giúp mk câu a,c,d ạ!!!!!
mk cảm ơn
A. behaved B. bored C. hoped D. tried
Gạch ngang ở dưới là ed nha(mk ko biết gõ, sorry)
Mn giúp mk câu này vs ạ,mk cảm ơn!
xem hộ mk mấy câu kia mk lm đúng chưa ạ . Giúp mk những câu mk chưa làm với ạ . Cảm ơn đã giúp mk ạ . Đề bài viết lại cho giống nghĩa cho giống nghĩa gốc ở câu a
8 How much do these apples cost?
9 THis is a blue car
11 Are there 40 classrooms in Phong's school?
13 How wide if the Great wall
Giúp mk câu 35,37,38 mk cảm ơn ạ
Câu 37: A
Câu 38: A
Câu 39: Cấu hình bậc hai
Giúp mk câu 2 vs ạ. Mk cảm ơn.
a) \(m_O=\dfrac{20.20}{100}=4\left(g\right)\)
=> \(n_{CaO}=n_O=\dfrac{4}{16}=0,25\left(mol\right)\)
\(\left\{{}\begin{matrix}\%m_{CaO}=\dfrac{0,25.56}{20}.100\%=70\%\\\%m_{Ca}=100\%-70\%=30\%\end{matrix}\right.\)
b) \(n_{Ca}=\dfrac{20.30\%}{40}=0,15\left(mol\right)\)
PTHH: Ca+ 2H2O --> Ca(OH)2 + H2
0,15-------------------->0,15
=> V = 0,15.22,4 = 3,36 (l)
\(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)
=> nFe = 0,3 (mol)
=> mFe = 0,3.56 = 16,8 (g)
=> \(m=\dfrac{16,8.100}{78,9474}=21,28\left(g\right)\)
c) Giả sử Fe3O4 bị khử thành Fe
Gọi số mol Fe3O4 pư là a (mol)
PTHH: Fe3O4 + 4H2 --> 3Fe + 4H2O
a--->4a----->3a
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,15}{4}\) => Hiệu suất tính theo H2
m = 23,2 - 232a + 168a = 21,28
=> a = 0,03 (mol)
=> \(\left\{{}\begin{matrix}n_{Fe_3O_4\left(pư\right)}=0,03\left(mol\right)\\n_{H_2\left(pư\right)}=0,12\left(mol\right)\end{matrix}\right.\)
\(H=\dfrac{n_{H_2\left(pư\right)}}{n_{H_2\left(bđ\right)}}=\dfrac{0,12}{0,15}.100\%=80\%\)
mn chuyển sang câu bị động giúp mk với ạ , mk cảm ơn ạ
a computer is used to do that job nowadays
He will be seen off at the airport by all his friends
Beer used to be drunk for breakfast in England years ago
Tea can not made with cold water.
the floor was being cleaned when i arrived
Should Julia be helped with the sewing ?
mn giúp mk câu 4,5 thu ạ , mk cảm ơn
4 There aren't any tomatoes left
5 My sister's favorite food is chicken
Mn giúp mk câu này vs, mk cảm ơn ạ
chuyển sang câu bị động giúp mk với ạ . mk cảm ơn
8. Her telephone number isn't known by me.
9. The children will be brought home by my students.
10. I was sent a present last week.
11. More information was given to us by her.
12. All the workers of the plan were being instructed by the chief engineer.
I hadn’t been told about it.
. Her telephone number isn’t known.
The children will be brought home by my students.
I was sent a present last week.
. More information was given us.
All the workers were being instructed of the plan by the chief engineer.
1 Her telephone number isn't known
2 The children will be brought home by my students
3 A present was sent to me last week
4 More information was given to us
giúp mk giải chi tiết 4 câu trên ạ
mk cảm ơn
\(5;;\sqrt{\left(x+5\right)\left(3x+4\right)}>4\left(x-1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4\left(x-1\right)\le0\\\left(x+5\right)\left(3x+4\right)\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}4\left(x-1\right)\ge0\\\left(x+5\right)\left(3x+4\right)\ge0\\\left(x+5\right)\left(3x+4\right)>16\left(x-1\right)^2\end{matrix}\right.\end{matrix}\right.\)
\(TH:\left\{{}\begin{matrix}4\left(x-1\right)\le0\\\left(x+5\right)\left(3x+4\right)\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left[{}\begin{matrix}x\le-5\\x\ge-\dfrac{4}{3}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x\in(-\infty;-5]\cup\left[-\dfrac{4}{3};1\right]\left(1\right)\)
\(TH:\left\{{}\begin{matrix}4\left(x-1\right)\ge0\\\left(x+5\right)\left(3x+4\right)\ge0\\\left(x+5\right)\left(3x+4\right)>16\left(x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x\le-5\\x\ge-\dfrac{4}{3}\end{matrix}\right.\\-\dfrac{1}{13}< x< 4\\\end{matrix}\right.\)\(\Rightarrow x\in[1;4)\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow x\in(-\infty;5]\cup[\dfrac{-4}{3};4)\)
\(6;;;;\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{49x^2+7x-42}< 181-14x\)
(đoạn 49x^2+7x+42 chắc bạn viết sai đề dấu"-" thành "+")
\(đk:\left\{{}\begin{matrix}7x+7\ge0\\7x-6\ge0\end{matrix}\right.\) \(\Leftrightarrow x\ge\dfrac{6}{7}\)
\(bpt\Leftrightarrow\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{\left(7x+7\right)\left(7x-6\right)}+14x+1< 182\left(1\right)\)
\(đặt:\sqrt{7x+7}+\sqrt{7x-6}=t>0\)
\(\Rightarrow t^2=14x+1+2\sqrt{\left(7x+7\right)\left(7x-6\right)}\)
\(\Rightarrow\left(1\right)\Leftrightarrow t^2+t< 182\Leftrightarrow-14< t< 13\)
\(\Rightarrow\sqrt{7x+7}+\sqrt{7x-6}< 13\Leftrightarrow14x+1+2\sqrt{\left(7x+7\right)\left(7x-6\right)}< 169\)
\(\Leftrightarrow2\sqrt{\left(7x+7\right)\left(7x-6\right)}< 168-14x\)
\(\Leftrightarrow\left\{{}\begin{matrix}168-14x\ge0\\\left(7x+7\right)\left(7x-6\right)\ge0\\4\left(7x+7\right)\left(7x-6\right)< \left(168-14x\right)^2\\\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le12\\\left[{}\begin{matrix}x\le-1\\x\ge\dfrac{6}{7}\end{matrix}\right.\\x< 6\\\end{matrix}\right.\)\(\Rightarrow\dfrac{6}{7}\le x< 6\)
\(7;\) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}< 2x+\dfrac{1}{2x}-1\left(đk:x>0\right)\)
\(\Leftrightarrow3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(x+\dfrac{1}{4x}\right)-1\left(1\right)\)
\(đặt:\sqrt{x}+\dfrac{1}{2\sqrt{x}}=t>0\)
\(\Leftrightarrow t^2=\sqrt{x}^2+2.\sqrt{x}.\dfrac{1}{2\sqrt{x}}+\left(\dfrac{1}{2\sqrt{x}}\right)^2=x+\dfrac{1}{4x}+1\)
\(\Rightarrow x+\dfrac{1}{4x}=t^2-1\)
\(\left(1\right)\Leftrightarrow3t< 2\left(t^2-1\right)-1\)
\(\Leftrightarrow2t^2-3t-3>0\Leftrightarrow\left[{}\begin{matrix}t< \dfrac{3-\sqrt{33}}{4}\\t>\dfrac{3+\sqrt{33}}{4}\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\dfrac{1}{2\sqrt{x}}>\dfrac{3+\sqrt{33}}{4}\)
\(\Leftrightarrow\dfrac{2x+1}{2\sqrt{x}}>\dfrac{3+\sqrt{33}}{4}\)
\(\Leftrightarrow\sqrt{x}< \dfrac{2\left(2x+1\right)}{3+\sqrt{33}}\Leftrightarrow\left\{{}\begin{matrix}x>0\\2\left(2x+1\right)\ge0\\x< \left[\dfrac{2\left(2x+1\right)}{3+\sqrt{33}}\right]^2\\\end{matrix}\right.\)
đến đây dễ dàng rồi như mấy ý trên bạn tự giải quyết để tìm ra x