Question Number 215276 by ppch145 last updated on 02/Jan/25 $$\mathrm{I}\:\mathrm{produced}\:\mathrm{the}\:\mathrm{drawing}\: \\ $$$$\mathrm{shown}\:\mathrm{below}\:\left(\mathrm{Q}.\:\mathrm{215275}\right)\: \\ $$$$\mathrm{using}\:\mathrm{this}\:\mathrm{great}\:\mathrm{app},\:\mathrm{but}\: \\ $$$$\mathrm{unfortunately}\:\mathrm{it}\:\mathrm{refused}\:\mathrm{to}\: \\ $$$$\mathrm{get}\:\mathrm{saved}\:\mathrm{after}\:\mathrm{clicking}\: \\ $$$$“\mathrm{Save}''\:\mathrm{button}\:\mathrm{several}\:\mathrm{times}.\: \\ $$$$\mathrm{Also},\:\mathrm{the}\:“\mathrm{Export}\:\mathrm{As}\:\mathrm{Image}''\: \\ $$$$\mathrm{button}\:\mathrm{is}\:\mathrm{not}\:\mathrm{working}.\:\mathrm{Does}\: \\…
Question Number 215277 by efronzo1 last updated on 02/Jan/25 $$\:\:\:\left(\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)\left(\mathrm{1}+\mathrm{sin}\:\mathrm{x}\right)\:=\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\:\:\:\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)\left(\mathrm{1}−\mathrm{sin}\:\mathrm{x}\right)\:=\:? \\ $$ Answered by MrGaster last updated on 02/Jan/25 $$\underset{\mathrm{Commutative}\:\mathrm{law}\:\mathrm{of}\:\mathrm{multiplication}} {\underbrace{\left(\mathrm{1}+\mathrm{cos}\:{x}\right)\left(\mathrm{1}+\mathrm{sin}\:{x}\right)\centerdot\left(\mathrm{1}−\mathrm{cos}\:{x}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\right)}}=\left(\frac{\mathrm{5}}{\mathrm{4}}\right)\left(\mathrm{1}−\mathrm{cos}\:{x}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\right) \\ $$$$\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}}…
Question Number 215272 by Rasheed.Sindhi last updated on 02/Jan/25 $$\begin{cases}{\overline {{abac}}=\left(\overline {{dc}}\right)^{\mathrm{2}} }\\{{d}=\frac{\overline {{ab}}}{{c}}}\\{{c}^{\mathrm{2}} =\overline {{ac}}}\end{cases} \\ $$$$\overline {{abac}}=? \\ $$ Answered by A5T last…
Question Number 215275 by ppch145 last updated on 02/Jan/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 215229 by Mingma last updated on 01/Jan/25 Answered by devdutt last updated on 01/Jan/25 $$\mathrm{1}.\:{f}\left(\frac{{k}}{{n}}\right)\:=\:\frac{{n}^{\mathrm{2025}} }{{n}^{\mathrm{2025}} +{k}^{\mathrm{2025}} } \\ $$$$ \\ $$$$\mathrm{2}.\:{f}\left(\frac{{k}}{{n}}\right)+{f}\left(\frac{{n}}{{k}}\right)\:=\:\mathrm{1},\: \\…
Question Number 215230 by Mingma last updated on 01/Jan/25 Answered by A5T last updated on 01/Jan/25 Commented by A5T last updated on 01/Jan/25 $$\mathrm{AC}=\mathrm{BC}=\mathrm{a};\:\mathrm{AB}=\mathrm{c};\:\:\angle\mathrm{CAB}=\theta;\:\mathrm{IE}\parallel\mathrm{AB} \\…
Question Number 215231 by Mingma last updated on 01/Jan/25 Answered by A5T last updated on 01/Jan/25 $$\mathrm{2024}=\mathrm{8}×\mathrm{253}=\mathrm{8}×\mathrm{11}×\mathrm{23} \\ $$$$\mathrm{2025}=\mathrm{45}^{\mathrm{2}} =\mathrm{3}^{\mathrm{4}} ×\mathrm{5}^{\mathrm{2}} ;\:\:\:\:\mathrm{2026}=\mathrm{2}×\mathrm{1013} \\ $$$$\Rightarrow\mathrm{smallest}\:\mathrm{perfect}\:\mathrm{square} \\…
Question Number 215256 by ajfour last updated on 01/Jan/25 Commented by ajfour last updated on 01/Jan/25 $${After}\:{this}\:{elastic}\:{collision}\:{on}\:{a}\:{flat} \\ $$$${frictionless}\:{surface}\:{Find}\:{after}\:{what} \\ $$$${time}\:{at}\:{what}\:{y}\:{coordinate}\:{does}\:{point} \\ $$$${A}\:{cross}\:{the}\:{y}-{axis}. \\ $$…
Question Number 215227 by Ghisom last updated on 01/Jan/25 $$\left(\mathrm{20}+\mathrm{25}\right)^{\mathrm{2}} =\mathrm{2025} \\ $$$$\mathrm{only}\:\mathrm{3}\:\mathrm{other}\:\mathrm{4}\:\mathrm{digit}\:\mathrm{numbers}: \\ $$$$\left(\mathrm{00}+\mathrm{01}\right)^{\mathrm{2}} =\mathrm{0001} \\ $$$$\left(\mathrm{30}+\mathrm{25}\right)^{\mathrm{2}} =\mathrm{3025} \\ $$$$\left(\mathrm{98}+\mathrm{01}\right)^{\mathrm{2}} =\mathrm{9801} \\ $$ Answered…
Question Number 215255 by Rasheed.Sindhi last updated on 01/Jan/25 $${a},{b},{c}\:{are}\:{pythagorean}\:{triples} \\ $$$${For}\:{the}\:{following}\:{values}: \\ $$$$\begin{array}{|c|c|}{{a}}&\hline{{b}}&\hline{\:{c}}\\{\mathrm{5}}&\hline{\mathrm{12}}&\hline{\mathrm{13}}\\\hline\end{array}\:\:\underset{\Rightarrow\mathrm{5}^{\mathrm{2}} =\mathrm{12}+\mathrm{13}=\mathrm{25}} {{a}^{\mathrm{2}} ={b}+{c}\:\:} \\ $$$$\begin{array}{|c|c|}{{a}}&\hline{{b}}&\hline{\:{c}}\\{\mathrm{45}}&\hline{\mathrm{1012}}&\hline{\mathrm{1013}}\\\hline\end{array}\underset{\:\:\:\:\:\:\:\Rightarrow\mathrm{45}^{\mathrm{2}} =\mathrm{1012}+\mathrm{1013}=\mathrm{2025}} {{a}^{\mathrm{2}} ={b}+{c}} \\ $$$$\Rightarrow\mathrm{45}^{\mathrm{2}} =\mathrm{1012}+\mathrm{1013}=\mathrm{2025}…