she didn't like chocolate 

she wouldn't see me the day after 

she was living in Paris for a few months

she had visited her parents at the weekend

she hadn't eaten sushi before

she hadn't travelled by underground before she had come to London

8 left - had bought - needed

9 had burnt - arrived

10 had done - was

11 had - heard - told

12 had lived - came

13 didn't talk - arrived - had already left

14 went - had stopped

15 thought - had won - didn't have

Oki:))

1 had finished - went

2 hadn't eaten - came

3 told - had never met

4 had done - went

5 went - got

6 had - begun - left

7 had been - started

Oki

a: ΔABC cân tại A

mà AH là đường cao

nên H là trung điểm của CB

=>CB=2CH

Xét ΔCKB vuông tại K và ΔCHA vuông tại H có

\(\hat{KCB}\) chung

Do đó: ΔCKB~ΔCHA

=>\(\frac{CB}{CA}=\frac{BK}{HA}=\frac{38.4}{32}=\frac65\)

=>\(\frac{2\cdot CH}{CA}=\frac65\)

=>\(\frac{CH}{CA}=\frac35\)

=>\(\frac{CH}{3}=\frac{CA}{5}=k\)

=>CH=3k; CA=5k

ΔAHC vuông tại H

=>\(HA^2+HC^2=AC^2\)

=>\(AC^2-CH^2=AH^2\)

=>\(\left(5k\right)^2-\left(3k\right)^2=32^2\)

=>\(16k^2=1024\)

=>\(k^2=64=8^2\)

=>k=8

=>\(CH=3\cdot8=24\left(\operatorname{cm}\right);CA=5\cdot8=40\left(\operatorname{cm}\right)\)

H là trung điểm của CB

=>\(CB=2\cdot CH=2\cdot24=48\left(\operatorname{cm}\right)\)

b: ΔBAC cân tại A

mà AH là đường cao

nên AH là đường trung trực của BC

=>O nằm trên đường trung trực của BC

=>OB=OC(1)

O nằm trên đường trung trực của AC

=>OA=OC(2)

Từ (1),(2) suy ra OA=OB=OC

Xét ΔABC có \(cosBAC=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)

\(=\frac{40^2+40^2-48^2}{2\cdot40\cdot40}=\frac{1600+1600-2304}{2\cdot1600}=\frac{896}{3200}=0,28=\frac{7}{25}\)

=>\(\sin BAC=\sqrt{1-\left(\frac{7}{25}\right)^2}=\frac{24}{25}\)

Xét ΔBAC có \(\frac{BC}{\sin BAC}=2R\)

=>\(2\cdot OA=48:\frac{24}{25}=50\)

=>OA=25(cm)

OA+OH=AH

=>OH=32-25=7(cm)

Đề đâu mà giúp :V

bài đâu?

1 because

2 although

3 While

4 Although

5 When

6 Although

7 if

8 while

9 although

10 because

Thanks nhá bro