Question Number 140407 by mnjuly1970 last updated on 07/May/21 $$\:\:\:\: \\ $$$$\:\:\:\:{prove}\:{that}\:\langle\:\mathrm{X}:=\mathbb{R}\:,\:\tau_{{e}} \:\rangle\:{is} \\ $$$$\:\:\:\:\:\:{a}\:{second}\:{topology}\:{space}\:. \\ $$$$\:\:\:\:\:\tau_{{e}} \:\:{is}\:\mathscr{E}{uclidian}\:{topology}\:{on}\:\mathbb{R}. \\ $$$$\:\:\:\mathrm{H}{int}\:::\:\:\mathcal{B}=\:\left\{\:\left({r}−\frac{\mathrm{1}}{{n}}\:,{r}+\frac{\mathrm{1}}{{n}}\right)\mid\:{r}\in\mathrm{Q}\:,\:{n}\in\mathbb{N}\right\} \\ $$$$\:\:\:{is}\:\:{a}\:{base}\:{for}\:\tau_{{e}\:} ….. \\ $$…
Question Number 9333 by j.masanja06@gmail.com last updated on 30/Nov/16 $$\mathrm{given}\:\mathrm{that}\:\mathrm{log}\:\mathrm{2}=\mathrm{0}.\mathrm{3}.\mathrm{show}\:\mathrm{that}\:\mathrm{log}\:\mathrm{5}=\mathrm{0}.\mathrm{7} \\ $$ Answered by mrW last updated on 30/Nov/16 $$\mathrm{2}×\mathrm{5}=\mathrm{10} \\ $$$$\mathrm{log}\:\left(\mathrm{2}×\mathrm{5}\right)=\mathrm{log}\:\mathrm{10}=\mathrm{1} \\ $$$$\mathrm{log}\:\mathrm{2}+\mathrm{log}\:\mathrm{5}=\mathrm{1} \\…
Question Number 9332 by tawakalitu last updated on 30/Nov/16 Answered by geovane10math last updated on 30/Nov/16 $${Answer}\:\mathrm{6}: \\ $$$${f}\left({x}\right)\:=\:\mathrm{4}{x}^{\mathrm{2}} \:−\:\mathrm{1}\:\:\:\:\:\:\:\:\:{g}\left({x}\right)\:=\:\mathrm{8}{x}\:+\:\mathrm{7} \\ $$$${f}\left(\mathrm{2}\right)\:=\:\mathrm{4}\left(\mathrm{2}\right)^{\mathrm{2}} \:−\:\mathrm{1}\:=\:\mathrm{4}\left(\mathrm{4}\right)\:−\:\mathrm{1}\:=\:\mathrm{15} \\ $$$${g}\left({f}\left(\mathrm{2}\right)\right)\:=\:{g}\left(\mathrm{15}\right)\:=\:\mathrm{8}\left(\mathrm{15}\right)\:+\:\mathrm{7}\:=\:\mathrm{120}\:+\:\mathrm{7}\:\%…
Question Number 140401 by mnjuly1970 last updated on 07/May/21 $$\:\:\: \\ $$$$\:\:\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\Phi:=\int_{\mathrm{0}} ^{\:\infty} {xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} {ln}\left({x}\right){dx}\:=\:{m}.\left(\:\pi\:\gamma\right) \\ $$$$\:\:\:\:\:\:{find}\:\:\:''\:\:{m}\:\:''\:…… \\ $$$$ \\ $$ Answered…
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Question Number 74863 by mrS last updated on 02/Dec/19 Commented by abdomathmax last updated on 03/Dec/19 $${we}\:{have}\:{S}=\sum_{{n}=\mathrm{1}} ^{\mathrm{45}} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\sum_{{p}=\mathrm{1}} ^{\left[\frac{\mathrm{45}}{\mathrm{2}}\right]} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}\right)^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\left[\frac{\mathrm{45}−\mathrm{1}}{\mathrm{2}}\right]} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 140399 by mnjuly1970 last updated on 07/May/21 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\xi}\::=\int_{\mathrm{0}} ^{\:\infty} \:\frac{{e}^{−{x}^{\mathrm{2}} } −{e}^{−{x}} }{{x}}\:{dx}\:=\:{k}.\gamma\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{find}\:\:''\:{k}\:\:''\:… \\ $$$$\:\:\:\:\:\:\:\:\:\:\gamma\::=\:\mathscr{E}{uler}\:{constant}…. \\ $$ Answered by…
Question Number 74860 by rajesh4661kumar@gmail.com last updated on 02/Dec/19 Answered by MJS last updated on 02/Dec/19 $$\mathrm{no}\:\mathrm{head} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{6}} =\frac{\mathrm{1}}{\mathrm{64}} \\ $$$$\mathrm{minimum}\:\mathrm{one}\:\mathrm{head} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{64}}=\frac{\mathrm{63}}{\mathrm{64}} \\…
Question Number 9325 by tawakalitu last updated on 30/Nov/16 Answered by RasheedSoomro last updated on 30/Nov/16 $$\left(\mathrm{B}\right)\:\mathrm{12}\: \\ $$$$\mathrm{25}\equiv−\mathrm{1}\left(\mathrm{mod}\:\mathrm{13}\right) \\ $$$$\left(\mathrm{25}\right)^{\mathrm{15}} \equiv\left(−\mathrm{1}\right)^{\mathrm{15}} \left(\mathrm{mod}\:\mathrm{13}\right) \\ $$$$\left(\mathrm{25}\right)^{\mathrm{15}}…
Question Number 74861 by aliesam last updated on 02/Dec/19 Answered by mind is power last updated on 02/Dec/19 $$\left(\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =−\mathrm{ln}\left(\mathrm{x}\right) \\ $$$$\left(\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} \right)\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}.\left(\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{x}\right)}\right)^{\frac{\mathrm{1}}{\mathrm{2}}}…