Math9#1

Math9#1

Assessment

Quiz

Created by

Маржан Жаксиликова

Mathematics

9th Grade

2 plays

Hard

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30 questions

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1.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Жазыңқы бұрыш: 

 α=90°\alpha=90\degree  

 0°<α<90°0\degree<\alpha<90\degree  

 α=360°\alpha=360\degree  

 90°<α<180°90\degree<\alpha<180\degree  

 α=180°\alpha=180\degree  

2.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Сүйір бұрыш: 

 α=90°\alpha=90\degree  

 0°<α<90°0\degree<\alpha<90\degree  

 α=360°\alpha=360\degree  

 90°<α<180°90\degree<\alpha<180\degree  

 α=180°\alpha=180\degree  

3.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Тік бұрыш: 

 α=90°\alpha=90\degree  

 0°<α<90°0\degree<\alpha<90\degree  

 α=360°\alpha=360\degree  

 90°<α<180°90\degree<\alpha<180\degree  

 α=180°\alpha=180\degree  

4.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Доғал бұрыш: 

 α=90°\alpha=90\degree  

 0°<α<90°0\degree<\alpha<90\degree  

 α=360°\alpha=360\degree  

 90°<α<180°90\degree<\alpha<180\degree  

 α=180°\alpha=180\degree  

5.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Толық бұрыш: 

 α=90°\alpha=90\degree  

 0°<α<90°0\degree<\alpha<90\degree  

 α=360°\alpha=360\degree  

 90°<α<180°90\degree<\alpha<180\degree  

 α=180°\alpha=180\degree  

6.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Қосындысы 180°180\degree-  қа тең бұрыштар . . . деп аталады. 

Сыбайлас

Вертикаль

Тік

Доғал

Толық

7.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Үшбұрыштың ішкі бұрыштарының қосындысы:

360°360\degree

90°90\degree

180°180\degree

270°270\degree

0°0\degree

8.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Үшбұрыштың сыртқы бұрыштарының қосындысы:

360°360\degree

90°90\degree

180°180\degree

270°270\degree

0°0\degree

9.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Косинустар теоремы:

c2=a2b22abcosγc^2=a^2-b^2-2ab\cos\gamma

c2=a2+b2+2abcosγc^2=a^2+b^2+2ab\cos\gamma

c2=a2+b2abcosαc^2=a^2+b^2-ab\cos\alpha

c2=a2+b2+abcosγc^2=a^2+b^2+ab\cos\gamma

c2=a2+b22abcosγc^2=a^2+b^2-2ab\cos\gamma

10.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Синустар теоремасы:

asinα=R\frac{\ a}{\sin\alpha}=R

bsinβ=2R\frac{b}{\sin\beta}=2R

asinα=bsin(90°β)\frac{a}{\sin\alpha}=\frac{b}{\sin\left(90\degree-\beta\right)}

csin(180°+α)=2R\frac{c}{\sin\left(180\degree+\alpha\right)}=2R

acos(90°+α)=csinγ\frac{a}{\cos\left(90\degree+\alpha\right)}=\frac{c}{\sin\gamma}

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