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Answers ( 118 )
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    A notebook costs: 16.80 : 8 = 2.10 ($)

    A pen costs: 15.25 : 5 = 3.05 ($)

    Anna buys: 2.10 x 4 + 3.05 x 8 = 8.40 + 24.40 = 32.80 ($) 

    Anna received: 2 x 20 - 32.80 = 7.20 ($)

    Benny buys: 5 x 2.10 + 10 x 3.05 = 10.50 + 30.50 = 41 ($)

    Benny received: 2 x 20 - 41 = -1 ($) 

    => Anna received more and more than: 7.20 + 1 = 8.20 ($)

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    Because 1 = 10 so 10 = 1 :D 

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    If x = 2 , we have x2 + 2x = 8 ( scrap because 8 isn't a prime number)

    If x = 3 , we have x2 + 2x = 17 ( satisfy )

    If x > 3 and x is a prime number then x =3k + 1 or 3k + 2 ( k is a natural number)

    +) x = 3k +1

    If x = 3k +1 then x2 divide to 3 has the remainder 1

    and 2x divide to 3 has the remainder 2

    => 2x + x2 divided to 3 

    But 2x + x2 > 3 

    So 2x + x2 isn't a prime number

    Do the similar to x = 3k+2

    The answer is x = 3

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    This is a problem in the newspaper:"Toán học tuổi thơ" that you must be solve it yourself, shouldn't give it to here!!

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    60 x 50 = 300 (cm2)

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    We have 3 straight points: (A;E;B);(B;G;C);(C;D;H);(A;I;D);(A;O;C);(I;O;G);(D;O;B);(E;O;H). So the middle point is: E;G;H;I;O

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    The number of red marbles is: 15-2-4 = 7(marbles)
    The quotion of the red marble is: 7/15

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    #35# = 35 x 1 x 2 x 3 = 35 x 6 = 210

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    Call the number of teams is n

    We have: The number of games is n(n+1) games

    But each game repeated 2 times 

    So the number of the games is \(\dfrac{n\left(n+1\right)}{2}\)

    There are 21 games

    So \(\dfrac{n\left(n+1\right)}{2}=21\Rightarrow n\left(n+1\right)=42\Rightarrow n=6\)

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    \(\left(x+2y\right)^2+\left(2x-y\right)^2-\left(5x+y\right)\left(x-y\right)-10\left(y+3\right)\left(y-3\right)\)

    \(=x^2+4xy+4y^2+4x^2-4xy+y^2-5x^2+5xy-xy+y^2-10\left(y^2-9\right)\)

    \(=-4y^2+4xy+90\)

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    We have:

    \(\left(3x-5\right)^{2016}\ge0\) with \(\forall x\in R\)

    \(\left(y^2-1\right)^{2008}\ge0\) with \(\forall y\in R\)

    \(\left(x-z\right)^{2100}\ge0\) with \(\forall z\in R\)

    \(\Rightarrow\left(3x-5\right)^{2006}+\left(y^2-1\right)^{2008}+\left(x-z\right)^{2100}\ge0\) with \(\forall x,y,z\in R\)

    Sign "=" occur 

    \(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\y^2-1=0\\x-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\\z=\dfrac{5}{3}\end{matrix}\right.\)

    \(\Rightarrow\left(x;y;z\right)\in\left\{\left(\dfrac{5}{3};1;\dfrac{5}{3}\right);\left(\dfrac{5}{3};-1;\dfrac{5}{3}\right)\right\}\)

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    Call the number of red marbles is r

    the number of blue marbles is b

    We have: r + b = 36

    and 2r + 6 = b 

    => r + b = r + 2r + 6 = 36

    => r = 10 

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    We have: 2@6 = ( 6 + 2 )( 6 - 2 ) = 8 . 4 = 32

    32@ -4 = ( 32 - 4 )( 32 + 4 ) = 28 . 32 = 896

    => 2. ((2@6)@-4) - 1 = 2. 896 - 1 = 1791

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    We have: The 4-people pizza's area is: \(8^2\Pi=200,96\left(inch^2\right)\)

    So each person can eat: \(\dfrac{200,96}{4}=50,24\left(inch^2\right)\)

    => 16 people eat: \(50,24.16=803,84\left(inch^2\right)\)

    => The suitalbe pizza has the diameter is: \(\sqrt{\dfrac{803,84}{3,14}}.2=32\left(inch\right)\)

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    This is geometry not algebraic men :) 

    Give you a :) 

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    Call the radius is r, the height is h

    We have: 

    Cylinder 1 has the volume is: \(\left(2r\right)^2.\Pi.h=\Pi.4r^2.h\)

    Cylinder 2 has the volume is: \(r^2.\Pi.k.h\)

    We have: \(\dfrac{C1}{C2}=1\)

    \(\Rightarrow\dfrac{4r^2.\Pi.h}{r^2.\Pi.k.h}=1\)

    \(\Rightarrow\dfrac{4}{k}=1\)

    \(\Rightarrow k=4\)

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    A package of Koka-Kola is: 12 x 8 = 96 (ounces)

    A package of Pepsy-Kola is: 16 x 6 = 96 (ounces)

    => The absolute different is 0, not right yet ^^

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    Call Yumi's age is a

    Rana's age is b

    Victoria's age is c.

    We have: a + b + c = 42 

    => ( a - 4 ) + ( b - 4 ) + c = 36

    But a - 4 + b - 4 = c 

    => a + b = c + 8 (1)

    We also have: a + b + c = 42

    => a + b = 42 - c (2)

    From (1) and (2) we have:

    42 - c = c + 8

    => 42 - 8 = 2c

    => 36 = 2c

    => c = 18

    So current Yumi's age is 18.

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    We have: 30 students have lived in Texas less than a year 

    => There are 5% students of 600 students at Goodnight Middle School in Texas have lived here less than 1 year.

    So there 65% students are not native Texas lived here less than a year and more than 10 years.

    But there are 85% are not native Texas 

    => The non-native students have lived more than 1 year but less than 10 year is 20% = 120(students)

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    Not plus more 42, because 42 = 40 + 2 

    And 2 appear 2 times and 40 appear 2 times => There must be 3(2+3+...+40)

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Questions ( 2 )
  • Given: \(a\ne b\ne c\) and \(\left(b-c\right)\sqrt[3]{1-a^3}+\left(c-a\right)\sqrt[3]{1-b^3}+\left(a-b\right)\sqrt[3]{1-c^3}=0\)

    Prove: \(\left(1-a^3\right)\left(1-b^3\right)\left(1-c^3\right)=\left(1-abc\right)^3\)

  • Do not use trigonometric formulas. Prove: Cos \(72^0\)x Cos \(36^0=\dfrac{1}{4}\)

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