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the-base-of-an-object-is-in-the-form-of-a-circle-with-radius-1-suppose-that-all-section-of-the-object-are-perpendicular-to-a-diameter-of-a-square-determine-the-volume-of-the-object-

Question Number 153918 by joki last updated on 12/Sep/21 $$\mathrm{the}\:\mathrm{base}\:\mathrm{of}\:\mathrm{an}\:\mathrm{object}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with} \\ $$$$\mathrm{radius}\:\mathrm{1}.\:\mathrm{suppose}\:\mathrm{that}\:\mathrm{all}\:\mathrm{section}\:\mathrm{of}\:\mathrm{the}\:\mathrm{object}\:\mathrm{are} \\ $$$$\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{a}\:\mathrm{diameter}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}.\:\mathrm{determine} \\ $$$$\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{object}? \\ $$ Answered by talminator2856791 last updated on 12/Sep/21…

Question-153896

Question Number 153896 by SANOGO last updated on 11/Sep/21 Answered by ARUNG_Brandon_MBU last updated on 12/Sep/21 $$\Omega=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{ln}{x}}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx},\:{x}=\frac{\mathrm{1}}{{u}}\Rightarrow{dx}=−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }{du} \\ $$$$\:\:\:\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}\mathrm{ln}{u}}{\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}}…

Question-153860

Question Number 153860 by SANOGO last updated on 11/Sep/21 Answered by puissant last updated on 11/Sep/21 $${Q}=\int_{\mathrm{0}} ^{\pi} {sin}^{\sqrt{\mathrm{3}}} {x}\:{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\sqrt{\mathrm{3}}} {x}\:{dx}…

write-to-geometry-form-3-5-4-5-i-help-plz-

Question Number 22769 by Bruce Lee last updated on 22/Oct/17 $$\boldsymbol{\mathrm{write}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{geometry}}\:\boldsymbol{\mathrm{form}} \\ $$$$\:\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{4}}{\mathrm{5}}\boldsymbol{\mathrm{i}} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{help}}\:\boldsymbol{\mathrm{plz}}\: \\ $$ Answered by $@ty@m last updated on…

Question-153800

Question Number 153800 by AliRaza123 last updated on 10/Sep/21 Answered by nadovic last updated on 10/Sep/21 $$\:\:\mathrm{Speed}\:\:=\:\:\frac{\mathrm{distance}}{\mathrm{time}} \\ $$$$\:\:\mathrm{600}\:\:=\:\:\frac{\mathrm{7200}}{{t}}\: \\ $$$$\:\:\:\:\:\:\:{t}\:\:=\:\:\frac{\mathrm{7200}}{\mathrm{600}}\: \\ $$$$\:\:\:\:\:\:\:{t}\:\:=\:\:\mathrm{12}\:{hours} \\ $$$$\:\:…