Giúp tôi với 30p nữa là nộp rồi, please:(((((
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I
1 B
2 A
3 C
4 A
5 B
6 D
7 C
8 A
II
1 are going to visit
2 went - had - have you been
3 was
5 do they get up
6 did they do - had
7 has lived
8 have visited
9 is playing
10 has played
11 won't go - will stay
12 watches - isn't watching - is playing
13 got - were
14 stayed - watched
15 has never read
16 won't go - will do
17 likes - is sleeping
18 inviting
19 is
20 is
III
1 B
2 A
3 D
4 B
5 A
6 A
7 C
8 B
9 C
10 C
11 A
12 B
13 B
14 D
15 C
16 B
17 D
18 B
19 D
20 D
21 A
22 A
23 C
Gọi \(M\left(x;y\right)\) là 1 điểm thuộc d \(\Rightarrow2x+y-4=0\) (1)
\(V_{\left(O;2\right)}\left(M\right)=M'\Rightarrow M'\in d'\)
\(\left\{{}\begin{matrix}x'=2x\\y'=2y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{x'}{2}\\y=\dfrac{y'}{2}\end{matrix}\right.\)
Thế vào (1):
\(2.\left(\dfrac{x'}{2}\right)+\dfrac{y'}{2}-4=0\Leftrightarrow2x'+y'-8=0\)
Vậy pt d' có dạng: \(2x+y-8=0\)
a.
\(\%_{Fe_{\left(Fe_2O_3\right)}}=\dfrac{56.2}{160}.100\%=70\%\)
\(\%_{O_{\left(Fe_2O_3\right)}}=100\%-70\%=30\%\)
b.
\(\%_{C_{\left(C_6H_{12}O_6\right)}}=\dfrac{12.6}{180}.100\%=7\%\)
\(\%_{H_{\left(C_6H_{12}O_6\right)}}=\dfrac{1.12}{180}.100\%=6,7\%\)
\(\%_{O_{\left(C_6H_{12}O_6\right)}}=100\%-7\%-6,7\%=86,3\%\)
c.
\(\%_{C_{\left(\left(C_6H_{10}O_5\right)_n\right)}}=\dfrac{12.6}{162n}.100\%=44,4n\%\)
\(\%_{H_{\left(\left(C_6H_{10}O_5\right)_n\right)}}=\dfrac{1.10}{162n}.100\%=6,2n\%\)
\(\%_{O_{\left(C_6H_{1o}O_5\right)}}=\dfrac{16.5}{162n}.100\%=49,4n\%\)
\(\Rightarrow49,4n\%=100\%-44,4n\%-6,2n\%\)
\(\Leftrightarrow n=1\)
\(\Rightarrow\left\{{}\begin{matrix}\%_C=44,4\%\\\%_H=6,2\%\\\%_O=49,4\%\end{matrix}\right.\)
d.
\(\%_{Na_{\left(NaCl\right)}}=\dfrac{23}{58,5}.100\%=39,3\%\)
\(\%_{Cl_{\left(NaCl\right)}}=100\%-39,3\%=60,7\%\)
a/ \(A=1+3+3^2+..........+3^{55}\)
\(\Leftrightarrow3A=3+3^2+...........+3^{55}+3^{56}\)
\(\Leftrightarrow3A-A=\left(3+3^2+........+3^{56}\right)-\left(1+3+....+3^{55}\right)\)
\(\Leftrightarrow2A=3^{56}-1\)
\(\Leftrightarrow A=\frac{3^{56}-1}{2}\)
help me please
1: Xét tứ giác BDEA có \(\widehat{BEA}=\widehat{BDA}=90^0\)
nên BDEA là tứ giác nội tiếp
hay B,D,E,A cùng thuộc 1 đường tròn