1) -0.5
2)90
3)-0.2-90=-90.5
4)-90.5*4.25= -384.625
5)0.074*0.5=0.037
6)-384.625+0.037= -384.588
Трекгольник ABC
AB=13
BC=7
AC-?
AB^2=BC^2+AC^2
13^2=7^2+AC
AC^2=AB^2-BC^2
AC^2=169-49
AC=корень из 120
Обозначим за половину отрезка AB x.
![x = \frac{1}{2}AB](https://tex.z-dn.net/?f=+x+%3D+%5Cfrac%7B1%7D%7B2%7DAB+)
Тогда:
![x - 9 = \frac{1}{4} * 2x ](https://tex.z-dn.net/?f=x+-+9+%3D++%5Cfrac%7B1%7D%7B4%7D+%2A+2x%0A)
, сократим в правой части на 2.
![x - 9 = \frac{1}{2}x](https://tex.z-dn.net/?f=x+-+9+%3D++%5Cfrac%7B1%7D%7B2%7Dx+)
, разделим всё выражение на 2.
![2x - 18 = x](https://tex.z-dn.net/?f=2x+-+18+%3D+x)
![x = 18](https://tex.z-dn.net/?f=x+%3D++18)
Весь путь от A до B:
2 * x = 2 * 18 = 36 км.
Х^2(Х-у)+х(Х-у)=(Х-у)(х^2+Х)
Ускорение - это вторая производная от пути по времени.
x(t) = 3t² + 9lnt + 7
![x'(t)=3(t^{2})'+9(lnt)'+7'=6t+9*\frac{1}{t}=6t+\frac{9}{t}\\\\x''(t)=6(t)'+9(\frac{1}{t})'=6-\frac{9}{t^{2}}\\\\6-\frac{9}{t^{2} }=2\\\\\frac{9}{t^{2}}=4\\\\t^{2}=\frac{9}{4}\\\\t=1,5](https://tex.z-dn.net/?f=x%27%28t%29%3D3%28t%5E%7B2%7D%29%27%2B9%28lnt%29%27%2B7%27%3D6t%2B9%2A%5Cfrac%7B1%7D%7Bt%7D%3D6t%2B%5Cfrac%7B9%7D%7Bt%7D%5C%5C%5C%5Cx%27%27%28t%29%3D6%28t%29%27%2B9%28%5Cfrac%7B1%7D%7Bt%7D%29%27%3D6-%5Cfrac%7B9%7D%7Bt%5E%7B2%7D%7D%5C%5C%5C%5C6-%5Cfrac%7B9%7D%7Bt%5E%7B2%7D+%7D%3D2%5C%5C%5C%5C%5Cfrac%7B9%7D%7Bt%5E%7B2%7D%7D%3D4%5C%5C%5C%5Ct%5E%7B2%7D%3D%5Cfrac%7B9%7D%7B4%7D%5C%5C%5C%5Ct%3D1%2C5)