some - they - where - go - with

1.some
2.they
3.where
4.go
5.with

a) Ta có:

\(\left. \begin{array}{l}\left. \begin{array}{l}SA \bot SB\\SA \bot SC\end{array} \right\} \Rightarrow SA \bot \left( {SBC} \right)\\SA \subset \left( {SAB} \right)\end{array} \right\} \Rightarrow \left( {SAB} \right) \bot \left( {SBC} \right)\)

b) Ta có:

\(\left. \begin{array}{l}\left. \begin{array}{l}SA \bot SB\\SA \bot SC\end{array} \right\} \Rightarrow SA \bot \left( {SBC} \right)\\SA \subset \left( {SCA} \right)\end{array} \right\} \Rightarrow \left( {SCA} \right) \bot \left( {SBC} \right)\)

c) Ta có:

\(\left. \begin{array}{l}\left. \begin{array}{l}SA \bot SB\\SB \bot SC\end{array} \right\} \Rightarrow SB \bot \left( {SCA} \right)\\SB \subset \left( {SAB} \right)\end{array} \right\} \Rightarrow \left( {SAB} \right) \bot \left( {SCA} \right)\)

\(\begin{array}{l}\left. \begin{array}{l} + )BC \bot AB\left( {hcn\,\,ABCD} \right)\\BC \bot SA\left( {SA \bot \left( {ABCD} \right)} \right)\\AB \cap SA = \left\{ A \right\}\end{array} \right\} \Rightarrow BC \bot \left( {SAB} \right);AM \subset \left( {SAB} \right) \Rightarrow BC \bot AM\\\left. \begin{array}{l} + )CD \bot AD\left( {hcn\,\,ABCD} \right)\\CD \bot SA\left( {SA \bot \left( {ABCD} \right)} \right)\\AD \cap SA = \left\{ A \right\}\end{array} \right\} \Rightarrow CD \bot \left( {SAD} \right);AN \subset \left( {SAD} \right) \Rightarrow CD \bot AN\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l} + )AM \bot SB\\AM \bot BC\\SB \cap BC = \left\{ B \right\}\end{array} \right\} \Rightarrow AM \bot \left( {SBC} \right);SC \subset \left( {SBC} \right) \Rightarrow SC \bot AM\\\left. \begin{array}{l} + )AN \bot SD\\AN \bot CD\\SD \cap CD = \left\{ D \right\}\end{array} \right\} \Rightarrow AN \bot \left( {SCD} \right);SC \subset \left( {SCD} \right) \Rightarrow SC \bot AN\\\left. \begin{array}{l} + )AM \bot SC\\AN \bot SC\\AM \cap AN = \left\{ A \right\}\end{array} \right\} \Rightarrow SC \bot \left( {AMN} \right)\end{array}\)

a)     Vì \(AB \bot \left( {BCD} \right) \Rightarrow AB \bot CD\)

Mà \(CD \bot BC\)\( \Rightarrow CD \bot \left( {ABC} \right)\)

Lại có \(BM \in \left( {ABC} \right)\)\( \Rightarrow CD \bot BM\)

b)    Ta có \(\left. \begin{array}{l}BM \bot CD\\BM \bot AC\end{array} \right\} \Rightarrow BM \bot \left( {ACD} \right)\)

Mà \(MN \in \left( {ACD} \right) \Rightarrow BM \bot MN\)

cleaner => cleanest

Wash-Bot is cleaner than Clean-Bot, but Trash-Bot is ironically the cleaner=>cleanest

Số kg bột mì mẹ dùng là

24 x 3/8=9 (kg)

Mẹ còn lại số kg bột mì là

24-9=15 (kg)

Đáp số:9 kg và 15 kg

a 9 kg bột mì : 24 : 8 x3 = 9 

b 15 kg bột mì : 24 - 9 = 15

a)     Ta có: \(\left. \begin{array}{l}OA \bot OB\\OA \bot OC\end{array} \right\} \Rightarrow OA \bot \left( {OBC} \right)\)

Mà \(BC \in \left( {OBC} \right) \Rightarrow OA \bot BC\)

b)    Ta có \(\left. \begin{array}{l}OA \bot OB\\OB \bot OC\end{array} \right\} \Rightarrow OB \bot \left( {OAC} \right)\)

Mà \(CA \in \left( {OAC} \right) \Rightarrow CA \bot OB\)

c)     Ta có \(\left. \begin{array}{l}OC \bot OB\\OA \bot OC\end{array} \right\} \Rightarrow OC \bot \left( {OAB} \right)\)

Mà \(AB \in \left( {OAB} \right) \Rightarrow AB \bot OC\)

                                   Giải

T D B A O

Ta có : \(\widehat{BOT}+\widehat{AOT}=180^0\) ( kề bù )

Mà \(\widehat{BOT}-\widehat{AOT}=20^0\)

nên \(\widehat{BOT}="\left(180^0+20^0\right)\div2=100^0\)

Tia OD là tia phân giác của \(\widehat{BOT}\)nên \(\widehat{BOD}=\frac{100^0}{2^0}=50^0\)

suy ra \(\widehat{AOD}=180^0-50^0=130^0\)

Vậy \(\widehat{AOD}=130^0\)

\(\widehat{AOT}+\widehat{BOT}=180\)và \(\widehat{BOT}-\widehat{AOT}=20\)suy ra \(\widehat{BOT}=100\)và \(\widehat{AOT}=80\)

có OD là phân giác góc BOT suy ra \(\widehat{TOD}=\frac{\widehat{BOT}}{2}=\frac{100}{2}=50\)suy ra

\(\widehat{AOD}=\widehat{AOT}+\widehat{TOD}=80+50=130\)