Rearrange the words and phrases to make meaningful sentences
1. an/pagoda/village/old/my/has.
- > My village has an old pagoda.
2.railway-station/not/small/a/that/town/has/got.
-> That town has not got a small railway- station.
3.one/Mr and Mrs. Brown/only/have/car/got.
-> Mr and Mrs Brown only have got one car.
4.building/school/new/has got/my/a.
-> My school has got a new building.
5.many/got/city/lakes/that/has/beautiful
-> That city has got many beautiful lakes.
6.Lodon/parks/has/famous/got/many
-> London has got many famous parks
7.any/city/museums/your/has/got?
-> Has your city got any museums?
8.art galleries/not/town/has/got/this/any
-> This town has not got any art galleries.
9.building/old/got/Hue/pagodas/and/many/has
-> Hue has got many old buildings and pagodas.
10.a/house/has/got/garage/your?
-> Has your house got a garage?

1. an/ pagoda/ village/old/ my has.

=> My village has an old pagoda.

2. railway-station/ not/small/ a/ that/ town/ has/ got.

=> That town has not got a small railway-station

3. one/ Mr. and Mrs. Brown/ only/ have/ car/ got.

=> Mr. and Mrs.Brown only have got one car.

4.building/ school/ new/ has got/ my/ a.

=> My new school has got a new building.

5. many/ got/ city/ lakes/ that/ has/ beautiful.

=> That city has got many beautiful lakes.

6. London/ park/ has/ famous/ got/ many

=> London has got many famous park.

7. any/ city/ museums/ your/ has/ got?

=> Has any museums in your city?

8. art gallaries/ not/ town/ has/ got/ this/ any

=>This town has not got any art gallaries.

9. buildings/ old/ got/ Hue/ pagodas/ and/ many/ has

=> Hue has got many old buildings and pagodas.

10. a/ house/ has/ got/ garage/ your?

=> Has your house got a garage?

\(n_{C_3H_6O_2}=0,05\left(mol\right)\)

\(n_{CO_2}=3n_{C_3H_6O_2}=3.0,05=0,15\left(mol\right)\)

\(\Rightarrow V_{CO_2}=22,4.0,15=3,36\left(l\right)\)

\(n_{H_2O}=\frac{6}{2}n_{C_3H_6O_2}=0,15\left(mol\right)\)

\(\Rightarrow m_{H_2O}=18.0,15=2,7\left(g\right)\)

a, \(x+34⋮x+1\) 

Ta có \(x+34=x+1+33\) 

mà \(x+1⋮x+1\)

 để \(x+34⋮x+1\) thì => \(33⋮x+1\) hay  x+1 \(\inƯ\left(33\right)\) 

\(Ư\left(33\right)=\left\{1;3;11;33\right\}\) 

Ta có bảng sau 

Vậy x \(\in\left\{0;2;10;32\right\}\) 

b, \(3x+13⋮x+1\) 

Ta có 3x+13 = 3(x+1) + 10 

mà \(3\left(x+1\right)⋮x+1\) 

để 3x+13\(⋮x+1\)  thì => \(10⋮x+1\) 

hay x+1 \(\in\) Ư(10)

Ư(10) = {1;2;510}

ta có bảng sau 

 Vậy x \(\in\left\{0;1;4;9\right\}\)

a/ \(\Delta ABC\) có AB=AC \(\Rightarrow\Delta ABC\) cân tại A \(\Rightarrow AD\perp BC\) (Trong tg cân phân giác đồng thời là đường cao và dường trung tuyến)

Xét tg vuông ADB và tg vuông ADC có 

\(AB=AC;\widehat{BAD}=\widehat{CAD}\Rightarrow\Delta ADB=\Delta ADC\) (2 tg vuông có cạnh huyền và 1 góc nhọn = nhau thì 2 tg đó bằng nhau)

b/ Ta có Ay//BC \(\Rightarrow\widehat{yAC}=\widehat{ACB}\) (góc so le trong)

Do \(\Delta ABC\) cân tại A \(\Rightarrow\widehat{ACB}=\widehat{ABC}\)

\(\Rightarrow\widehat{yAC}=\widehat{ABC}\) (cùng \(=\widehat{ACB}\) )

c/ Ta có \(AD\perp BC;Cx\perp BC\) => AD//Cx (cùng vuông góc với BC)

d/ Ta có AD//Cx (cmt); Ay//BC => AKCD là hình bình hành

AC và DK là hai đường chéo hình bình hành AKCD => AC và DK cắt nhau tại trung điểm mỗi đường giả sử là điểm I'

=> I' là trung điểm của AC mà I cũng là trung điểm của AC => I trùng I'

=> I là trung điểm của DK

a)15 x (-2) x (-5) x (-6)

= -30 x 30

= -900

b)4 x 7 x (-11) x (-2)

= 28 x -11 x -2

= 28 x 22

=616

mua luôn 16 cuộn dây điện?!

so qua cam con hai duoc la:

45:9=5(qua cam)

so qua cam cam ca hai me con hai duoc la:

45+5=50 (qua cam)

Dap so 50 qua cam