Răspuns:
1. [tex]\frac{1}{2}-\frac{1}{3}-\frac{1}{6}=\frac{3-2-1}{6}=\frac{0}{6}=0[/tex] (A)
2.
[tex]a=\sqrt{3}\\b=\sqrt{2}\\a^2+b^2=(\sqrt{3})^2+(\sqrt{2})^2=3+2=5[/tex] (C)
3.
[tex]x^2-4=0\\x^2=4\\x={}_+^-2[/tex]
Dat fiind ca ne cere namar natural raspunsul este 2 (B)
4. a+b=20
a-b=10
2a=30
a=15
b=20-15=5
Cel mai mare numar este 15 (C)
5. p=cazuri favorabile/cazuri posibile
cazuri favorabile={2,4,6}=3 numere
Cazuri posibile={1,2,3,4,5,6}=6 numere
p=3/6=1/2 (B)
6. linie mijlocie= (baza mare+baza mica)/2=[tex]\frac{11+9}{2}=\frac{20}{2}=10[/tex] (B)
7.
[tex]tgB=\frac{sinB}{cosB}=\frac{\frac{AC}{BC}}{\frac{AB}{BC}}[/tex]
AC=6
AB=4
[tex]BC^2=AC^2+AB^2=4^2+6^2=16+36=54\\BC^2=52\\BC=\sqrt{52}=2\sqrt{13}[/tex]
[tex]tgB=\frac{\frac{6}{2\sqrt{13}}}{\frac{4}{2\sqrt{13}}}=\frac{6}{2\sqrt{13}}*\frac{2\sqrt{13}}{4}=\frac{6}{4}=\frac{3}{2}[/tex] (B)
8.
AC=12
BD=10
AC,BD diagonalele rombului
[tex]A_{ABCD}=\frac{AC*BD}{2}=\frac{12*10}{2}=6*10=60 cm^2[/tex] (C)
9.[tex]sin30cos30-sin60cos60=\frac{1}{2}*\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}*\frac{1}{2}=0[/tex] (A)
Explicație pas cu pas:
