2piR^2H=27pir^2(10) eching

$$2 \pi R ^ { 2 } H = 27 \pi r ^ { 2 } ( 10 )$$

H uchun yechish

$$\left\{\begin{matrix}H=135\times \left(\frac{r}{R}\right)^{2}\text{, }&R\neq 0\\H\in \mathrm{R}\text{, }&r=0\text{ and }R=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\pi ni ikki tarafidan bekor qilish.

$$2R^{2}H=27r^{2}\times 10$$

270 hosil qilish uchun 27 va 10 ni ko'paytirish.

$$2R^{2}H=270r^{2}$$

Ikki tarafini 2R^{2} ga bo‘ling.

$$\frac{2R^{2}H}{2R^{2}}=\frac{270r^{2}}{2R^{2}}$$

2R^{2} ga bo'lish 2R^{2} ga ko'paytirishni bekor qiladi.

$$H=\frac{270r^{2}}{2R^{2}}$$

270r^{2} ni 2R^{2} ga bo'lish.

$$H=\frac{135r^{2}}{R^{2}}$$

R uchun yechish

$$\left\{\begin{matrix}R=3\sqrt{\frac{15}{H}}r\text{; }R=-3\sqrt{\frac{15}{H}}r\text{, }&H>0\\R=0\text{, }&r=0\text{ and }H\neq 0\\R\in \mathrm{R}\text{, }&r=0\text{ and }H=0\end{matrix}\right,$$