12 x-7 при x=0,05 3-1,5x при x=-1/3 (2x-y ) : (x-3y - вопрос №41609111270053151323113 от anastassia7890 18.11.2020 14:32

Question

Алгебра: 12 x-7 при x=0,05 3-1,5x при x=-1/3 (2x-y ) : (x-3y ) при x-1 ,y=1/3

anastassia7890 · Answer

1) 12x-7 при x=0,05 12·0,05-7=0,06-7=-6,94

2)3-1,5x при x=-1/3
3-1,5·(-1/3)=3-(3/2)·(-1/3)=3-(-1/2)=3,5

3)(2x-y ) :(x-3y ) при x=-1 ,y=1/3

2х-у=2·(-1)-(1/3)=-2целых 1/3=-7/3
x-3у=-1-3·(1/3)=-1-1=-2

(2x-y ) :(x-3y )=(-7/3):(-2)=(7/3)·(1/2)=7/6