Giúp em câu 14 với ạ
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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!
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