1.
35 : (-7) = - 5
(-5,2) : (- 0,4) = 13
(-2,613) : (-0,26) = 10,05
-5/12 : 4 = -5/12 * 1/4 = -5/48
2.
-27/8 = -3 3/8
-39/ -11 = 3 6/11
129/-5 = -25 4/5
-46/13 = -3 7/13
51/27 = 1 24/27
3.
3 2/5 = 17/5
-(2 7/8) = -23/8
-14 6/11 = - 160/11
35 1/4 = 141/4
4.
-2 0 5
----------!-----!-----!-----!-----!-----!-----!-----!-----------
[ -2 ; 5]
Применены : формула объёма цилиндра, свойства правильного треугольника
1)100:2=50(т)
2)50+9,8=59,8(т) - на 2-ом.
3)50-9,8=40,2(т) - на 1-ом.
4tg4x*1/2*sinx*cosx-tg4x=0
2tg4x*1/2sin2x-tg4x=0
tg4x*sin2x-tg4x=0
tg4x*(sin2x-1)=0
tg4x=0⇒4x=πn⇒x=πn/4,n∈z
0≤πn/4≤π
0≤n≤4
n=0⇒x=0
n=1⇒x=π/4
n=2⇒x=π/2
n=3⇒x=3π/4
n=4⇒x=π
sin2x-1=0
sin2x=1⇒2x=π/2+2πk⇒x=π/4+πk
0≤π/4+πk≤π
0≤1+4k≤4
-1≤4k≤3
-1/4≤k≤3/4
k=0⇒x=π/4
2 * cos(x)^2 + (2 - sqrt(2)) * sin(x) + sqrt(2) - 2 = 0
cos(x)^2 = 1 - sin(x)^2
sin(x) = t
2 * (1 - t^2) + (2 - sqrt(2)) * t + sqrt(2) - 2 = 0
2 - 2*t^2 + 2t - sqrt(2)*t + sqrt(2) - 2 = 0
-2*t^2 + t*(2 - sqrt(2)) + sqrt(2) = 0
2*t^2 - t*(2 - sqrt(2)) - sqrt(2) = 0
t1,2 = (sqrt(2) - 2 +- sqrt(6 - 4sqrt(2) + 8sqrt(2)))/4
t1,2 = (sqrt(2) - 2 +- sqrt(6 + 4sqrt(2))/4
sqrt(6 + 4sqrt(2)) = 2 + sqrt(2)
t1,2 = (sqrt(2) - 2 +- (2 + sqrt(2)))/4
t1 = 2sqrt(2)/4 = sqrt(2)/2
t2 = -1
sin(x) = -1
x = 3p/2 + 2*pi*k
sin(x) = sqrt(2)/2
x = pi/4 + 2*pi*k
x = 5*pi/4 + 2*pi*k
Ответ:
x = 3p/2 + 2*pi*k
x = pi/4 + 2*pi*k
x = 5*pi/4 + 2*pi*k
k - любое целое