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dan4s
@dan4s
July 2022
1
17
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Срочно помогите. 4sin^2 ( 120)° - 2cos(600)° + квадратный корень из 27 * tg (660)°
tg (660)° - не под корнем!!!
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mymurkin
4sin²120-2cos(600-360*2)+√27*tg(660-360*2)=4sin²120-2cos(-120)+3√3tg(-60)=4*(√3/2)²-2*1/2-3√3*√3=4*3/4-1-9=3-1-9=-7
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