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Danilove5
Danilove5
25.04.2020 05:36 •  Алгебра

7. Вычислите площадь фигуры, ограниченную линиями ,y=x^2 y=2-x

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Ответ:
Lara138009
Lara138009
11.09.2020 18:11

The towel on the wall… It’s our ancient custom. There was not a single Ukrainian house , which would not be decorates with towels. A house without a towel was compared with a family without children.  A towel symbolized not only aesthetic taste , it was a face of a house and of a hostess. The towels proved that the hostess was tidy and hard-working.  

A newly-born child was handed with the towel, guests were met with bread and salt on the towel. A son , a husband, a father were seen off with the towel.  While getting married the  newly-formed family was standing on a towel too. Bread and salt on an embroidered towel were a symbol of hospitality of the Ukrainian people. To take the towel , to kiss bread symbolized unity, peace, love , understanding among people. This custom has a good tradition in our country.

Towel were used in different ceremonies and in everyday life as well: for wiping a face and hands, for drying dishes, for decorating the houses, for presenting dear people. The ornaments and colours of towels were different in different regions. A lot of songs about the embroidered towel have been created in Ukraine. Объяснение:

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Ответ:
FJFJKD99
FJFJKD99
23.01.2023 10:11
1)
sin2x -cos²x =0 ;
2sinx*cosx -cos²x =0 ;
cosx(2sinx-cosx) =0 ;
[cosx =0 ; 2sinx-cosx=0⇒x=π/2+πn , x =arcctq2 ; n∈Z.

2) 
cos2x +cos²x =0 ;
cos²x - sin²x+cos²x =0 ;
sin²x =0 ⇒sinx =0 ;
x =πn , n∈Z.

3).
2cos⁴x+3cos²x-2=0 ;
* * * замена переменной  t = cos²x ; 0≤ t ≤ 1 * * *
2t²+3t-2=0 ; * * * D =3² -4*2*(-2) =25 =5² * * *
t₁ = (-3 -5)/4 = -2  не удов. 0≤ t ≤ 1.
t₂ =(-3+5)/4 =1/2⇒cos²x =1/2⇔(1+cos2x)/2 =1/2⇔cos2x=0 ⇒
2x =π/2+ πn , n∈Z ;
x = π/4+ (π/2)*n , n∈Z.

4).
2cos²x+5sinx-4=0 ;
2(1-sin²x)+5sinx-4=0 ;
2sin²x-5sinx+2=0  ;  * * * D =5² -4*2*2 =25 =3² * * *
sinx = (5+3)/4 =2  не умеет решения ;
sinx = (5-3)/4 =1/2 ⇒    x =(-1)^n *(π/6) + πn , n∈Z .

5).   2cos^2x(3p/2-x)-5sin(p/2-x)-4=0 ;
2cos²(3π/2-x)-5sin(π/2-x)-4=0 ;
2sin²x -5cosx -4 = 0 ;
2(1-cos²x) -5cosx -4 = 0 ;
2cos²x +5cosx +2 = 0 ; * * *D =5² -4*2*2 =25 =3²  * * *
cos²x +(2+1/2)cosx +1 = 0 ⇒[cosx =2 ; cosx =1/2 .
cosx =1/2 ;

x =±π/3 +2πn , n∈Z .
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