In spite of his broken leg, he managed to get out of his car.

Unless anyone put the coal in, the fire will go out.

The man who looks worried is being questioned by the police.

Despite the heavy rain, they go on collecting the trash along the beach.

If you practise speaking English every day, you can improve your English.

We have cooked for the party for four hours.

In spite of the bad weather, he climbed to the top of the mountain.

Câu 10:

Gọi chiều rộng là a

Chiều dài là a+5

Theo đề, ta có: a(a+5)=70

\(\Leftrightarrow a^2+5a-70=0\)

\(\text{Δ}=5^2-4\cdot1\cdot\left(-70\right)=305>0\)

Do đó:Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}a_1=\dfrac{-5-\sqrt{305}}{2}\left(loại\right)\\a_2=\dfrac{-5+\sqrt{305}}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: Chiều rộng và chiều dài lần lượt là \(\dfrac{\sqrt{305}-5}{2}\left(m\right);\dfrac{\sqrt{305}+5}{2}\left(m\right)\)

Câu 10:

(1) CH3COOH + Na -> CH3COONa + 1/2 H2

(2) CH3COOH + NaOH -> CH3COONa + H2O

(3) CH3COOH + C2H5OH \(⇌\) CH3COOC2H5 + H2O (đk trên mũi tên phản ứng là H⁺ và nhiệt độ)

Câu 10.

Từ thông qua khung dây:

\(e_c=\left|\dfrac{\Delta\Phi}{\Delta t}\right|=\left|\dfrac{\Delta\Phi}{10}\right|=0,05\Rightarrow\Delta\Phi=0,5Wb\)

Độ lớn cảm ứng từ:

\(\Delta\Phi=BS\cdot cos\alpha\Rightarrow B=\dfrac{\Delta\Phi}{S}=\dfrac{0,5}{100\cdot10^{-4}}=50T\)

10.

ĐKXĐ: \(\left\{{}\begin{matrix}2x+10\ge0\\-x-5\ge0\\-2x-7\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x\le-5\\x\le-\dfrac{7}{2}\end{matrix}\right.\)

\(\Rightarrow\) Không tồn tại x thỏa mãn

Vậy BPT đã cho vô nghiệm

12.

\(\Leftrightarrow\dfrac{3}{x+3}-\dfrac{3x+4}{x\left(x+4\right)}\ge0\Leftrightarrow\dfrac{-x-12}{x\left(x+3\right)\left(x+4\right)}\ge0\)

\(\Rightarrow\left[{}\begin{matrix}-12\le x< -4\\-3< x< 0\end{matrix}\right.\)

\(x=\left\{-12;-11;...-5;-2;-1\right\}\) có 10 số nguyên

1 lived

2 was watching - rang

3 had gone

4 were- driving

5 had moved

6 hated

7 had already gone

8 saw - were sitting

9 had happened

10 was crossing

Câu 7 : nCO2 = 0,02 (mol) ; nH2O = 0,03 (mol)

=> nH2O > nCO2 -> ancol no ; đơn chức 

PTHH : \(C_nH_{2n+2}O+\dfrac{3}{2}nO_2\rightarrow nCO_2+\left(n+1\right)H_2O\) 

                                                  0,02            0,03      /mol

Ta có : \(\dfrac{0,02}{n}=\dfrac{0,03}{n+1}\)  \(\Leftrightarrow n=2\)  CTPT của X là : \(C_2H_6O\)