Question Number 206473 by hardmath last updated on 15/Apr/24 $$\mathrm{If}\:\:\:\mathrm{4}^{\boldsymbol{\mathrm{p}}} \:=\:\mathrm{5} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{2}^{\mathrm{3}\boldsymbol{\mathrm{p}}} \:=\:? \\ $$ Answered by A5T last updated on 15/Apr/24 $$\mathrm{2}^{\mathrm{2}{p}} =\mathrm{5}\Rightarrow\mathrm{2}^{\mathrm{3}{p}}…
Question Number 206475 by hardmath last updated on 15/Apr/24 $$\mathrm{If}\:\:\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{12}}\:−\:\mathrm{5b}^{\mathrm{3}} \:=\:−\mathrm{30} \\ $$$$\mathrm{Find}:\:\:\:\frac{\mathrm{2}}{\mathrm{45}}\:\mathrm{a}^{\mathrm{2}} \:−\:\frac{\mathrm{8}}{\mathrm{3}}\:\mathrm{b}^{\mathrm{3}} \:=\:? \\ $$ Answered by A5T last updated on 15/Apr/24…
Question Number 206443 by Safojon last updated on 14/Apr/24 Commented by necx122 last updated on 15/Apr/24 getting a more detailed solution will be excellent for us. I still tried going through the solution already done but I couldn't understand the approaches readily. Please, whoever can expound would be a help. Thanks. Commented by A5T last updated on 15/Apr/24 $${Let}\:\boldsymbol{{u}}=\left({a},{b},{c}\right);\boldsymbol{{v}}=\left({x},{y},{z}\right)…
Question Number 206425 by hardmath last updated on 13/Apr/24 $$\mathrm{If}\:\:\:\mathrm{cos}\boldsymbol{\alpha}\:=\:\frac{\mathrm{3}}{\mathrm{5}}\:\:\:\left(\mathrm{0}<\boldsymbol{\alpha}<\frac{\boldsymbol{\pi}}{\mathrm{2}}\right) \\ $$$$\mathrm{Find}:\:\:\:\frac{\mathrm{tan}^{\mathrm{2}} \:\left(\mathrm{45}°\:+\:\frac{\boldsymbol{\alpha}}{\mathrm{2}}\right)}{\mathrm{3}}\:=\:? \\ $$ Answered by MM42 last updated on 13/Apr/24 $${tan}^{\mathrm{2}} {a}=\frac{\mathrm{1}−{cos}\mathrm{2}{a}}{\mathrm{1}+{cos}\mathrm{2}{a}} \\…
Question Number 206391 by hardmath last updated on 13/Apr/24 $$\mathrm{Find}: \\ $$$$\int_{−\mathrm{3}} ^{\:−\mathrm{2}} \:\left(\mid\mathrm{x}\mid\:+\:\mid\mathrm{x}\:−\:\mathrm{4}\mid\right)\:\mathrm{dx}\:=\:? \\ $$ Answered by MM42 last updated on 13/Apr/24 $${I}=\int_{−\mathrm{3}} ^{−\mathrm{2}}…
Question Number 206365 by hardmath last updated on 12/Apr/24 $$\mathrm{Number}\:\mathrm{series}: \\ $$$$\mathrm{a}_{\mathrm{3}} \:=\:\mathrm{2a}\:+\:\mathrm{b}\:−\:\mathrm{6} \\ $$$$\mathrm{a}_{\mathrm{9}} \:=\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{5} \\ $$$$\mathrm{a}_{\mathrm{15}} \:=\:\mathrm{3a}\:+\:\mathrm{b}\:−\:\mathrm{7} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{a}\:=\:? \\ $$ Commented by…
Question Number 206332 by MATHEMATICSAM last updated on 12/Apr/24 $$\mathrm{If}\:\mathrm{log}\left({a}\:+\:{b}\:+\:{c}\right)\:=\:\mathrm{log}{a}\:+\:\mathrm{log}{b}\:+\:\mathrm{log}{c}\: \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{log}\left(\frac{\mathrm{2}{a}}{\mathrm{1}\:−\:{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}{b}}{\mathrm{1}\:−\:{b}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}{c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} }\right)\:=\: \\ $$$$\mathrm{log}\left(\frac{\mathrm{2}{a}}{\mathrm{1}\:−\:{a}^{\mathrm{2}} }\right)\:+\:\mathrm{log}\left(\frac{\mathrm{2}{b}}{\mathrm{1}\:−\:{b}^{\mathrm{2}} }\right)\:+\:\mathrm{log}\left(\frac{\mathrm{2}{c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} }\right). \\ $$ Answered…
Question Number 206328 by mnjuly1970 last updated on 12/Apr/24 $$ \\ $$$$\:\:\:\:\:\:\:{f}\left({x}\right)=\:\lfloor\:\frac{\:\mathrm{1}+\:{x}\:+{x}^{\mathrm{2}} }{{x}+\:{x}^{\:\mathrm{2}} }\:\rfloor\:{is}\:{given}. \\ $$$$\:\:\:\:\:\:\:\Rightarrow\begin{cases}{\:\:{D}_{{f}} \:=\:?\:\left({domain}\:\right)}\\{\:\:\:{R}_{\:{f}} \:=\:?\left(\:{range}\right)}\end{cases} \\ $$$$ \\ $$ Answered by TheHoneyCat…
Question Number 206367 by hardmath last updated on 12/Apr/24 $$\mathrm{Find}: \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{24}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{60}}\:\:+\:…\:+\:…\:\:\frac{\mathrm{1}}{\mathrm{720}}\:=\:? \\ $$ Commented by TheHoneyCat last updated on 12/Apr/24 Can you please explain what the "..." are? Commented by TonyCWX08…
Question Number 206363 by hardmath last updated on 12/Apr/24 $$\mathrm{If}\:\:\:\mathrm{0}<\mathrm{a}<\mathrm{1} \\ $$$$\mathrm{Compare}: \\ $$$$\frac{\mathrm{1}}{\mathrm{a}−\mathrm{1}}\:\:,\:\:\frac{\mathrm{a}}{\mathrm{a}−\mathrm{1}}\:\:,\:\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{a}}\:\:,\:\:\frac{\mathrm{a}}{\mathrm{1}−\mathrm{a}}\:\:,\:\:\frac{\mathrm{a}}{\mathrm{2a}} \\ $$ Commented by TheHoneyCat last updated on 12/Apr/24 Commented by…