Ответ:
Геометрмческие построения . Деление отрезка пополам.
C^5+c^8/c^5+3
c^5+c^8/c^8
c^5/1
0,2^5=0.00032
A² - x² + 4x - 4 = a² - (x² - 4x + 4) = a² - (x - 2)² = (a - (x - 2))·(a + (x - 2)) =
= (a - x + 2)·(a + x - 2)
![180\textdegree<\alpha <270\textdegree~~~\Rightarrow ~~~90\textdegree<\dfrac {\alpha}2 <135\textdegree](https://tex.z-dn.net/?f=180%5Ctextdegree%3C%5Calpha+%3C270%5Ctextdegree~~~%5CRightarrow+~~~90%5Ctextdegree%3C%5Cdfrac+%7B%5Calpha%7D2+%3C135%5Ctextdegree)
Угол <em>α/2</em> во второй четверти, значит, значение косинуса отрицательное.
![\cos^2 \dfrac{\alpha }2 = \dfrac{1+\cos \alpha }2=\dfrac{1-\frac45}2=\dfrac1{10}\\\\ \cos \dfrac {\alpha }2 = -\sqrt {\dfrac 1{10}}](https://tex.z-dn.net/?f=%5Ccos%5E2+%5Cdfrac%7B%5Calpha+%7D2+%3D+%5Cdfrac%7B1%2B%5Ccos+%5Calpha+%7D2%3D%5Cdfrac%7B1-%5Cfrac45%7D2%3D%5Cdfrac1%7B10%7D%5C%5C%5C%5C+%5Ccos+%5Cdfrac+%7B%5Calpha+%7D2+%3D+-%5Csqrt+%7B%5Cdfrac+1%7B10%7D%7D)
Ответ: ![\boldsymbol {\cos \dfrac {\alpha }2 = -\sqrt {0,1}}](https://tex.z-dn.net/?f=%5Cboldsymbol+%7B%5Ccos+%5Cdfrac+%7B%5Calpha+%7D2+%3D+-%5Csqrt+%7B0%2C1%7D%7D)
D=a3-a2=0.7,
S13= ((2a1+d(n-1))/2)*n, тогда
S13= ((2*(-0.5)+0.7*12)/2)*13= 3.7*13=48.1