Question Number 220286 by Tawa11 last updated on 10/May/25 Commented by Tawa11 last updated on 10/May/25 $$\mathrm{perimeter}\:\mathrm{of}\:\mathrm{semi}\:\mathrm{circle}\:\mathrm{interms}\:\mathrm{of}\:\pi \\ $$ Commented by fantastic last updated on…
Question Number 220249 by Spillover last updated on 10/May/25 Answered by Spillover last updated on 12/May/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 220248 by Spillover last updated on 10/May/25 Answered by Spillover last updated on 12/May/25 Answered by Spillover last updated on 12/May/25 Answered by…
Question Number 220244 by Spillover last updated on 10/May/25 Answered by Spillover last updated on 12/May/25 Answered by Spillover last updated on 12/May/25 Answered by…
Question Number 220250 by Spillover last updated on 10/May/25 Answered by mr W last updated on 10/May/25 $${as}\:{proved}\:{in}\:{Q}\mathrm{220231}: \\ $$$${R}={a}+{b} \\ $$$${S}=\frac{\pi}{\mathrm{2}}\left({R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)=\pi{ab}…
Question Number 220245 by Spillover last updated on 10/May/25 Answered by Spillover last updated on 12/May/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 220246 by Spillover last updated on 10/May/25 Answered by Spillover last updated on 11/May/25 Answered by mr W last updated on 10/May/25 Commented…
Question Number 220243 by Spillover last updated on 10/May/25 Answered by mr W last updated on 11/May/25 Commented by mr W last updated on 11/May/25…
Question Number 220307 by Tawa11 last updated on 10/May/25 Answered by fantastic last updated on 10/May/25 Commented by fantastic last updated on 10/May/25 $${Here}\:\Box{ABCD}\:={DC}×{AN}\:.{As}\:{the}\:{opposite}\:{sides}\:{of}\:{a}\:{parallelogram}\:{is}\:{equal}\:{so}\:{AB}={DC} \\…
Question Number 220264 by Tawa11 last updated on 10/May/25 Answered by SdC355 last updated on 10/May/25 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{2}{h}+\mathrm{9}\right)\left(\mathrm{2}{h}+\mathrm{11}\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}{h}+\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{2}{h}+\mathrm{11}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{11}}−\frac{\mathrm{1}}{\mathrm{13}}+\frac{\mathrm{1}}{\mathrm{13}}−\frac{\mathrm{1}}{\mathrm{15}}+….\right)…