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1st way :
9920 = (992)10 = 980110 < 999910
2nd way :
9920 = 9910 x 9910 ; 999910 = 9910 x 10110
Since 9910 < 10110, 9920 < 999910
Phan Huy Toàn selected this answer.
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Phan Huy Toàn 12/08/2017 at 10:37
Answer:
3^223>3^222=(3^2)^111=9^111
2^322<2^333=(2^3)111=8^111
Form(1) and (2) deduced:2^332<8^111<9^111<3^223
Dress:2^332<3^223
yourself home
phanhuytien selected this answer. -
Flash Shit :3 12/08/2017 at 10:47
There are something not truth here !!! :V
Both of you , phanhuytien and Phan Huy Toàn , are you kidding me ???
Thôi , nói tiếng việt cho dễ
Bọn bây đùa nhau ak , 2 bài phải nói là rất khác :V , khác như là cá mè một lứa
Chép bài nhau :V , trong bài còn có chỗ ta hơi thắc mắc , "Dress" ở đó nghĩa là j , là "váy, đầm" ak , lên google mà trans á , trans được thì trans , ko trans được thì thôi , ở web này thì ko cần mấy bọn dog nhả xương rồi gặm đâu :v
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Ok , compare :))
We have : \(2^{332}< 2^{333}=\left(2^3\right)^{111}< \left(3^2\right)^{111}=3^{222}< 3^{223}\)
That is too difficult , right =))
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Vũ Thị Hương Giang 09/03/2017 at 21:28
a) 4xy3 + \(\dfrac{1}{3}\)xy8 + \(\dfrac{2}{5}\)xy3 + \(-\dfrac{1}{3}\)xy8
= 4xy3 + \(\dfrac{2}{5}\)xy3 + \(\dfrac{1}{3}xy^8\) + \(-\dfrac{1}{3}xy^8\)
= \(\dfrac{14}{3}xy^3+0=\dfrac{14}{3}xy^3\)
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a) \(4xy^3+\dfrac{1}{3}xy^8+\dfrac{2}{3}xy^3+\left(-\dfrac{1}{3}\right)xy^8\)
\(=\left(4+\dfrac{2}{3}\right)xy^3+\left[\left(-\dfrac{1}{3}\right)+\dfrac{1}{3}\right]xy^8\)
\(=\dfrac{14}{3}xy^3+0\)
\(=\dfrac{14}{3}.\dfrac{2}{3}.\left(-\dfrac{1}{2}\right)^3=-\dfrac{7}{18}\)
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FA FIFA Club World Cup 2018 16/01/2018 at 21:58
a) 4xy3+13xy8+23xy3+(−13)xy8
=(4+23)xy3+[(−13)+13]xy8
=143xy3+0
=143.23.(−12)3=−71
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Vũ Mạnh Hùng 06/12/2017 at 19:28
we need: 28*3=84
we have:84:25=3(remaining 9)
so we need a packet for these 9 sheets of graph of papers
3+1=4 (packets)
answer:4 packets
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The number of sheets of graph of paper that the students need is :
3 x 28 = 84
We have : 84 : 25 = 3 with remainder 9
So we need a packet for these 9 sheets of graph of paper.
Then the answer is : 3 + 1 = 4
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KEITA FC 8C 27/12/2017 at 19:10
we have : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{6xy}=\dfrac{1}{6}\Rightarrow6y+6x+1=xy\)
\(\Rightarrow\left(x-6\right)\left(y-6\right)=37\)
suppose : \(x\ge y\) then \(\left\{{}\begin{matrix}x-6=-1\\y-6=-37\end{matrix}\right.\) or \(\left\{{}\begin{matrix}x-6=37\\y-6=1\end{matrix}\right.\)
Case type, the latter case \(x=43\) , \(y=7\)
Selected by MathYouLike
Min Hoang moderators
16/03/2017 at 10:16
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Do Not Ask Why 23/04/2017 at 21:22
Last number is: 998
The first number: 100
Số có ba chữ số mà chia hết cho 2 là :
( 998 - 100 ) : 2 + 1 = 450 ( numbers )
Đ/s : 450 numbers
Easy you
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Nguyễn Thanh Huyền 23/04/2017 at 22:54
The first number is : 100
Last number is : 998
Number with three digits that share for 2 is :
( 998 - 100 ) : 2 + 1 = 450 ( numbers )
Đ/s : 450 numbers.
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Phan Huy Toàn 12/08/2017 at 10:14
Answer:
9^200=(3^2)^200=3^400
Dress 3^400=9^200
phanhuytien selected this answer. -
phanhuytien 12/08/2017 at 10:17
Answer;
3^400=(3^4)^1000=81^1000
9^2000=(9^2)^1000=81^1000
Dress 3^400=9^200
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I'm sorry , i'm wrong at question c !
We have :
25 = 32 = 2x
=> x = 5
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a) 23 = 2x
<=> \(\dfrac{2^3}{2^x}=1\)
<=> 23 - x = 1
<=> 3 - x = 0
<=> x = 3
b) 25 = 5x
<=> 52 = 5x
=> x = 2
c) 32 = 2x
<=> 24 = 2x
<=> x = 4
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