βΟΒ²Γ45-71/7+β^Β²43*(912Β²)
$1 + \frac{\tan(20^\circ)}{\cot(65,5^\circ)}$
limxβ2β
(x β 1)(x β 2) \ (x + 1) =
(x+4)Β²=32+8x
0.24 cubed multiplied by 10 to the power of negative 3
find the alue of p+4 when p =5
-
5
factorize 6x^2-7x+2
618,426,157,909,362,144,706,560 + 1,209,045,452,278,921,297,920 + 1,182,704,543,241,134,080 + 1,156,938,030,508,920 =
βΉβββ β β΄βββ =
$m^{2}$ =m+13
