1) 31-6=25мм
2)P=(31+25)*2=112мм
3)S=31*25=775
<span>y = x</span>²<span>(1 - 2x) = х</span>² - 2х³
y' = (х<span>² - 2х³</span>)' = 2x - 6x²
1. А(0;0), В(10;7), С(7;10) - треугольник. Найдём длины его сторон.
![AB=\sqrt{(10-0)^2+(7-0)^2}=\sqrt{100+49}=\sqrt{149}\\AC=\sqrt{(7-0)^2+(10-0)^2}=\sqrt{49+100}=\sqrt{149}\\BC=\sqrt{(7-10)^2+(10-7)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt2](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%2810-0%29%5E2%2B%287-0%29%5E2%7D%3D%5Csqrt%7B100%2B49%7D%3D%5Csqrt%7B149%7D%5C%5CAC%3D%5Csqrt%7B%287-0%29%5E2%2B%2810-0%29%5E2%7D%3D%5Csqrt%7B49%2B100%7D%3D%5Csqrt%7B149%7D%5C%5CBC%3D%5Csqrt%7B%287-10%29%5E2%2B%2810-7%29%5E2%7D%3D%5Csqrt%7B9%2B9%7D%3D%5Csqrt%7B18%7D%3D3%5Csqrt2)
AB=BC, значит треугольник равнобедренный. Его площадь равна половине произведения основания и высоты. Высота AD делит сторону BC пополам. Найдём координаты точки D, как середины отрезка
![D\left(\frac{7+10}2;\frac{10+7}2\right)=(8,5;\;8,5)](https://tex.z-dn.net/?f=D%5Cleft%28%5Cfrac%7B7%2B10%7D2%3B%5Cfrac%7B10%2B7%7D2%5Cright%29%3D%288%2C5%3B%5C%3B8%2C5%29)
Найдём длину высоты AD
![AD=\sqrt{(8,5-0)^2+(8,5-0)^2}=\sqrt{2\cdot8,5^2}=8,2\sqrt2](https://tex.z-dn.net/?f=AD%3D%5Csqrt%7B%288%2C5-0%29%5E2%2B%288%2C5-0%29%5E2%7D%3D%5Csqrt%7B2%5Ccdot8%2C5%5E2%7D%3D8%2C2%5Csqrt2)
Площадь ABC
![S=\frac12\cdot3\sqrt2\cdot8,5\sqrt2=\frac12\cdot3\cdot8,5\cdot2=25,5](https://tex.z-dn.net/?f=S%3D%5Cfrac12%5Ccdot3%5Csqrt2%5Ccdot8%2C5%5Csqrt2%3D%5Cfrac12%5Ccdot3%5Ccdot8%2C5%5Ccdot2%3D25%2C5)
кв.ед.
![2.\;\cos\frac{\pi(x-7)}3=\frac12\\\frac{\pi(x-7)}3=\frac\pi3+2\pi n,\;n\in\mathbb{Z}\\\frac\pi3(x-7)=\frac\pi3+2\pi n\\x-7=1+6n\\x=8+6n\\8+6n<0\\6n<-8\\n<-1\frac13\\n\in\mathbb{Z}\Rightarrow n\leq-2\\x=8+6\cdot(-2)=8-12=-4](https://tex.z-dn.net/?f=2.%5C%3B%5Ccos%5Cfrac%7B%5Cpi%28x-7%29%7D3%3D%5Cfrac12%5C%5C%5Cfrac%7B%5Cpi%28x-7%29%7D3%3D%5Cfrac%5Cpi3%2B2%5Cpi+n%2C%5C%3Bn%5Cin%5Cmathbb%7BZ%7D%5C%5C%5Cfrac%5Cpi3%28x-7%29%3D%5Cfrac%5Cpi3%2B2%5Cpi+n%5C%5Cx-7%3D1%2B6n%5C%5Cx%3D8%2B6n%5C%5C8%2B6n%3C0%5C%5C6n%3C-8%5C%5Cn%3C-1%5Cfrac13%5C%5Cn%5Cin%5Cmathbb%7BZ%7D%5CRightarrow+n%5Cleq-2%5C%5Cx%3D8%2B6%5Ccdot%28-2%29%3D8-12%3D-4)
![3.\;\frac{33}{4\sqrt{33}}=\frac{\sqrt{33}\cdot\sqrt{33}}{4\sqrt{33}}=\frac{\sqrt{33}}4\\tgA=\frac{BC}{AC}\\\frac{\sqrt{33}}{4}=\frac{BC}{4}\\BC=\sqrt{33}\\AB=\sqrt{BC^2+AC^2}=\sqrt{33+\frac{33}{16}}=\sqrt{\frac{528+33}{16}}=\frac{\sqrt{561}}{4}\\4.\;36\sqrt6tg\frac\pi6\sin\frac\pi4=36\sqrt6\cdot\frac1{\sqrt3}\cdot\frac1{\sqrt2}=\frac{36\sqrt6}{\sqrt6}=36](https://tex.z-dn.net/?f=3.%5C%3B%5Cfrac%7B33%7D%7B4%5Csqrt%7B33%7D%7D%3D%5Cfrac%7B%5Csqrt%7B33%7D%5Ccdot%5Csqrt%7B33%7D%7D%7B4%5Csqrt%7B33%7D%7D%3D%5Cfrac%7B%5Csqrt%7B33%7D%7D4%5C%5CtgA%3D%5Cfrac%7BBC%7D%7BAC%7D%5C%5C%5Cfrac%7B%5Csqrt%7B33%7D%7D%7B4%7D%3D%5Cfrac%7BBC%7D%7B4%7D%5C%5CBC%3D%5Csqrt%7B33%7D%5C%5CAB%3D%5Csqrt%7BBC%5E2%2BAC%5E2%7D%3D%5Csqrt%7B33%2B%5Cfrac%7B33%7D%7B16%7D%7D%3D%5Csqrt%7B%5Cfrac%7B528%2B33%7D%7B16%7D%7D%3D%5Cfrac%7B%5Csqrt%7B561%7D%7D%7B4%7D%5C%5C4.%5C%3B36%5Csqrt6tg%5Cfrac%5Cpi6%5Csin%5Cfrac%5Cpi4%3D36%5Csqrt6%5Ccdot%5Cfrac1%7B%5Csqrt3%7D%5Ccdot%5Cfrac1%7B%5Csqrt2%7D%3D%5Cfrac%7B36%5Csqrt6%7D%7B%5Csqrt6%7D%3D36)