29 : \(\left(10\cdot2\right)+\left(0+1+2+...+9\right)\)

So we have this expression :

(0 + 1 +...+ 9) - The numbers 0-9

\(10\cdot1+\left(0+1+...+9\right)\)- The numbers 10-19

\(10\cdot2+\left(0+1+...+9\right)\)- The numbers 20-29

\(10\cdot3+\left(0+1+...+9\right)\)- The numbers 30-39

\(10\cdot4+\left(0+1+...+9\right)\) -The numbers 40-49

\(10\cdot5+\left(0+1+...+9\right)\) -The numbers 50-59

\(10\cdot6+\left(0+1+...+9\right)\) -The numbers 60-69

\(10\cdot7+\left(0+1+...+9\right)\) -The numbers 70-79

\(10\cdot8+\left(0+1+...+9\right)\) -The numbers 80-89

\(10\cdot9+\left(0+1+...+9\right)\)  - The numbers 90-99.

Calculate all of these expressions we have the sum of the digits from 1-99 is 900.

So the sum of the digits from 1 - 100 is 900 + 0 + 0 + 1 = 901.

Answer : 901.

The sum of 1 + 7 + 13 + 19 +...+253 + 259 :

\(S=\left(259+1\right)\cdot\dfrac{\left(\dfrac{259-1}{6}+1\right)}{2}=5720\)

So the remainder is : \(\dfrac{5720}{6}=Q=953,R=2\)

Answer : 2.

\(P=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

Condition : \(x\ge1\)

a) We have \(x+2\sqrt{x-1}=\left(x-1\right)+2+1\sqrt{x-1}+1=\left(\sqrt{x-1}+1\right)^2\)

So P = \(\sqrt{\left(1+\sqrt{x-1}\right)^2}+\sqrt{\left(1-\sqrt{x-1}\right)^2}\)

\(\Rightarrow\left|1+\sqrt{x-1}\right|+\left|1-\sqrt{x-1}\right|\)

+ With \(1-\sqrt{x-1}\ge0\Rightarrow x\le2:\)

\(P=1+\sqrt{x-1}+1-\sqrt{x-1}=2\)

+ With \(1-\sqrt{x-1}< 0\Rightarrow x>2:\)

\(P=1+\sqrt{x-1}+\sqrt{x-1}-1=2\sqrt{x-1}\)

Sum up : \(\left[{}\begin{matrix}1\le x\le2\Rightarrow P=2\\x>2\Rightarrow P=2\sqrt{x-1}\end{matrix}\right.\)

b) We have inequality \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)

So : \(P=\left|1+\sqrt{x-1}\right|+\left|1-\sqrt{x-1}\right|\ge\left|\sqrt{\left(1-\sqrt{x-1}\right)+\left(1-\sqrt{x-1}\right)}\right|\)

\(\Rightarrow P\ge2\Leftrightarrow1\le x\le2\)

So Pmin = 2 when \(1\le x\le2\).

Factoring : \(2^{58}+1=\left(2^{29}+2^{15}+1\right)\cdot\left(2^{29}-2^{15}+1\right)\)

Because : \(\left\{{}\begin{matrix}2^{29}+2^{15}+1=536903681\\2^{29}-2^{15}+1=536838145=5\cdot107367629\end{matrix}\right.\)

So \(2^{58}+1=\left(2^{29}+2^{15}+1\right)\cdot\left(2^{29}-2^{15}+1\right)=536903681\cdot107367629\cdot5\)

\(\rightarrow\dfrac{2^{58}+1}{5}=\dfrac{536903681\cdot107367629\cdot5}{5}=536903681\cdot107367629\)

So the number \(\dfrac{2^{58}+1}{5}\) is composite.

How about this ?

\(A=1+9+9^2+...+9^{50}\)

\(\Rightarrow9A=9+9^2+9^3+...+9^{50}+9^{51}\)

\(\rightarrow9A-A=8A=\left(9+9^2+9^3+...+9^{50}+9^{51}\right)-\left(1+9+9^2+...+9^{50}\right)\)

\(\Rightarrow8A=9^{51}-1\Rightarrow A=\dfrac{9^{51}-1}{8}\)

We find that \(9^{2k},9^{2k+1},...\)always have the last digit is 1 or 9, or we can simply understand that whether 9 has index number is odd => the last digit is 9, and whether 9 has index number is even => the last number is 1.

\(9^{51}\) has the last digit is 9 (Cause 51 is odd number)

So : \(A=\dfrac{\overline{B9}-1}{8}=\dfrac{\overline{B8}}{8}=\overline{C1}\)

Therefore, the last digit of \(1+9+...+9^{50}\) is 1.

Answer : 1.

Also, we use Venn to this.

Pretend there are 100 americans.

Call x is the number of Americans have type a and type b in their blood

Similarly, we have : \(43+15-x+46=100\Rightarrow x=4\)

So the percent is \(\dfrac{4}{100}\cdot100\%=4\%\).

Answer : 4%.