Datrys T=2pisqrt{l/98}quad[2]

$$T = 2 \pi \sqrt { \frac { l } { 98 } } \quad [ 2 ]$$

Datrys ar gyfer l

$$l=\frac{49\times \left(\frac{T}{\pi }\right)^{2}}{8}$$

$$T\geq 0$$

Camau ar gyfer Datrys Hafaliadau Radical

Lluosi 2 a 2 i gael 4.

$$T=4\pi \sqrt{\frac{l}{98}}$$

Cyfnewidiwch yr ochrau fel bod yr holl dermau newidiol ar yr ochr chwith.

$$4\pi \sqrt{\frac{l}{98}}=T$$

Rhannu’r ddwy ochr â 4\pi .

$$\frac{4\pi \sqrt{\frac{1}{98}l}}{4\pi }=\frac{T}{4\pi }$$

Mae rhannu â 4\pi yn dad-wneud lluosi â 4\pi .

$$\sqrt{\frac{1}{98}l}=\frac{T}{4\pi }$$

Sgwariwch ddwy ochr yr hafaliad.

$$\frac{1}{98}l=\frac{T^{2}}{16\pi ^{2}}$$

Lluosi’r ddwy ochr â 98.

$$\frac{\frac{1}{98}l}{\frac{1}{98}}=\frac{T^{2}}{\frac{1}{98}\times 16\pi ^{2}}$$

Mae rhannu â \frac{1}{98} yn dad-wneud lluosi â \frac{1}{98}.

$$l=\frac{T^{2}}{\frac{1}{98}\times 16\pi ^{2}}$$

Rhannwch \frac{T^{2}}{16\pi ^{2}} â \frac{1}{98} drwy luosi \frac{T^{2}}{16\pi ^{2}} â chilydd \frac{1}{98}.

$$l=\frac{49T^{2}}{8\pi ^{2}}$$

Datrys ar gyfer T

$$T=\frac{2\pi \sqrt{2l}}{7}$$

$$l\geq 0$$