All Questions
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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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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
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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 = -11
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ichigodangyeu123 22/04/2017 at 20:58
What grade are you in?Uchiha Sasuke
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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
=> \(\dfrac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}+\dfrac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\dfrac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}=1\)
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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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Thái Ninh Nguyễn Phạm 18/07/2018 at 08:44
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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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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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
=> \(\dfrac{x}{y}=\dfrac{6,28}{6}=\dfrac{3,14}{3}=\dfrac{\Pi}{3}\)
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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
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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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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.
Lê Quốc Trần Anh Coordinator
25/06/2018 at 03:10
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Alone 25/06/2018 at 06:54
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 \(S_{ADC}=6\sqrt{3}\)\(\Rightarrow\dfrac{AD.CH}{2}=6\sqrt{3}\Rightarrow AD.CH=12\sqrt{3}\Rightarrow CH=\dfrac{12\sqrt{3}}{2\sqrt{7}}=\dfrac{6\sqrt{21}}{7}\)
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Nguyễn Mạnh Hùng 25/06/2018 at 07:40
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.
Mr. Bee moderators
25/06/2018 at 02:53
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Alone 25/06/2018 at 07:10
\(S^o=\)\(\angle\)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
=\(180^o+180^o+180^o+180^o=720^o\)
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Sory, I mistyped. It should've been \(720^o\) not \(540^o.\)
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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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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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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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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 \(\left(x;y;z\right)=\left(24;9;30\right)\)