РЕШЕНИЕ
1)
Полная длина окружности по формуле
L = 2*π*R = 20π
Полный угол окружности = 360°
Длина дуги в α = 150° из пропорции
Lд(150) = L*α/360 = 20*π*(5/12) = (8 1/3)*π ≈ 8.33*3.14 ≈ 26.16 см - длина дуги - ОТВЕТ
2)
Площадь круга по формуле
S = π*R² = 100π
Площадь сектора по формуле
Sc(150) = α/360*π*R² = 5/12*100*π = (41 2/3)*π ≈129.46 см² площадь сектора - ОТВЕТ
![6cos^2x-7cosx-5=0](https://tex.z-dn.net/?f=6cos%5E2x-7cosx-5%3D0+)
делаем замену:
![a=cosx,\ a \in [-1;1]](https://tex.z-dn.net/?f=a%3Dcosx%2C%5C+a+%5Cin+%5B-1%3B1%5D)
получим:
![6a^2-7a-5=0](https://tex.z-dn.net/?f=6a%5E2-7a-5%3D0)
решим это уравнение:
![D=49+20*6=169=13^2 \\a_1= \frac{7+13}{12} = \frac{20}{12} \notin [-1;1] \\a_2= \frac{7-13}{12} =- \frac{6}{12} =- \frac{1}{2} \in [-1;1]](https://tex.z-dn.net/?f=D%3D49%2B20%2A6%3D169%3D13%5E2%0A%5C%5Ca_1%3D+%5Cfrac%7B7%2B13%7D%7B12%7D+%3D+%5Cfrac%7B20%7D%7B12%7D+%5Cnotin+%5B-1%3B1%5D%0A%5C%5Ca_2%3D+%5Cfrac%7B7-13%7D%7B12%7D+%3D-+%5Cfrac%7B6%7D%7B12%7D+%3D-+%5Cfrac%7B1%7D%7B2%7D+%5Cin+%5B-1%3B1%5D)
обратная замена:
![cosx=- \frac{1}{2} \\x_{1,2}=\pm arccos(- \frac{1}{2})+2\pi n \\x_{1,2}=\pm \frac{2\pi}{3}+2\pi n,\ n \in Z](https://tex.z-dn.net/?f=cosx%3D-+%5Cfrac%7B1%7D%7B2%7D%0A%5C%5Cx_%7B1%2C2%7D%3D%5Cpm+arccos%28-+%5Cfrac%7B1%7D%7B2%7D%29%2B2%5Cpi+n%0A%5C%5Cx_%7B1%2C2%7D%3D%5Cpm++%5Cfrac%7B2%5Cpi%7D%7B3%7D%2B2%5Cpi+n%2C%5C+n+%5Cin+Z+)
проводим отбор корней на промежутке
![[-\pi;2\pi]](https://tex.z-dn.net/?f=%5B-%5Cpi%3B2%5Cpi%5D)
Для этого решим следующие неравенства, при условии что n целое число:
![-\pi\leq \frac{2\pi}{3}+2\pi n\leq 2\pi \\-1\leq \frac{2}{3}+2n\leq 2 \\-1- \frac{2}{3} \leq 2n\leq 2- \frac{2}{3} \\- \frac{5}{3} \leq 2n \leq \frac{4}{3} \\- \frac{5}{6} \leq n \leq \frac{2}{3} \\n=0;\ x_1= \frac{2\pi}{3} +2\pi *0= \frac{2\pi}{3} \\-\pi\leq -\frac{2\pi}{3}+2\pi n\leq 2\pi \\-1\leq - \frac{2}{3}+2n\leq 2 \\ \frac{2}{3} -1 \leq 2n\leq 2+ \frac{2}{3} \\- \frac{1}{3} \leq 2n \leq \frac{8}{3} \\- \frac{1}{6} \leq n \leq \frac{4}{3}](https://tex.z-dn.net/?f=-%5Cpi%5Cleq+%5Cfrac%7B2%5Cpi%7D%7B3%7D%2B2%5Cpi+n%5Cleq+2%5Cpi%0A%5C%5C-1%5Cleq++%5Cfrac%7B2%7D%7B3%7D%2B2n%5Cleq+2%0A%5C%5C-1-+%5Cfrac%7B2%7D%7B3%7D++%5Cleq+2n%5Cleq+2-+%5Cfrac%7B2%7D%7B3%7D+%0A%5C%5C-+%5Cfrac%7B5%7D%7B3%7D+%5Cleq+2n+%5Cleq++%5Cfrac%7B4%7D%7B3%7D+%0A%5C%5C-+%5Cfrac%7B5%7D%7B6%7D+%5Cleq+n+%5Cleq+%5Cfrac%7B2%7D%7B3%7D+%0A%5C%5Cn%3D0%3B%5C+x_1%3D+%5Cfrac%7B2%5Cpi%7D%7B3%7D+%2B2%5Cpi+%2A0%3D+%5Cfrac%7B2%5Cpi%7D%7B3%7D%0A%5C%5C-%5Cpi%5Cleq+-%5Cfrac%7B2%5Cpi%7D%7B3%7D%2B2%5Cpi+n%5Cleq+2%5Cpi%0A%5C%5C-1%5Cleq+-+%5Cfrac%7B2%7D%7B3%7D%2B2n%5Cleq+2+%0A%5C%5C+%5Cfrac%7B2%7D%7B3%7D+-1+%5Cleq+2n%5Cleq+2%2B+%5Cfrac%7B2%7D%7B3%7D+%0A%5C%5C-+%5Cfrac%7B1%7D%7B3%7D+%5Cleq+2n+%5Cleq++%5Cfrac%7B8%7D%7B3%7D+%0A%5C%5C-+%5Cfrac%7B1%7D%7B6%7D+%5Cleq+n+%5Cleq++%5Cfrac%7B4%7D%7B3%7D+)
![n=0;\ x_2= -\frac{2\pi}{3} +2\pi *0=-\frac{2\pi}{3} \\n=1;\ x_3= -\frac{2\pi}{3} +2\pi= \frac{6\pi-2\pi}{3}= \frac{4\pi}{3}](https://tex.z-dn.net/?f=n%3D0%3B%5C+x_2%3D+-%5Cfrac%7B2%5Cpi%7D%7B3%7D+%2B2%5Cpi+%2A0%3D-%5Cfrac%7B2%5Cpi%7D%7B3%7D%0A%5C%5Cn%3D1%3B%5C+x_3%3D+-%5Cfrac%7B2%5Cpi%7D%7B3%7D+%2B2%5Cpi%3D+%5Cfrac%7B6%5Cpi-2%5Cpi%7D%7B3%7D%3D+%5Cfrac%7B4%5Cpi%7D%7B3%7D++)
Ответ: корни уравнения:
![x_{1,2}=\pm \frac{2\pi}{3}+2\pi n,\ n \in Z](https://tex.z-dn.net/?f=x_%7B1%2C2%7D%3D%5Cpm+%5Cfrac%7B2%5Cpi%7D%7B3%7D%2B2%5Cpi+n%2C%5C+n+%5Cin+Z)
; корни на отрезке:
![\pm \frac{2\pi}{3} ;\ \frac{4\pi}{3}](https://tex.z-dn.net/?f=%5Cpm++%5Cfrac%7B2%5Cpi%7D%7B3%7D+%3B%5C++%5Cfrac%7B4%5Cpi%7D%7B3%7D+)
1)42:2*10=210см
30,8:2*10=154кг
6:2*10=30 литров
2)9:3*4=12см
67,5:3*4= 89,2
18:3*4=24
2^(2х-3)•(2^2+2+1)=448
2^(2х-3)•7=448
2^(2х-3)=448:7
2^(2х-3)=64
2^(2х-3)=2^6
2х-3=6
2х=6+3
2х=9
Х=9:2=4,5