Giúp e bài này với được không ạ ?
em xin cám ơn !
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Hello! John.
Glad to see you again. Tam, this is my cousin, Peter. It's his first to visit to your country.
How do you do? Welcome to VN
Thank you. Nice to meet you, Tam
Can I help you with your suitcases, John?
Thanks. I can manage.
OK. Now we're going to the hotel in the center of the city by taxi
That would be nice. How far is it from the airport to the center of the city?
It's about a half-hour drive
Look, John! What a lot of motorbikes in the streets!
Oh, yeah. That surprised me by the time I first came to VN
Motorbikes are our main means of transport. I go to school every day by bike.
Would you mind taking me around the city by bike?
No, of course not. But I have to ask someone else to get Peter, too
Peter, do you mind if Tam's friend gives us a ride around the city?
No, I don't mind. But I feel a little bit scared because the traffic is so heavy
OK. So we'll go on a sightseeing tour by bikes at weekend
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Câu 4:
4.1/ Ta có: \(n_{NaCl}=2,5.0,4=1\left(mol\right)\)
\(\Rightarrow m_{NaCl}=1.58,5=58,5\left(g\right)\)
4.2/ Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
_____0,2___________0,2____0,2 (mol)
b, \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, \(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)
Bạn tham khảo nhé!
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1. bag
2. bunch
3. slice
4. box
5. piece
6. carton
7. can
8. bowl
9. jar
10. bottle
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\(\dfrac{2020}{2019}>\dfrac{2019}{2020}\Rightarrow0< a< 1\)
\(log_ba< 1\Rightarrow b>1\)
\(P=log_b^2a+log_b^22-\dfrac{m^2log_2b}{log_2a}+2\left(log_ba-2log_b2\right)-\dfrac{4^{ab^2}-2m.2^{ab^2}}{log_ba}\)
\(=log_b^2a+log_b^22+2log_ba-4log_b2-\dfrac{4^{ab^2}-2m.2^{ab^2}+m^2}{log_ba}\)
\(=\left(log_ba+1\right)^2+\left(log_b2-2\right)^2+\dfrac{\left(2^{ab^2}-m\right)^2}{-log_ba}-5\ge-5\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}log_ba=-1\\log_b2=2\\2^{ab^2}=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{\sqrt{2}}\\b=\sqrt{2}\\m=2^{ab^2}=2^{\sqrt{2}}\end{matrix}\right.\)
Sau khi tính lại thì không có đáp án nào đúng :(
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Câu 96: D. AB > CD (do AB là đường kính; CD là dây).
Câu 97: A. IC = ID (do CD \(\perp\) AB; CD là dây; AB là đường kính).
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\(\Delta'=\left(m-1\right)^2-\left(2m-4\right)=\left(m-2\right)^2+1>0;\forall m\)
\(\Rightarrow\) Pt đã cho luôn có 2 nghiệm pb với mọi m
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m-2\\x_1x_2=2m-4\end{matrix}\right.\) (1)
a. Pt có 2 nghiệm đối nhau khi:
\(x_1+x_2=0\Leftrightarrow2m-2=0\Rightarrow m=1\)
b. Trừ vế cho vế của (1) ta được:
\(x_1+x_2-x_1x_2=2m-2-\left(2m-4\right)\)
\(\Leftrightarrow x_1+x_2-x_1x_2=2\)
Đây là hệ thức liên hệ 2 nghiệm ko phụ thuộc m
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cái này áp dụng hệ thức lượng thôi bạn
AH=căn 6^2-4,8^2=3,6cm
=>AC=6^2/3,6=10cm
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