205÷5=41 (км/ч)-против течения реки
41+3=44 (км/ч)-по течению реки
44×4=176(км)-за 4 часа
Ответ: 383
В столбик мама подскажет как
A) y=4x*e^2x
y' = 4*(x*e^2x)' =
4*( x' * e^2x + x*(e^2x)' ) =
4*( 1 * e^2x + x*(e^2x)*(2x)' ) =
4*( e^2x + x*(e^2x)*2 ) =
4 * e^2x + x*(e^2x)*8 =
4 * e^2x + 8 * x * e^2x
Б) y=ln (tg^2 (x/6))
y' = 1/tg^2 (x/6) * (tg^2 (x/6))' =
1/tg^2 (x/6) * 2*(tg^(2-1) (x/6)) * (tg (x/6))' =
1/tg^2 (x/6) * 2*(tg^1 (x/6)) * (tg (x/6))' =
1/tg^2 (x/6) * 2*(tg (x/6)) * (1/cos^2 (x/6)) * (x/6)' =
1/tg (x/6) * 2 * (1/cos^2 (x/6)) * (x/6)' =
1/tg (x/6) * 2 * (1/(cos^2 (x/6)) ) * (1/6) =
(2/6) * 1/tg (x/6) * (1/(cos^2 (x/6)) )=
(1/3) * 1/tg (x/6) * (1/(cos^2 (x/6)) )=
(1/3) * (cos (x/6)/sin (x/6)) * (1/(cos^2 (x/6)) )=
(1/3) * (1/sin (x/6)) * (1/(cos (x/6)) )=
(1/3) * (1/( (sin (x/6)) * (cos (x/6)) ) )=
(1/3) * ( 1/( (1/2)*sin(2x/6) ) ) =
(1/3) * (2/(sin(2x/6) ) ) =
(1/3) * (2/(sin(x/3) ) ) =
(2/3) * (1/(sin(x/3) ) ) =
2 / (3*sin(x/3))
(11/24+1/6)×1 3/5=(11/24+ 4/24)×8/5=15/24×8/5=3/3=1