Х×14×10=280
140х=280
х=280÷140
х=2
163+37х+18х=328
163+55х=328
55х=328-163
55х=165
х=165÷55
х=3
(573+а)-173=617
573-173+а=617
400+а=617
а=617-400
а=217
Запишем формулу площади круга и выразим из неё радиус:
![S= \pi r^2\\ r^2= \frac{S}{ \pi }\\ r= \sqrt{ \frac{S}{ \pi }} = \sqrt{ \frac{7,065}{3.14} } \approx \sqrt{2.25} = 1.5](https://tex.z-dn.net/?f=S%3D+%5Cpi+r%5E2%5C%5C%0Ar%5E2%3D+%5Cfrac%7BS%7D%7B+%5Cpi+%7D%5C%5C%0Ar%3D+%5Csqrt%7B+%5Cfrac%7BS%7D%7B+%5Cpi+%7D%7D++%3D+%5Csqrt%7B+%5Cfrac%7B7%2C065%7D%7B3.14%7D+%7D+%5Capprox++%5Csqrt%7B2.25%7D+%3D++1.5++)
Ответ: r = 1.5 м
1 час 10 мин=1¹/₆ часа=7/6 часа.
Пусть скорость, с которой велосипедист ехал от турбазы
до города -х. ⇒
(10/x)+10/(x+5)=7/6
10*(x+5)+10*x=(7/6)*x*(x+5)
10x+50+10x=(7/6)*(x²+5x)
20x+50=7*(x²+5x)/6 |×6
120x+300=7x²+35x
7x²-85x-300=0 D=15625 √D=125
x₁=15 x₂=-20/7 ∉
Ответ: скорость, с которой велосипедист ехал от турбазы
до города 15 км/ч.
![\cos(3x)\cos(2x) - \sin(x)\sin(6x) = \cos(7x)\\\cos(3x)\cos(2x) - \sin(x)\sin(6x) = \cos(x + 6x)\\\cos(3x)\cos(2x) - \sin(x)\sin(6x) = \cos(x)\cos(6x) - \sin(x)\sin(6x)\\\cos(3x)\cos(2x) = \cos(x)\cos(6x)\\\frac{1}{2}(\cos(x) + \cos(5x)) = \frac{1}{2}(\cos(x) + \cos(7x))\\\cos(5x) = \cos(7x)\\5x = \pm7x + 2\pi k, k \in \mathbb{Z}\\\left \{ {{x = \pi n, n \in \mathbb{Z}} \atop {x=\frac{\pi m}{6}, m \in \mathbb{Z}}} \right. \\x = \frac{\pi m}{6}, m \in \mathbb{Z}](https://tex.z-dn.net/?f=%5Ccos%283x%29%5Ccos%282x%29+-+%5Csin%28x%29%5Csin%286x%29+%3D+%5Ccos%287x%29%5C%5C%5Ccos%283x%29%5Ccos%282x%29+-+%5Csin%28x%29%5Csin%286x%29+%3D+%5Ccos%28x+%2B+6x%29%5C%5C%5Ccos%283x%29%5Ccos%282x%29+-+%5Csin%28x%29%5Csin%286x%29+%3D+%5Ccos%28x%29%5Ccos%286x%29+-+%5Csin%28x%29%5Csin%286x%29%5C%5C%5Ccos%283x%29%5Ccos%282x%29+%3D+%5Ccos%28x%29%5Ccos%286x%29%5C%5C%5Cfrac%7B1%7D%7B2%7D%28%5Ccos%28x%29+%2B+%5Ccos%285x%29%29+%3D+%5Cfrac%7B1%7D%7B2%7D%28%5Ccos%28x%29+%2B+%5Ccos%287x%29%29%5C%5C%5Ccos%285x%29+%3D+%5Ccos%287x%29%5C%5C5x+%3D+%5Cpm7x+%2B+2%5Cpi+k%2C+k+%5Cin+%5Cmathbb%7BZ%7D%5C%5C%5Cleft+%5C%7B+%7B%7Bx+%3D+%5Cpi+n%2C+n+%5Cin+%5Cmathbb%7BZ%7D%7D+%5Catop+%7Bx%3D%5Cfrac%7B%5Cpi+m%7D%7B6%7D%2C+m+%5Cin+%5Cmathbb%7BZ%7D%7D%7D+%5Cright.+%5C%5Cx+%3D+%5Cfrac%7B%5Cpi+m%7D%7B6%7D%2C+m+%5Cin+%5Cmathbb%7BZ%7D)
Корни из второго уравнения системы включают в себя корни первой.