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Lâm Phan Bình Nguyên
04/08/2017 at 20:35
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1.A mathematics competition consists of 10 questions.3 marks are awarded for answering a question correctly.2 marks will be deduced for a wrong answer.1 mark will be deducted for not answering a question.

 

How many students must have taken part in order to have at least four students of the same score?

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    Lê Quốc Trần Anh Coordinator 07/08/2017 at 08:51

    There is 3 activities in the competition is: +3 marks, -2 marks and -1 marks.

    So there must be: \(\left(10.3\right).4=120\left(students\right)\)taken the test to satisfy the question.


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Uchiha Sasuke
21/04/2017 at 08:40
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If f (x) = px7 + qx3 + rx - 4 and f (-7) = 3 then the value of  f (7) is...

Help me please!gianroi

 

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    Phan Thanh Tinh Coordinator 21/04/2017 at 08:48

    f(7) = p.77 + q.73 + r.7 - 4 = -p.(-7)7 - q.(-7)3 - r.(-7) - 4

    = -[p.(-7)7 + q.(-7)3 + r.(-7)] - 4 = -[f(-7) + 4] - 4

    = -(3 + 4) - 4 = -7 - 4 = -11

    Selected by MathYouLike
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    Khánh Lê Minh 23/04/2017 at 09:06

    f(7) = p.77 + q.73 + r.7 - 4 = -p.(-7)7 - q.(-7)3 - r.(-7) - 4

    = -[p.(-7)7 + q.(-7)3 + r.(-7)] - 4 = -[f(-7) + 4] - 4

    = -(3 + 4) - 4 = -7 - 4 = -11banhqua

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    ichigodangyeu123 22/04/2017 at 20:58

    What grade are you in?Uchiha Sasuke 


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Carter
18/04/2017 at 15:09
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4
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 Simplify the expression (where a, b, and c are different real numbers) :

\(\dfrac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}+\dfrac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\dfrac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}\)

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    An Duong 21/04/2017 at 07:27

    Let \(f\left(x\right)=\dfrac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}+\dfrac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\dfrac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}-1\)

    We have:

      \(f\left(a\right)=0+1+0-1=0\)

    Similarity \(f\left(b\right)=0,f\left(c\right)=0\)

    f(x) is a degree of 2 and have 3 different solotutions (a, b, c) .

    => f(x) = 0

    => \(\dfrac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}+\dfrac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\dfrac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}=1\)

    Selected by MathYouLike
  • ...
    Sarah Marianna 01/08/2019 at 04:36

    Let f(x)=(x−a)(x−b)(c−a)(c−b)+(x−b)(x−c)(a−b)(a−c)+(x−c)(x−a)(b−c)(b−a)−1f(x)=(x−a)(x−b)(c−a)(c−b)+(x−b)(x−c)(a−b)(a−c)+(x−c)(x−a)(b−c)(b−a)−1

    We have:

      f(a)=0+1+0−1=0f(a)=0+1+0−1=0

    Similarity f(b)=0,f(c)=0f(b)=0,f(c)=0

    f(x) is a degree of 2 and have 3 different solotutions (a, b, c) .

    => f(x) = 0

    => (x−a)(x−b)(c−a)(c−b)+(x−b)(x−c)(a−b)(a−c)+(x−c)(x−a)(b−c)(b−a)=1

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    Sarah Marianna 01/08/2019 at 04:35

    Let f(x)=(x−a)(x−b)(c−a)(c−b)+(x−b)(x−c)(a−b)(a−c)+(x−c)(x−a)(b−c)(b−a)−1f(x)=(x−a)(x−b)(c−a)(c−b)+(x−b)(x−c)(a−b)(a−c)+(x−c)(x−a)(b−c)(b−a)−1

    We have:

      f(a)=0+1+0−1=0f(a)=0+1+0−1=0

    Similarity f(b)=0,f(c)=0f(b)=0,f(c)=0

    f(x) is a degree of 2 and have 3 different solotutions (a, b, c) .

    => f(x) = 0

    => (x−a)(x−b)(c−a)(c−b)+(x−b)(x−c)(a−b)(a−c)+(x−c)(x−a)(b−c)(b−a)=1                                                             Study Well !


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Mr Puppy
05/07/2018 at 09:06
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15
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Five pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treatrous and selfish (especially the captain). The captain always proposes a distribution of the loot. All pirates vote on the proposal and if at least half of the crew (which includes himself) approve, the loot will be divided as proposal, as no pirates would be willing to take on the captain without superior force on their side. Otherwise he will face a mutiny: all pirates will turn against him and make him walk the plank. The pirates will start over again with the next senior pirate as captain. What's the maximum number of coins the captain can keep without risking his life?

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    Thái Ninh Nguyễn Phạm 18/07/2018 at 08:44

    hihi mình nha!

    Năm tên cướp biển đã thu được 100 đồng tiền vàng và phải phân chia chiến lợi phẩm. Các hải tặc đều cực kỳ thông minh, treatrous và ích kỷ (đặc biệt là đội trưởng). Đội trưởng luôn đề nghị một bản phân phối của loot. Tất cả các hải tặc biểu quyết về đề nghị này và nếu có ít nhất một nửa số thuyền viên (trong đó bao gồm bản thân) phê duyệt, các chiến lợi phẩm sẽ được phân chia như đề nghị, như là không có hải tặc sẽ sẵn sàng để đưa vào các đội trưởng mà không cần lực lượng vượt trội về phía họ. Nếu không anh sẽ phải đối mặt với một cuộc nổi loạn: tất cả các tên cướp biển sẽ quay lưng lại với anh và làm cho anh ta bước đi trên tấm ván. Các hải tặc sẽ bắt đầu lại một lần nữa với cướp biển cao cấp tiếp theo là đội trưởng. số lượng tối đa của đồng tiền các đội trưởng có thể giữ mà không mạo hiểm cuộc sống của mình là gì?

    Mr Puppy selected this answer.
  • ...
    imphon 01/04/2019 at 13:53

    The captain gives himself 98 gold coins and give the third and the fifth 1 coin each.

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    Gia Bảo Bùi 24/03/2020 at 13:02

    The captain can keep 98 coins. Click thí link for more:https://www.youtube.com/watch?v=Mc6VA7Q1vXQ


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Huỳnh Anh Phương
09/07/2018 at 11:13
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3
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Find x; y , know that x and y is postive ỉterger to be satisfied 2 calculations:

1) xy = yx

2) \(x\times y=8\)

 
 

Find x

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    Lê Quốc Trần Anh Coordinator 13/07/2018 at 04:10

    Because x,y are positive integers and \(x\times y=8\)

    => \(\left\{{}\begin{matrix}x=1;y=8\\x=2;y=4\\x=4;y=2\\x=8;y=1\end{matrix}\right.\)

    Because \(x^y=y^x\)

    => \(\left\{{}\begin{matrix}x=2;y=4\\x=4;y=2\end{matrix}\right.\)

    Huỳnh Anh Phương selected this answer.
  • ...
    Kalila Thao Van 30/01/2019 at 06:05

    i think the first x and y are 2 . the second i don't do it


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Lê Quốc Trần Anh Coordinator
24/06/2018 at 04:07
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3
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Suppose a regular hexagon has a perimeter equal to the circumference of a circle. What is the ratio of a side of the hexagon to the radius of the circle? Express your answer as a common fraction in terms of π

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    Nguyễn Mạnh Hùng 24/06/2018 at 08:36

    Call the hexagon's side is x

    the circle's radius is y

    => The hexagon's perimeter is 6x

    The circle's perimeter is 6,28y

    If 6x = 6,28y

    => \(\dfrac{x}{y}=\dfrac{6,28}{6}=\dfrac{3,14}{3}=\dfrac{\Pi}{3}\)

    Selected by MathYouLike
  • ...
    trương trí tuệ 15/04/2019 at 08:32

    Call the hexagon's side is x

    the circle's radius is y

    => The hexagon's perimeter is 6x

    The circle's perimeter is 6,28y

    If 6x = 6,28y

    => xy=6,286=3,143=Π3

  • ...
    trương trí tuệ 15/04/2019 at 08:31

    Call the hexagon's side is x

    the circle's radius is y

    => The hexagon's perimeter is 6x

    The circle's perimeter is 6,28y

    If 6x = 6,28y

    => xy=6,286=3,143=Π3


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Lê Quốc Trần Anh Coordinator
25/06/2018 at 03:07
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1
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Let T be a positive integer whose only digits are 0s and 1s. If X = T ÷ 12, and X is an integer, what is the smallest possible value of X?

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    Nguyễn Mạnh Hùng 25/06/2018 at 08:07

    We have:

    12 = 3 x 4 

    So T \(⋮12\) <=> T \(⋮3and4\)

    T \(⋮3\Leftrightarrow\)The sum of all the number of T \(⋮3\)

    \(\Leftrightarrow\left(1+1+1+...+1+0+...+0\right)⋮3\)

    But T's sum must be different than 0 

    => The smallest sum must be 3 = 1 + 1 + 1 + 0 + 0 + ... + 0

    T divided by 4 <=> Two-last digit must be divided by 4

    => Two last digits must be 00

    => T = 11100

    It's too big :) 

    Lê Quốc Trần Anh selected this answer.

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Lê Quốc Trần Anh Coordinator
25/06/2018 at 03:10
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Equilateral triangle ABC has a side length of 6 units. Point D lies on segment BC such that DC = 2(BD). What is the length of the altitude of triangle ADC from point C? Express your answer as a common fraction in simplest radical form.

Mathcounts

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    Alone 25/06/2018 at 06:54

    A B C D H 6 K

    Draw AK is the altitude of triangle ABC(\(K\in BC\)),the altitude of triangle ADC from point C cut AD at I

    So \(BK=CK=\dfrac{1}{2}BC\)

    Applying Pitago's theorem we have:

     \(AK^2+CK^2=AC^2\Rightarrow AK^2=AC^2-CK^2=AC^2-\left(\dfrac{1}{2}BC\right)^2=6^2-3^2=27\)

    \(\Rightarrow AK=3\sqrt{3}\)\(\Rightarrow S_{ADC}=\dfrac{AK.CD}{2}=\dfrac{3\sqrt{3}.4}{2}=6\sqrt{3}\)

    Because BD+DK=BK \(\Rightarrow DK=BK-BD=3-2=1\)

    Applying Pitago's theorem in triangle ADK we have:

     \(AD^2=AK^2+DK^2=27+1^2=28\)\(\Rightarrow AD=2\sqrt{7}\)

    But \(S_{ADC}=6\sqrt{3}\)\(\Rightarrow\dfrac{AD.CH}{2}=6\sqrt{3}\Rightarrow AD.CH=12\sqrt{3}\Rightarrow CH=\dfrac{12\sqrt{3}}{2\sqrt{7}}=\dfrac{6\sqrt{21}}{7}\)

    Selected by MathYouLike

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Lê Quốc Trần Anh Coordinator
25/06/2018 at 03:11
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  In trapezoid ABCD, bases AB and CD are 13 and 39 units, respectively. Legs BC and DA are 24 and 10 units, respectively, and sides BC and DA lie on lines that are perpendicular to each other. What is the area of ABCD? 

Mathcounts

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    Nguyễn Mạnh Hùng 25/06/2018 at 07:40

    13 units 39 units 24 units 10 units E A B D C a b

    Call \(AD\perp BC=\left\{E\right\}\)

    We have: \(a^2+b^2=13^2=169\)(Pythagorean theorem in triangle EAB)

    We also have: 

    \(\left(a+10\right)^2+\left(b+24\right)^2=39^2=1521\)

    \(\Rightarrow a^2+20a+100+b^2+48b+576=1521\)

    \(\Rightarrow a^2+20a+b^2+48b=1521-576-100=845\)

    Because \(a^2+b^2=169\)

    \(\Rightarrow20a+48b=845-169=676\)

    \(\Rightarrow5a+12b=169\)

    \(a^2+b^2=169\)

    => \(\left\{{}\begin{matrix}a^2=5a\\b^2=12b\end{matrix}\right.\)

    \(\Rightarrow a=5;b=12\)

    => DE = 15

    BE = 12 

    => \(S_{DBE}=\dfrac{15.12}{2}=90\left(units^2\right)\)
    \(S_{ABE}=\dfrac{5.12}{2}=30\left(units^2\right)\)

    \(\Rightarrow S_{ABCD}=90-30=60\left(units^2\right)\)

    Lê Quốc Trần Anh selected this answer.

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Uchiha Sasuke
14/06/2018 at 11:29
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There are 2017 points on the plane. The area of any triangle with verticles does not exceed 1. Assume that in any case, all these points can be placed in a triangle whose area is a positive integer K. Find the least value of K.

Triangle


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Mr. Bee moderators
25/06/2018 at 02:53
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2
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A polygon is a closed figure whose sides are straight lines. The above figure shows a six-sided polygon (a hexagon). Show that the sum of the angles of a hexagon, \(S^o\), is \(S^o= 540^o\).

Then, show that the generalized formula for the sum of the angles of an n-sided polygon is \(S^o= 180^o(n-2)\).

PolygonIGCSE

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    Alone 25/06/2018 at 07:10

    A B C D E F

    \(S^o=\)\(\angle\)BAF + \(\angle\)ABC + \(\angle\)BCD + \(\angle\)CDE + \(\angle\)DEF + \(\angle\)EFA

      =\(\angle\)AFB + \(\angle\)BAF + \(\angle\)ABF + \(\angle\)CBF + \(\angle\)BCF + \(\angle\)DCF + \(\angle\)CDF + \(\angle\)EDF + \(\angle\)DEF + \(\angle\)EFD + \(\angle\)CFD + \(\angle\)BFC

     =\(180^o+180^o+180^o+180^o=720^o\)

    Selected by MathYouLike
  • ...
    Mr. Bee moderators 25/06/2018 at 07:31

    Sory, I mistyped. It should've been \(720^o\) not \(540^o.\)


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Lê Quốc Trần Anh Coordinator
22/06/2018 at 07:36
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A bird collection has exactly four types of birds (eagles, doves, crows and sparrows). The eagles and doves make up 60% of the collection, and the doves and crows make up 20% of the collection. If the 18 crows in the collection represent 5% of the total number of birds, how many of the birds are sparrows? 


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Lê Quốc Trần Anh Coordinator
22/06/2018 at 07:36
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A 20-foot-high rectangular room has a floor that measures 18’ by 15’. Its doorway measures 3’ by 12’, and its only window measures 7’ by 10’. How many square feet of wall space does the room have?  


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Lê Quốc Trần Anh Coordinator
22/06/2018 at 07:36
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2
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A square and a circle overlap such that a vertex of the square is at the center of the circle. The 4-inch radius of the circle is one-half the length of a side of the square. What is the area of the portion of the square region that is outside the circular region? Express your answer in terms of π.

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    huynh anh phuong 02/08/2019 at 14:01

    OMG !I don't khow what do you say!

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    A.R.M.Y 09/08/2018 at 15:55

    I don't know I'm sorry but i'm grade 4 =)


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Lê Quốc Trần Anh Coordinator
11/06/2018 at 02:11
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Prove that the product of three consecutive integers is not a square number.


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Lê Quốc Trần Anh Coordinator
13/06/2018 at 02:05
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2
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Given \(a^3+b^3=2\) Prove that \(a+b\le2\)

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    Nguyễn Mạnh Hùng 13/06/2018 at 09:35

    We have:

    \(a^3+b^3=\left(a+b\right)\left(a^2+ab+b^2\right)=2\) (*)

    Other way:

    \(a^2+ab+b^2=a^2+2.\dfrac{1}{2}b+\left(\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\)

    \(=\left(a+\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\)

    Because \(\left(a+\dfrac{1}{2}b\right)^2\ge0\) with \(\forall a,b\)

    \(\dfrac{3}{4}b^2\ge0\) with \(\forall b\)

    So \(\left(a+\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\ge0\) with \(\forall a,b\)

    or \(a^2+ab+b^2\ge0\) with \(\forall a,b\) (**)

    From (*) and (**) we have  \(a+b\le2\)

    Your ex is so hard to do :) 

    Lê Quốc Trần Anh selected this answer.
  • ...
    Lê Thành 23/06/2018 at 05:27

    We have:

    a3+b3=(a+b)(a2+ab+b2)=2a3+b3=(a+b)(a2+ab+b2)=2 (*)

    Other way:

    a2+ab+b2=a2+2.12b+(12b)2+34b2a2+ab+b2=a2+2.12b+(12b)2+34b2

    =(a+12b)2+34b2=(a+12b)2+34b2

    Because (a+12b)2≥0(a+12b)2≥0 with ∀a,b∀a,b

    34b2≥034b2≥0 with ∀b∀b

    So (a+12b)2+34b2≥0(a+12b)2+34b2≥0 with ∀a,b∀a,b

    or a2+ab+b2≥0a2+ab+b2≥0 with ∀a,b∀a,b (**)

    From (*) and (**) we have  a+b≤2a+b≤2

    Your ex is so hard to do :) 


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Lê Quốc Trần Anh Coordinator
13/06/2018 at 02:07
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Prove that with a,b,c > 0: \(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{a}\)


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Lê Quốc Trần Anh Coordinator
09/06/2018 at 03:41
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In June, Casey counted the months until he would turn 16, the minimum age at which he could obtain his driver’s license. If the number of months Casey counted until his birthday was 45, in what month would Casey turn 16? 


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Lê Quốc Trần Anh Coordinator
09/06/2018 at 04:36
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Given \(P\left(x\right)+3P\left(2\right)=5x^2\left(\forall x\right)\). Calculate P(3)

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    Alone 13/06/2018 at 10:16

    With x=2 then \(4P\left(2\right)=5.2^2=20\Rightarrow P\left(2\right)=5\)

    With x=3 then \(P\left(3\right)+3P\left(2\right)=5.3^2=45\Rightarrow P\left(3\right)=45-5.3=30\)

    Lê Quốc Trần Anh selected this answer.

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Lê Quốc Trần Anh Coordinator
10/06/2018 at 06:31
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Find x,y,z satisfys: \(\dfrac{x}{8}=\dfrac{y}{3}=\dfrac{z}{10}\) and xy + yz + zx = 1206

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    Alone 14/06/2018 at 01:07

    \(\dfrac{x}{8}=\dfrac{y}{3}=\dfrac{z}{10}\Rightarrow\dfrac{x^2}{64}=\dfrac{xy}{24}=\dfrac{yz}{30}=\dfrac{zx}{80}=\dfrac{xy+yz+zx}{24+30+80}=\dfrac{1206}{134}=9\)

    \(\Rightarrow x^2=64.9=8^2.3^2\Rightarrow x=\pm24\)

    With x=24 then y=24:8.3=9;z=30

    With x=-24 then y=-9;z=-30

    Lê Quốc Trần Anh selected this answer.
  • ...
    Nguyễn Mạnh Hùng 15/06/2018 at 01:35

    We have:

    \(\dfrac{x}{8}=\dfrac{y}{3}=\dfrac{z}{10}\)

    \(\Rightarrow\dfrac{x^2}{64}=\dfrac{xy}{24}=\dfrac{yz}{30}=\dfrac{xz}{80}\)

    Apply the same sequence properties, we have:

    \(\dfrac{x^2}{64}=\dfrac{xy}{24}=\dfrac{yz}{30}=\dfrac{xz}{80}=\dfrac{xy+yz+xz}{24+30+80}=\dfrac{1206}{134}=9\)

    So \(x^2=9.64=576\)\(\Rightarrow x=24\)

    \(\dfrac{xy}{24}=9\Rightarrow xy=24.9=216\Rightarrow y=9\)

    \(\dfrac{yz}{30}=9\Rightarrow yz=270\Rightarrow z=30\)

    So \(\left(x;y;z\right)=\left(24;9;30\right)\)


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