C

40:

a: =2,5*4*20,21

=20,21*10

=202,1

b: =0,25*0,4*1,25*8

=10*0,1

=1

9:

a: =>0,4y=6

=>y=6:0,4=15

b: 105:y=7,46-3,26

=>105:y=4,2

=>y=105/4,2=25

c: y+1,59=12*2,5

=>y+1,59=30

=>y=28,41

d: 3,75:y=3:4

=>3,75:y=0,75

=>y=3,75:0,75=5

\(VD5\)

\(\overrightarrow{AB}=\left(2;6\right)\\ =>AB=\sqrt{2^2+6^2}=2\sqrt{10}\\ \overrightarrow{AC}=\left(15;5\right)\\=>AC=\sqrt{15^2+5^2}=5\sqrt{10}\)

Giả sử \(AD\) là tia phân giác của góc \(A\)

\(\Rightarrow\dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac{2\sqrt{10}}{5\sqrt{10}}=\dfrac{2}{5}\)

mà \(\overrightarrow{BD},\overrightarrow{DC}\) cùng phương

\(\Rightarrow\overrightarrow{BD}=\dfrac{2}{5}\overrightarrow{DC}\)

Giả sử điểm \(D(x;y)\)

\(=>\overrightarrow{BD}=\left(x+4;y-3\right)\\ \dfrac{2}{3}\overrightarrow{DC}=\dfrac{2}{3}.\left(9-x;2-y\right)=\left(6-9x;\dfrac{4}{3}-\dfrac{2}{3}y\right)\)

\(\Rightarrow\left\{{}\begin{matrix}x+4=6-9x\\y-3=\dfrac{4}{3}-\dfrac{2}{3}y\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=\dfrac{13}{5}\end{matrix}\right.\\ =>D\left(\dfrac{1}{5};\dfrac{13}{5}\right)\)

\(\Rightarrow\overrightarrow{AD}=\left(\dfrac{31}{5};\dfrac{28}{5}\right)\) là vt chỉ phương

\(\Rightarrow\overrightarrow{n}=\left(-\dfrac{28}{5};\dfrac{31}{5}\right)\) là vt pháp tuyến

Đường thẳng \(AD\) đi qua \(A\left(-6;-3\right)\) và nhận \(\overrightarrow{n}\) là vt pháp tuyến có pt

\(-\dfrac{28}{5}\left(x+6\right)+\dfrac{31}{5}\left(y+3\right)=0\\ =>-\dfrac{28}{5}x+\dfrac{31}{5}y-15=0\)

tam giác ABC có AD là tia phân giác :

định lí talet ta có :\(\dfrac{AB}{AC}\)=\(\dfrac{BD}{DC}\)=\(\dfrac{2}{5}\)

mà D nằm trên đoạn BD => \(\overrightarrow{BD}\)=\(\dfrac{2}{5}\)\(\overrightarrow{DC}\)

BD = ( x + 4; y- 3 )  = \(\dfrac{2}{5}\)DC = \(\dfrac{2}{5}\)( 9- x ; 2 - y)

\(\left\{{}\begin{matrix}x+4=\dfrac{18}{5}-\dfrac{2}{5}x\\y-3=\dfrac{4}{5}-\dfrac{2}{5}y\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=-\dfrac{2}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)

→vecto \(\overrightarrow{u}\)=\(\overrightarrow{AD}\) = (\(\dfrac{40}{7}\);\(\dfrac{40}{7}\))

→\(\overrightarrow{n}\)=(-\(\dfrac{40}{7}\);\(\dfrac{40}{7}\))

→phương trình đường thẳng là :

\(\dfrac{-40}{7}\)x + \(\dfrac{40}{7}\)y - \(\dfrac{120}{7}\) = 0

↔-x + y -3 = 0 

a: Xét ΔAHE có 

AM là đường cao

AM là đường trung tuyến

DO đó: ΔAHE cân tại A

hay AH=AE

b: Xét ΔAKI và ΔAHI có

AK=AH

\(\widehat{KAI}=\widehat{HAI}\)

AI chung

Do đó: ΔAKI=ΔAHI

Suy ra: \(\widehat{AKI}=\widehat{AHI}=90^0\)

hay IK//AB

c: Ta có: IK=IH

mà IK<IC

nên IH<IC

1. are visiting
2. don't go
3. are not going to work
4. Does Mary have
5. meet

are visting

don't go

won't work

Does Mary have

meet

1. are visiting 

2. don't go

3. aren't going to work

4. Does Mary have

5. meet

\(\left(1\right)FeCl_2+2NaOH\to Fe\left(OH\right)_2\downarrow +2NaCl\\ \left(2\right)Fe\left(OH\right)_2\xrightarrow{t^0}FeO+H_2O\\ \left(3\right)FeO+C\to Fe+CO\uparrow \\ \left(4\right)2Fe+3Cl_2\xrightarrow[]{t^o}2FeCl_3\\ \left(5\right)FeCl_3+3NaOH\to Fe\left(OH\right)_3\downarrow +3NaCl\\ \left(6\right)2Fe\left(OH\right)_3\xrightarrow{t^o}Fe_2O_3+3H_2O\\ \left(7\right)Fe_2O_3+3CO\xrightarrow{t^o}2Fe+3CO_2\\ \left(8\right)Fe+2HCl\to FeCl_2+H_2\)