Recapitulare matematică clasa a VIII-a – semestrul I

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English

Algebră și geometrie – semestrul I – clasa a VIII-a

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Question 1 of 10

1. Question
1 points
$\sqrt{3 - 2 \sqrt{2}} + \sqrt{6 - 4 \sqrt{2}} = {(\sqrt{2})}^{x} .$ Atunci $\frac{x}{\sqrt{2}} =$
- 1
- Alt răspuns
- $\frac{\sqrt{2}}{2}$
- 2
Correct

Incorrect
Question 2 of 10

2. Question
1 points
$C a r d \{x \in ℤ^{*} | | x | \leq \sqrt{101}\} =$
- 100
- 20
- 202
- 21
Correct

Incorrect
Question 3 of 10

3. Question
1 points
VABCDEF piramidă hexagonală regulată, AB=10 cm, VA=20 cm. Atunci d(A, VD)=
- 10
- $10 \sqrt{3}$
- 20
- $20 \sqrt{3}$
Correct

Incorrect
Question 4 of 10

4. Question
1 points
Dacă $a < b$ , atunci $(- \infty, b) \cap [a, b + 1) =$
- $(a, b)$
- $[a, b)$
- $(a, b]$
- $[a, b]$
Correct

Incorrect
Question 5 of 10

5. Question
1 points
${(\sqrt{7 - 4 \sqrt{3}} + \sqrt{7 + 4 \sqrt{3}})}^{2} =$
- 14
- 16
- $14 + 8 \sqrt{3}$
- 15
Correct

Incorrect
Question 6 of 10

6. Question
1 points
Numărul $a = \frac{1}{\sqrt{3} - 2} + \frac{1}{\sqrt{3} + 2}$ aparține intervalului:
- $(- \infty, π)$
- Alt răspuns
- $(- 3, 3)$
- $(π, + \infty)$
Correct

Incorrect
Question 7 of 10

7. Question
1 points
Piramida regulată are fețele laterale triunghiuri dreptunghice. Alegeți varianta incorectă:
- $A_{△ A T E} = \frac{M A^{2} \sqrt{3}}{2}$
- $m (∠ M A T) = 45^{\circ}$
- $M A ⊥ (A T E)$
- $M A ⊥ (M T E)$
Correct

Incorrect
Question 8 of 10

8. Question
1 points
Dacă $M A ⊥ (A B C)$ , AB=3 cm, BC= 5 cm, MA= 18 cm, atunci d(M, BC)=
- AB
- AC
- BC
- Celelalte răspunsuri sunt greșite
Correct

Incorrect
Question 9 of 10

9. Question
1 points
$\sqrt{n (n + 1) (n + 2) (n + 3) + 1} =$
- $n^{2} + 3 n$
- $n^{2} + 1$
- $\sqrt{{(n + 1)}^{3}}$
- $n^{2} + 3 n + 1$
Correct

Incorrect
Question 10 of 10

10. Question
1 points
ABCDEFG cub, iar I este mijlocul lui [HG]. Atunci d(M, AB)=
- BC
- AG
- $2 \sqrt{A B}$
- $A B \sqrt{2}$
Correct

Incorrect

Examples of questions from "Recapitulare matematică clasa a VIII-a – semestrul I"

Numărul $a = \frac{1}{\sqrt{3} - 2} + \frac{1}{\sqrt{3} + 2}$ aparține intervalului:
$\sqrt{3 - 2 \sqrt{2}} + \sqrt{6 - 4 \sqrt{2}} = {(\sqrt{2})}^{x} .$ Atunci $\frac{x}{\sqrt{2}} =$
Piramida regulată are fețele laterale triunghiuri dreptunghice. Alegeți varianta incorectă: