4 cry a lot anymore

5 used to study at this school

6 used to walk to school

7 is a famous singer

8 have very long hair anymore

9 used to be good boys

10 used to get up early

11 work on a farm anymore

12 take much morning exercises anymore

13 use to smoke a lot

14 use to spend a lot of money

Câu 13:

ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC^2=6^2+8^2=100\)

=>\(BC=10\left(cm\right)\)

Xét ΔABC vuông tại A có AH là đường cao

nên \(AH\cdot BC=AB\cdot AC\)

=>\(AH\cdot10=6\cdot8=48\)

=>AH=48/10=4,8(cm)

Xét ΔABC vuông tại A có AH là đường cao

nên \(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BH=\dfrac{6^2}{10}=3,6\left(cm\right)\\CH=\dfrac{8^2}{10}=6,4\left(cm\right)\end{matrix}\right.\)

Xét ΔHAB vuông tại H có HM là đường cao

nên \(BM\cdot BA=BH^2\)

=>\(BM\cdot6=3,6^2\)

=>BM=2,16(cm)

Xét ΔHAC vuông tại H có HN là đường cao

nên \(AN\cdot AC=AH^2\)

=>\(AN\cdot8=4,8^2\)

=>AN=2,88(cm)

ΔABN vuông tại A

=>\(AB^2+AN^2=BN^2\)

=>\(BN^2=2.88^2+6^2=44,2944\)

=>\(BN=\sqrt{44,2944}=\dfrac{6\sqrt{769}}{25}\left(cm\right)\)

Xét tứ giác AMHN có \(\widehat{AMH}=\widehat{ANH}=\widehat{MAN}=90^0\)

nên AMHN là hình chữ nhật

=>AH=MN=4,8(cm)

Xét ΔMBN có \(cosBMN=\dfrac{MB^2+MN^2-NB^2}{2\cdot MB\cdot MN}\)

\(=\dfrac{4,8^2+2,16^2-\dfrac{27684}{625}}{2\cdot4,8\cdot2,16}=\dfrac{-10368}{625}:\dfrac{2592}{125}=-\dfrac{4}{5}\)

=>\(sinBMN=\sqrt{1-\left(-\dfrac{4}{5}\right)^2}=\dfrac{3}{5}\)

Xét ΔBMN có \(\dfrac{NB}{sinBMN}=2R\)

=>\(2R=\dfrac{6\sqrt{769}}{25}:\dfrac{3}{5}=\dfrac{6\sqrt{769}}{25}\cdot\dfrac{5}{3}=\dfrac{2}{5}\sqrt{769}\)

=>\(R=\dfrac{\sqrt{769}}{5}\)

=>Chọn A

6) Does your apartment have a pool?
7) Does your house have a garden?
8) Does your house have a basement?
9) Are there 4 rooms in your house?
10) Is there a clock in your class?
11) Harry Potter and the Sorcerer's Stone is written by J.K.Rowling.
12) Treasure Island is written by Robert Louis Stevenson.
13) My class has twenty people.
14) My team has 4 people.
Chúc bạn học tốt!

13.A

14.B

15.A

13, A
14, C
15, A

13. A

14. B

15. A

a) 

P1:

\(n_{Br_2}=\dfrac{80.20\%}{160}=0,1\left(mol\right)\)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

             0,1<--0,1

=> \(n_{C_2H_4\left(P_1\right)}=0,1\left(mol\right)\)

=> \(m_{C_3H_8\left(P_1\right)}=\dfrac{12,2}{2}-0,1.28=3,3\left(g\right)\)

=> \(n_{C_3H_8\left(P_1\right)}=\dfrac{3,3}{44}=0,075\left(mol\right)\)

=> \(V=\left(0,1.2+0,075,2\right).22,4=7,84\left(l\right)\)

\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1}{0,1+0,075}.100\%=57,143\%\\\%V_{C_3H_8}=\dfrac{0,075}{0,1+0,075}.100\%=42,857\%\end{matrix}\right.\)

b) P₂ \(\left\{{}\begin{matrix}C_2H_4:0,1\left(mol\right)\\C_3H_8:0,075\left(mol\right)\end{matrix}\right.\)

Bảo toàn C: \(n_{CO_2}=0,425\left(mol\right)\) => \(n_{BaCO_3}=0,425\left(mol\right)\)

Bảo toàn H: \(n_{H_2O}=0,5\left(mol\right)\)

Xét \(\Delta m=m_{CO_2}+m_{H_2O}-m_{BaCO_3}=0,425.44+0,5.18-0,425.197=-56,025\left(g\right)\)

=> khối lượng dd sau pư giảm 56,025 gam