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We have :
x < -0.8 \(\Rightarrow\) x + 0,8 < 0 \(\Rightarrow\)|x + 0.8| = -(x + 0.8)
x < -0.8 \(\Rightarrow\) x < 25 \(\Rightarrow\) |x - 25| = 25 - x
\(\Rightarrow\) A = -(x + 0.8) - (25 - x) + 1.9
A = -x - 0.8 - 25 + x + 1.9
A = -23.9
Kayasari Ryuunosuke selected this answer. -
→இے๖ۣۜQuỳnh 23/03/2017 at 05:50We have : x < -0.8 ⇒ x + 0,8 < 0 ⇒|x + 0.8| = -(x + 0.8)
x < -0.8 ⇒ x < 25 ⇒ |x - 25| = 25 - x
Candle A = -(x + 0.8) - (25 - x) + 1.9
= -x + 0,8 - 25 + x + 1,9
= -x + x - 0,8 - 25 + 1,9
= -25,8 + 1,9
A = -23,8
A = -x - 0.8 - 25 + x + 1.9
A = -23.9
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→இے๖ۣۜQuỳnh 22/03/2017 at 20:04
We have : x < -0.8 ⇒ x + 0,8 < 0 ⇒|x + 0.8| = -(x + 0.8)
x < -0.8 ⇒ x < 25 ⇒ |x - 25| = 25 - x
Candle A = -(x + 0.8) - (25 - x) + 1.9
= -x + 0,8 - 25 + x + 1,9
= -x + x - 0,8 - 25 + 1,9
= -25,8 + 1,9
A = -23,8
A = -x - 0.8 - 25 + x + 1.9
A = -23.9
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Vũ Hà Vy Anh 18/05/2017 at 22:201dm=10cm
the area of shaded triangle :
10x36:2=180(cm2)
Maths selected this answer. -
Fujitora Ishito 19/05/2017 at 09:45\(\text{1dm=10cm }\)
\(\text{ The area of shaded triangle : 10x36:2=180(cm2)}\)
\(Answer:180cm^2\)
Good Maths
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Phan Minh Anh 21/05/2017 at 14:53180 cm2
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We have :
\(\left(x-1\right)^4=\left(x-1\right)^6\)
\(\Rightarrow\left(x-1\right)^6-\left(x-1\right)^4=0\)
\(\left(x-1\right)^4.\left[\left(x-1\right)^2-1\right]=0\)
So \(\left(x-1\right)^4=0\) or \(\left(x-1\right)^2-1=0\)
\(\Leftrightarrow x-1=0\) or \(\left(x-1\right)^2=1\)
\(\Leftrightarrow x=1\) or \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
So , x = {0,1,2}
Selected by MathYouLike
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(2x - 5)(2x + 5) = 5
\(\Leftrightarrow\left(2x\right)^2-5^2=5\)
\(\Leftrightarrow4x^2=5+25=30\)
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Dao Trong Luan 30/08/2017 at 17:28\(\left(3-4x\right)^3=-125\)
\(\Leftrightarrow\left(3-4x\right)^3=\left(-5\right)^3\)
=> 3 - 4x = -5
=> 4x = 3 - \(\left(-5\right)\)
=> 4x = 8
=> x = 2
Lê Quốc Trần Anh selected this answer.
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We notice that every term of the sum is divided by 6 with remainder 1 (since each term is 6 less than the next term and 1 is the first term)
The sum has : (253 - 1) : 6 + 1 = 43 (terms)
We have : \(1\times43\equiv43\equiv1\left(mod6\right)\) . So, the answer is 1
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The answer is : \(1029:\left(45+53\right)=10.5\) (hours)
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\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=3^2\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=9\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2\left(a+b+c\right)}{abc}=9\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2abc}{abc}=9\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=9\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=7\)
Alone selected this answer. -
FA Liên Quân Garena 02/01/2018 at 12:331a+1b+1c=3
⇒(1a+1b+1c)2=32
⇒1a2+1b2+1c2+2(1ab+1bc+1ca)=9
⇒1a2+1b2+1c2+2(a+b+c)abc=9
⇒1a2+1b2+1c2+2abcabc=9
⇒1a2+1b2+1c2+2=9
⇒1a2+1b2+1c2=7
