vectori

vectori

9th - 12th Grade

9 Qs

quiz-placeholder

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vectori

vectori

Assessment

Quiz

Mathematics

9th - 12th Grade

Practice Problem

Hard

Created by

Doriana Dorca

Used 14+ times

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9 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Fie ABC un triunghi oarecare. Dacă M este mijlocul laturii BC, atunci: 

 AM=12(AB+BC)\overrightarrow{AM}=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{BC}\right)  

 AM=12(AC+CB)\overrightarrow{AM}=\frac{1}{2}\left(\overrightarrow{AC}+\overrightarrow{CB}\right)  

 AM=12(AB+AC)\overrightarrow{AM}=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)  

 AM=12(BM+AC)\overrightarrow{AM}=\frac{1}{2}\left(\overrightarrow{B}M+\overrightarrow{AC}\right)  

2.

FILL IN THE BLANK QUESTION

1 min • 1 pt

 Fie ABC  un triunghi oarecare. Atunci 

 AB+BC=\overrightarrow{AB}+\overrightarrow{BC}=  

3.

FILL IN THE BLANK QUESTION

1 min • 1 pt

 Fie ABCD un paralelogram. Atunci   CD+CB=\overrightarrow{CD}+\overrightarrow{CB}=  


4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Daca  ABCD este un paralelogram cu  ACBD=OAC\cap BD=O atunci  BO+CO=BC\overrightarrow{BO}+\overrightarrow{CO}=\overrightarrow{BC}   

A

F

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Daca  ABCD este un patrat de latura 2 , atunci modulul vectorului    AB+AC\overrightarrow{AB}+\overrightarrow{AC}  este  5\sqrt{5}  .

A

F

6.

MULTIPLE SELECT QUESTION

45 sec • 1 pt

Se dau vectorii   v1=2i3j\ \overrightarrow{v_1}=2\overrightarrow{i}-3\overrightarrow{j}  și  v2=i+2j\overrightarrow{v_2}=-\overrightarrow{i}+2\overrightarrow{j}  

 v1+v2=i+j\overrightarrow{v_1}+\overrightarrow{v_2}=\overrightarrow{i}+\overrightarrow{j}  

 v1=5\left|\overrightarrow{v_1}\right|=5  

 (v2)=i2j\overline{\left(\overrightarrow{v_2}\right)}=-i-2\overrightarrow{j}  

 3v14v2 3\overrightarrow{v_1}-4\overrightarrow{v_2}\   are coordonatele (10,17)\left(10,-17\right)  

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Se considera patrulaterul ABCD în care punctele E si F sunt mijloacele diagonalelor AC și respectiv BD. Atunci AB+AD+CB+CD=\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}=  

 4EF4\overrightarrow{EF}  

 4FE4\overrightarrow{FE}  

 2FE2\overrightarrow{FE}  

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