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The bathroom needs to be tilted: \(\left(12^2\right):\left(1^2\right)=144\left(tiles\right)\)
Because the bathroom has 4 edges (the bathroom's shape is square) so the probability is: \(\dfrac{4}{144\cdot\left(1441\right)}=\dfrac{4}{20592}=\dfrac{1}{5148}\)
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\(4x12y=19\Leftrightarrow12y=4x+19\Leftrightarrow y=\dfrac{1}{3}x+\dfrac{19}{12}\)
The xcoordinates of the intersection points of the graphs satisfy :
\(x^23x+3=\dfrac{1}{3}x+\dfrac{19}{12}\Leftrightarrow x^2\dfrac{10}{3}x+\dfrac{17}{12}=0\)
Since there are 2 intersection points, we can apply the Vieta's formulas. The answer is \(\dfrac{10}{3}\)

When the water level rises 1 in, the amount of water in the tank increases by : 10 x 15 x 1 = 150 (in^{3})
The volume of each sphere is : \(\dfrac{4}{3}\pi\left(\dfrac{1}{6}:2\right)^3=\dfrac{1}{1296}\pi\) (in^{3})
The answer is : \(150:\dfrac{1}{1296}\pi\approx61900\) (spheres)

The price of a box of pens now costs: 100%  10% 90%
The percentage of the pens in the box now is: 100%  25% = 75%
The percent change in the cost per pen is: \(\left(90\%:75\%\right)100\%=20\%\)
ANSWER: 20%
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The sum of the factors of 120 which are different from 3 and 8 is :
4 + 5 + 6 + 10 + 12 + 15 + 20 + 24 + 30 + 40 + 60 + 120 = 346
So, the answer is 346

Let O be the center of the circle, then \(\widehat{AOC}=2\widehat{ABC}=2.30^0=60^0\)
The answer is : \(12\pi.\dfrac{60}{360}=2\pi\) (in)
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Kaya Renger Coodinator
16/09/2017 at 21:25
Denote :
\(B=\dfrac{a}{b+c+d}+\dfrac{b}{c+d+a}+\dfrac{c}{a+b+d}+\dfrac{d}{a+b+c}\)
\(C=\dfrac{b+c}{b+c+d}+\dfrac{c+d}{c+d+a}+\dfrac{a+d}{a+b+d}+\dfrac{a+b}{a+b+c}\)
\(D=\dfrac{c+d}{b+c+d}+\dfrac{d+a}{c+d+a}+\dfrac{a+b}{a+b+d}+\dfrac{b+c}{a+b+c}\)
\(E=\dfrac{b+d}{b+c+d}+\dfrac{a+c}{c+d+a}+\dfrac{b+d}{a+b+d}+\dfrac{a+c}{a+b+c}\)
\(F=\dfrac{b+c+d}{a}+\dfrac{c+d+a}{b}+\dfrac{a+b+d}{c}+\dfrac{a+b+c}{d}\)
We have :
\(C+D+E=8\)
Applying the AMGM inequality, we have :
\(B+C=\dfrac{a+b+c}{b+c+d}+\dfrac{b+c+d}{c+d+a}+\dfrac{a+c+d}{a+b+d}+\dfrac{a+b+d}{a+b+c}\ge4\)
\(B+D=\dfrac{a+c+d}{b+c+d}+\dfrac{a+b+d}{a+c+d}+\dfrac{a+b+c}{a+b+d}+\dfrac{b+c+d}{a+b+c}\ge4\)
\(B+E=\dfrac{a+b+d}{b+c+d}+\dfrac{a+b+c}{c+d+a}+\dfrac{b+c+d}{a+b+d}+\dfrac{a+c+d}{a+b+c}\ge4\)
(the equalities happen only when a = b = c = d)
\(\Rightarrow3B+C+D+E\ge12\Rightarrow3B\ge4\Rightarrow B\ge\dfrac{4}{3}\)
(the equality happens only when a = b = c = d)
\(F=\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{d}{a}+\dfrac{c}{b}+\dfrac{d}{b}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{d}{c}+\dfrac{a}{d}+\dfrac{b}{d}+\dfrac{c}{d}\ge12\)
(the equality happens only when a = b = c = d)
\(\Rightarrow A=B+F\ge12+\dfrac{4}{3}=\dfrac{40}{3}\)
(the equality happens only when a = b = c = d)
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Shirayuki Akagami 02/12/2017 at 13:05
mk ko biết làm kiểu kia xin bn thông cảm

Shirayuki Akagami 02/12/2017 at 13:03
B=ab+c+d+bc+d+a+ca+b+d+da+b+cB=ab+c+d+bc+d+a+ca+b+d+da+b+c
C=b+cb+c+d+c+dc+d+a+a+da+b+d+a+ba+b+cC=b+cb+c+d+c+dc+d+a+a+da+b+d+a+ba+b+c
D=c+db+c+d+d+ac+d+a+a+ba+b+d+b+ca+b+cD=c+db+c+d+d+ac+d+a+a+ba+b+d+b+ca+b+c
E=b+db+c+d+a+cc+d+a+b+da+b+d+a+ca+b+cE=b+db+c+d+a+cc+d+a+b+da+b+d+a+ca+b+c
F=b+c+da+c+d+ab+a+b+dc+a+b+cdF=b+c+da+c+d+ab+a+b+dc+a+b+cd
We have :
C+D+E=8C+D+E=8
Applying the AMGM inequality, we have :
B+C=a+b+cb+c+d+b+c+dc+d+a+a+c+da+b+d+a+b+da+b+c≥4B+C=a+b+cb+c+d+b+c+dc+d+a+a+c+da+b+d+a+b+da+b+c≥4
B+D=a+c+db+c+d+a+b+da+c+d+a+b+ca+b+d+b+c+da+b+c≥4B+D=a+c+db+c+d+a+b+da+c+d+a+b+ca+b+d+b+c+da+b+c≥4
B+E=a+b+db+c+d+a+b+cc+d+a+b+c+da+b+d+a+c+da+b+c≥4B+E=a+b+db+c+d+a+b+cc+d+a+b+c+da+b+d+a+c+da+b+c≥4
(the equalities happen only when a = b = c = d)
⇒3B+C+D+E≥12⇒3B≥4⇒B≥43⇒3B+C+D+E≥12⇒3B≥4⇒B≥43
(the equality happens only when a = b = c = d)
F=ba+ca+da+cb+db+ab+ac+bc+dc+ad+bd+cd≥12F=ba+ca+da+cb+db+ab+ac+bc+dc+ad+bd+cd≥12
(the equality happens only when a = b = c = d)
⇒A=B+F≥12+43=403⇒A=B+F≥12+43=403
(the equality happens only when a = b = c = d)
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