Giúp mik với mik dò
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Tham khảo
1. Someone will clean the room today. =>THE ROOM WILL BE CLEANED TODAY
2. Someone might steal the car. =>THE CAR MIGHT BE STOLEN
3. They had to cut down that tree.=>THAT TREE HAD TO BE CUT DOWN
4. They're going to demolish the old houses.=>THE OLD HOUSES ARE GOING TO BE DEMOLISHED
5. We can't restore the picture. =>THE PICTURE CAN'T BE RESTORED
6. You must make an appointment in advance.=>AN APPOINTMENT MUST BE MADE IN ADVANCE
7. I don't want people to make me a fool. =>I DON'T WANT TO BE MADE A FOOL
8. Someone has to look after the garden.=>THE GARDEN HAS TO BE LOOKED AFTER
9. He wants everybody to serve him. =>HE WANTS TO BE SERVED
10. They're going to interview him next week=>HE IS GOING TO BE INTERVIEWED NEXT WEEK
Tham khảo
1. Someone will clean the room today. =>THE ROOM WILL BE CLEANED TODAY
2. Someone might steal the car. =>THE CAR MIGHT BE STOLEN
3. They had to cut down that tree.=>THAT TREE HAD TO BE CUT DOWN
4. They're going to demolish the old houses.=>THE OLD HOUSES ARE GOING TO BE DEMOLISHED
5. We can't restore the picture. =>THE PICTURE CAN'T BE RESTORED
6. You must make an appointment in advance.=>AN APPOINTMENT MUST BE MADE IN ADVANCE
7. I don't want people to make me a fool. =>I DON'T WANT TO BE MADE A FOOL
8. Someone has to look after the garden.=>THE GARDEN HAS TO BE LOOKED AFTER
9. He wants everybody to serve him. =>HE WANTS TO BE SERVED
10. They're going to interview him next week=>HE IS GOING TO BE INTERVIEWED NEXT WEEK
\(n_{NaOH}=0.015\cdot2=0.03\left(mol\right)\)
\(n_{H_2SO_4}=0.015\cdot1.5=0.0225\left(mol\right)\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)
\(0.03..........0.015...........0.015\)
\(n_{H_2SO_4\left(dư\right)}=0.0225-0.015=0.0075\left(mol\right)\)
\(C_{M_{Na^+}}=\dfrac{0.015\cdot2}{0.015+0.015}=1\left(M\right)\)
\(C_{M_{H^+}}=\dfrac{0.0075\cdot2}{0.015+0.015}=0.5\left(M\right)\)
\(C_{M_{SO_4^{2-}}}=\dfrac{0.015+0.0075}{0.015+0.015}=0.75\left(M\right)\)
\(\left|cosx\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{1}{2}\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x=\pm\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{3}+k2\pi\)
\(\left|cosx\right|=\dfrac{1}{2}\)
\(\Leftrightarrow cos^2x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1+cos2x}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow cos2x=-\dfrac{1}{2}\)
\(\Rightarrow2x=\pm\dfrac{2\pi}{3}+k2\pi\)
\(\Rightarrow x=\pm\dfrac{\pi}{3}+k\pi\)
Thì là vs nơi ơ, ăn uống hợp vệ sinh, rửa tay trc khi ăn và sau khi đại tiện... nhìu lắm bn ak ;)
Câu 3:
a: \(BD=\sqrt{BC^2-DC^2}=4\left(cm\right)\)
b: \(\widehat{A}=180^0-2\cdot70^0=40^0< \widehat{B}\)
nên BC<AC=AB
c: Xét ΔEBC vuông tại E và ΔDCB vuông tại D có
BC chung
\(\widehat{EBC}=\widehat{DCB}\)
Do đó:ΔEBC=ΔDCB
d: Xét ΔOBC có \(\widehat{OBC}=\widehat{OCB}\)
nên ΔOBC cân tại O
Câu 2
a) Thay y = -2 vào biểu thức đã cho ta được:
2.(-2) + 3 = -1
Vậy giá trị của biểu thức đã cho tại y = -2 là -1
b) Thay x = -5 vào biểu thức đã cho ta được:
2.[(-5)² - 5] = 2.(25 - 5) = 2.20 = 40
Vậy giá trị của biểu thức đã cho tại x = -5 là 40