Question Number 202630 by ajfour last updated on 30/Dec/23 Commented by Frix last updated on 31/Dec/23 $$\left[\mathrm{I}\:\mathrm{forgot}\:\mathrm{to}\:\mathrm{check}\:\Rightarrow\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{positive}\:\mathrm{root}\right. \\ $$$$\left.\mathrm{is}\:\mathrm{false}\right] \\ $$ Commented by Frix…
Question Number 202562 by cherokeesay last updated on 29/Dec/23 Answered by ajfour last updated on 29/Dec/23 Commented by ajfour last updated on 29/Dec/23 Happy New Year! Commented…
Question Number 202056 by mr W last updated on 19/Dec/23 Commented by mr W last updated on 19/Dec/23 $${the}\:{lake}\:{with}\:{center}\:{at}\:{O}\:{has}\:{a}\:{radius} \\ $$$${r}\:\left({r}=\mathrm{1}\:{km}\right).\:{the}\:{shortest}\:{distances}\: \\ $$$${from}\:{the}\:{villages}\:{A}\:{and}\:{B}\:{to}\:{the}\: \\ $$$${lake}\:{are}\:{a}\:{and}\:{b}\:{respectively}\:\left({a}=\mathrm{4}\:{km},\right.…
Question Number 202001 by necx122 last updated on 18/Dec/23 $${If}\:{a}\:{circle}\:{of}\:{radius}\:{r}\:{is}\:{inscribed}\:{in} \\ $$$${a}\:{triangl}\:{ABC}.\:{Express}\:{r}\:{in}\:{terms}\:{of} \\ $$$${a},{b}\:{and}\:{c}\:{only} \\ $$ Answered by AST last updated on 18/Dec/23 $${r}=\frac{\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}}{{s}}\:{where}\:{s}=\frac{{a}+{b}+{c}}{\mathrm{2}} \\…
Question Number 202003 by 31282329g last updated on 18/Dec/23 $${what}\:{is}\:{meant}\:{by}\:\xi \\ $$ Commented by mr W last updated on 18/Dec/23 $${it}\:{can}\:{mean}\:{nothing}\:{or}\:{every}\:{thing}, \\ $$$${depenting}\:{on}\:{where}\:{you}\:{are}\:{using}\:{it}. \\ $$…
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Question Number 201979 by mr W last updated on 17/Dec/23 Commented by mr W last updated on 19/Dec/23 $${Q}\mathrm{155657}\:{reposted}\:{for}\:{alternative} \\ $$$${solutions}. \\ $$ Answered by…
Question Number 201914 by Anonim_X last updated on 15/Dec/23 Commented by mr W last updated on 16/Dec/23 $${see}\:{Q}\mathrm{198828} \\ $$ Terms of Service Privacy Policy…
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Question Number 201907 by mnjuly1970 last updated on 15/Dec/23 Answered by Rasheed.Sindhi last updated on 15/Dec/23 $$\bullet{Radius}\:{of}\:{Semi}-{circle}\:{R}=\:\frac{\mathrm{10}}{\mathrm{2}}=\mathrm{5}\:{cm} \\ $$$$\bullet{let}\:{radous}\:{of}\:{greater}\:{quarter}-{circle}={r}_{\mathrm{1}} \\ $$$$\left({R}+{r}_{\mathrm{1}} \right)^{\mathrm{2}} =\mathrm{10}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} =\mathrm{125}…