1) 150,2* 15:100=22,53
2)4-12в
3)2^2-4a+4
4)2x^y+4
5)a(x-y+2)
Найдем пересечение линий:
9/x^2 = -4x+13
![x \neq 0 \\ 4x^{3}-13 x^{2} +9=0 \\ x=1 \\ 4x^{3}-13 x^{2} +9 / (x-1)=4x^2-9x-9 \\ 4x^2-9x-9=0 \\ D=81-4*4*(-9)=225 \\ x_{1}=3 \\ x_{2}=- \frac{3}{4}](https://tex.z-dn.net/?f=x+%5Cneq+0+%5C%5C+4x%5E%7B3%7D-13+x%5E%7B2%7D++%2B9%3D0+%5C%5C+x%3D1+%5C%5C+4x%5E%7B3%7D-13+x%5E%7B2%7D++%2B9+%2F+%28x-1%29%3D4x%5E2-9x-9+%5C%5C+4x%5E2-9x-9%3D0+%5C%5C+D%3D81-4%2A4%2A%28-9%29%3D225+%5C%5C+x_%7B1%7D%3D3+%5C%5C++x_%7B2%7D%3D-+%5Cfrac%7B3%7D%7B4%7D+++)
Т.к. нас интересует первая четверть, то подходят две абциссы
x1=1 и х2=3
Далее используем интеграл )
![\int\limits^3_1 {-4x+13 - \frac{9}{x^2} } \, dx =(-2x^2+13x+ \frac{9}{x})= \\ (-18+39+3)-(-2+13+9)=24-20=4](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E3_1+%7B-4x%2B13+-++%5Cfrac%7B9%7D%7Bx%5E2%7D+%7D+%5C%2C+dx+%3D%28-2x%5E2%2B13x%2B+%5Cfrac%7B9%7D%7Bx%7D%29%3D+%5C%5C+%28-18%2B39%2B3%29-%28-2%2B13%2B9%29%3D24-20%3D4+)
Ответ: 4
Y ' - y*ctgx = 0
y' = dy/dx ==>
dy/dx = y*ctgx
dy/y = ctgx*dx
∫ dy/y = <span>∫ ctgx*dx
</span>ln y = ln (sinx)
y = sinx + C