Question Number 147921 by abdurehime last updated on 24/Jul/21 Commented by abdurehime last updated on 25/Jul/21 $$\mathrm{please}\:\mathrm{help}\:\mathrm{me} \\ $$ Terms of Service Privacy Policy Contact:…
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Question Number 16845 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/Jun/17 Commented by prakash jain last updated on 28/Jun/17 $$\left.\mathrm{1}\right) \\ $$$$\mathrm{cos}\:{A}+\mathrm{cos}\:{B}\mathrm{cos}\:{C} \\ $$$${C}=\mathrm{90}°,{A}=\mathrm{45}°,{B}=\mathrm{45}° \\ $$$$\mathrm{cos}\:{A}+\mathrm{cos}\:{B}\centerdot\mathrm{cos}\:{C}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\…
Question Number 147918 by abdurehime last updated on 24/Jul/21 Commented by abdurehime last updated on 24/Jul/21 Commented by abdurehime last updated on 24/Jul/21 $$\mathrm{please}\:\mathrm{help}\:\mathrm{me} \\…
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Question Number 82378 by ajfour last updated on 20/Feb/20 Commented by ajfour last updated on 20/Feb/20 $$\mathrm{If}\:\mathrm{all}\:\mathrm{5}\:\mathrm{regions}\:\mathrm{have}\:\mathrm{equal}\:\mathrm{areas}, \\ $$$$\mathrm{find}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{rectangle},\:\mathrm{given} \\ $$$$\mathrm{radius}=\mathrm{1}. \\ $$ Answered by…
Question Number 147915 by tabata last updated on 24/Jul/21 Commented by tabata last updated on 24/Jul/21 $${how}\:{can}\:{solve}\:{this}\:{please}\:? \\ $$ Commented by tabata last updated on…
Question Number 147914 by mnjuly1970 last updated on 24/Jul/21 $$ \\ $$$$\:\:{f}\:\left({x}\:\right):=\:\mathrm{3}{x}\:+\left[\:\frac{{x}}{\mathrm{4}}\:\right]\:\:,\:\mathrm{D}_{\:{f}} \:=\left[\:\mathrm{0},\infty\right) \\ $$$$\:\:\:\:\:\:\:\:\:{f}^{\:−\mathrm{1}} \left(\:{x}\:\right)=\:? \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 16840 by Sai dadon. last updated on 27/Jun/17 $$\mathrm{6}/\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right) \\ $$ Commented by RasheedSoomro last updated on 27/Jun/17 $$\mathrm{6}/\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right) \\ $$$$=\mathrm{6}/\left\{\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right)\right\} \\ $$$$=\mathrm{6}/\left(\mathrm{2}×\mathrm{3}\right)…