А = s''(t)
s'(t) = 10sin(t)*cos(t)
s''(t) = 10cos(2t)
cos(2t) <= 1 для любого t ==> 10cos(2t) <= 10 ==> a(max) = 10
Sсектора=П*R^2/360 *a
=> П*64/360 *80=128П/9
1) (150)^(3/2):(6^(3/2)=(150/6)^(3/2)=(25)^(3/2)=((5^2))^(3/2)=5^3=125
3)(2/3)^(-2)-(1/27)^(1/3)+3*(589)^0=(3/2)^2-∛(1/27)+3*1=9/4-1/3+3=5+1/4-1/3=5-1/12=4+11/12
4) (∛128+∛1/4):∛2=∛128/∛2+∛1/4/∛2=∛128/2+∛1/4/2=∛64+∛1/8=4+1/2
5) (12^(2/3)*(3^(7/3)):4^(-1/3)=(3^(2/3)*4^(2/3)*(3^(7/3)*4^(1/3)=27*4=108
3(^2/3)*3^(7/3)=3^(2/3+7/3)=3^(9/3)=3^3=27
4^(2/3)*4^(1/3)=4^(2/3+1/3)=4^(3/3)=4^1=4
12^(2/3)=(3*4)^(2/3)=3^(2/3)*4^(2/3)
1/(4^(-1/3)=4^(1/3)
Х-4/х-2/х+3+3-4/х=0. х-2•4/х-2/х=-6. . х-2/х-2/х=-6. х-х=-6 0=-6 уравнение не имеет корней
Ответ:
4 наверное но это не точно