Tell x is the number to find

=> x(x+2) = 783

=> x = 27 or x = -29

But x is a positive => x = 27

Tell x is the number to find

=> x(x+2) = 783

=> x = 27 or x = -29

But x is a positive => x = 27

Tell x is the number to find

=> x(x+2) = 783

=> x = 27 or x = -29

But x is a positive => x = 27

The perimeter of the hexagon is: 16 x 6 = 96 (cm)
 Because the circumference of the hexagon equal the circumference of the star, the perimeter of the star is 96 cm
 Answer: 96cm

Sum of odd-numbered digits ( - ) Sum of even-numbered digits or vice versa is divisible by 11

We have:

(2 + D + D) - (3 + 6 + 9) is divisible by 11

<=> 2 + 2D - 18 is divisible by 11

<=> 2D - 16 is divisible by 11

<=> 2 (D - 8) is divisible by 11

So Greatest common divisor (2 ; 11) = 1

=> D - 8 is divisible by 11

0 \(\le D\le9\)

=> D = 8

So The answer is : 386,892

The number of large clips that weigh the same as 12 small clips is:

       12 x 2/3 = 8(large clips)

The number of tacks weigh the same amount as 12 small clips is:

       8 x 2 = 16(tacks)

2/3 = 4/6

Replace x = 2 into the given equations,we have :

\(\left\{{}\begin{matrix}2p+q=91\\p-q=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2p+p-2=91\\q=p-2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3p=93\\q=p-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}p=31\\q=29\end{matrix}\right.\)=> pq = 899

Replace x = 2 into the given equations,we have :

{2p+q=91p−q=2⇒{2p+p−2=91q=p−2

⇒{3p=93q=p−2⇒{p=31q=29

=> pq = 899

\(PO=PM+MO=8+5=13\)

\(PX=PY=\sqrt{PO^2-OX^2}=\sqrt{13^2-5^2}=12\)

\(PX.XO=PO.XH\left(=2.area\left(XPO\right)\right)\)

=> \(XH=\dfrac{PX.XO}{PO}=\dfrac{12.5}{13}=\dfrac{60}{13}\)

\(XY=2.XH=2.\dfrac{60}{13}=\dfrac{120}{13}\)

Phanthanhtinh help me translator please..

Solution and contest

The answer is \(\dfrac{120}{13}\),but I don't know the solution.

Is this problem from USC Mathematics Contest 3/12/1994 ?

Visit here : http://www.math.sc.edu/math-contest-1994-answer-key

a) We have:

\(3\left(2x-1\right)^2\ge0\forall x\)

\(7\left(3y+5\right)^2\ge0\forall y\)

\(\Rightarrow\left\{{}\begin{matrix}3\left(2x-1\right)^2=0\\7\left(3y+5\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(2x-1\right)^2=0\\\left(3y+5\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-1=0\\3y+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{5}{3}\end{matrix}\right.\)

Answer: \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{5}{3}\end{matrix}\right.\)

b) \(x^2+y^2-2x+10y+26=0\)

\(\left(x^2-2x+1\right)+\left(y^2+10y+25\right)=0\)

\(\left(x-1\right)^2+\left(y+5\right)^2=0\)

We have:

\(\left(x-1\right)^2\ge0\forall x\)

\(\left(y+5\right)^2\ge0\forall x\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+5\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-5\end{matrix}\right.\)

Answer: \(\left\{{}\begin{matrix}x=1\\y=-5\end{matrix}\right.\)

We have : 3(2x - 1)² + 7(3y + 5)² = 0

<=> 3(4x² - 1) + 7(9y² + 25) = 0

<=> 12x² - 3 + 63y²  + 175 = 0 

<=> 12x² + 63y² + 172 = 0 

<=> 12x² + 63y² = -172 (Meaningless)

Because : 12x² \(\ge0\) ; 63y2​ ​​ \(\ge0\)

What is the grade

\(\dfrac{x^2-3x+2}{x^2+4x+5}=\dfrac{1}{2}\Rightarrow2\left(x^2-3x+2\right)=x^2+4x+5\)

\(\Leftrightarrow2x^2-6x+4=x^2+4x+5\)

\(\Leftrightarrow x^2-10x-1=0\)

=> We have two solutions : \(\left[{}\begin{matrix}x=5+\sqrt{26}\\x=5-\sqrt{26}\end{matrix}\right.\)

We have : x2−3x+2x2+4x+5=12

⇒(x2−3x+2).2=x2+4x+5

⇔2x2−6x+4=x2+4x+5

⇔2x2−x2−6x−4x=5−4

⇔x2−10x−1=0

⇔x2−10x+25−26=0

​ ​⇔(x−5)2=26

=> x - 5 = ​ ​​√26

=> x = √26+5

Searching4You was right

Oh, I'm wrong. My solution is: 4.3.2 (not 3.3.2). Thanks Phan Thanh Tinh for your answer.

Lê Quốc Trần Anh, I think there are : 4.3.2.1 = 24 (choices) for abcd

My answer is : \(\dfrac{1}{24}=4.1\left(6\right)\%\)

How did you think about the way to find the number of choices for abcd ? I can't understand 3.3.2

Call the digits of the mobile device: abcd \(\left(a\ne b\ne c\ne d\right)\) and \(a,b,c,d\in\left\{0;1;2;...;9\right\}\)

So there'll be: \(3.3.2=18\left(ways\right)\)for the number abcd in random order.

So the probability is: \(\dfrac{1}{18}\simeq5,56\%\)