Question Number 79404 by jagoll last updated on 25/Jan/20 $$\mathrm{for}\:\mathrm{every}\:\mathrm{real}\:\mathrm{number}\:\mathrm{a}\:,\:\mathrm{b}\: \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{4a}−\mathrm{6b}=\mathrm{2}.\: \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{and}\: \\ $$$$\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{expression}\: \\ $$$$\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{8a}−\mathrm{10b}+\mathrm{41}}\:? \\…
Question Number 13706 by tawa tawa last updated on 22/May/17 $$\mathrm{The}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{a}\:\mathrm{right}\:\mathrm{circular}\:\mathrm{cone}\:\mathrm{is}\:\mathrm{5}\:\mathrm{litres}\:.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{volumes}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two} \\ $$$$\mathrm{parts}\:\mathrm{into}\:\mathrm{which}\:\mathrm{the}\:\mathrm{cone}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{a}\:\mathrm{plane}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{base}\:,\:\mathrm{One}\:\mathrm{third} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{way}\:\mathrm{down}\:\mathrm{from}\:\mathrm{the}\:\mathrm{vertex}\:\mathrm{to}\:\mathrm{the}\:\mathrm{base}.\:\mathrm{Give}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{the}\:\mathrm{nearest} \\ $$$$\mathrm{ml}. \\ $$ Answered by ajfour last updated on…
Question Number 78941 by M±th+et£s last updated on 21/Jan/20 $${Q}.{find}\:{the}\:{sum} \\ $$$${S}=\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{2}!}+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}!}+\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{4}!}+…. \\ $$$${then}\:{find}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{4}} }{{n}!} \\ $$ Answered by mind…
Question Number 13307 by tawa tawa last updated on 18/May/17 Answered by ajfour last updated on 18/May/17 $${A}=\pi{RL}−\pi{rl}+\mathrm{2}\pi{rh} \\ $$$$\:\:\:\:=\pi\left(\mathrm{9}{cm}\right)\left(\frac{\mathrm{12}}{\mathrm{5}}×\mathrm{9}{cm}\right)−\pi\left(\mathrm{4}{cm}\right)\left(\frac{\mathrm{12}}{\mathrm{5}}×\mathrm{4}{cm}\right)+\mathrm{2}\pi\left(\mathrm{4}{cm}\right)\left(\mathrm{10}{cm}\right) \\ $$$$\:\:\:\:=\left[\pi\left(\frac{\mathrm{12}}{\mathrm{5}}\right)\left(\mathrm{81}−\mathrm{16}\right)+\mathrm{80}\pi\right]\:{cm}^{\mathrm{2}} \\ $$$$\:\:\:=\mathrm{236}\pi\:{cm}^{\mathrm{2}} .…
Question Number 13206 by tawa tawa last updated on 16/May/17 $$\mathrm{An}\:\mathrm{open}\:\mathrm{box}\:\mathrm{whose}\:\mathrm{shape}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cuboid}\:\mathrm{has}\:\mathrm{dimensions}\:\mathrm{9}\:\mathrm{cm}\:\mathrm{by}\:\mathrm{7}\:\mathrm{cm}\:\mathrm{by}\:\mathrm{6}\:\mathrm{cm}. \\ $$$$\mathrm{Find} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{The}\:\mathrm{outer}\:\mathrm{surface}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{box} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{The}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{box} \\ $$ Terms of Service Privacy Policy Contact:…
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Question Number 131465 by bramlexs22 last updated on 05/Feb/21 $${Let}\:{a},{b}\:{and}\:{c}\:{be}\:{three}\:{positive}\:{real}\:{numbers} \\ $$$$.\:{Prove}\:{that}\:{a}+{b}+{c}\:\leqslant\:\frac{{a}^{\mathrm{2}} +{bc}}{{b}+{c}}+\frac{{b}^{\mathrm{2}} +{ca}}{{c}+{a}}+\frac{{c}^{\mathrm{2}} +{ab}}{{a}+{b}} \\ $$ Answered by EDWIN88 last updated on 05/Feb/21 $${According}\:{to}\:{the}\:{identity}\:\frac{{a}^{\mathrm{2}}…
Question Number 99 by sagarwal last updated on 25/Jan/15 $$\mathrm{The}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{an}\:\mathrm{isosceles}\:\mathrm{right}−\mathrm{angled} \\ $$$$\mathrm{triangle}\:\mathrm{is}\:\mathrm{2}{p}.\:\mathrm{Find}\:\mathrm{out}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}. \\ $$ Answered by 123456 last updated on 14/Dec/14 $$\mathrm{lets}\:\Delta\mathrm{A}{BC}\:\mathrm{be}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{lenght}\:{a},{b},{c} \\ $$$$\mathrm{by}\:\mathrm{it}\:\mathrm{isosceles}\:\mathrm{mean}\:\mathrm{that}\:\mathrm{two}\:\mathrm{or}\:\mathrm{more}\:\mathrm{angle}\:\mathrm{and}\:\mathrm{lenght}\:\mathrm{are}\:\mathrm{equal} \\…
Question Number 12933 by chux last updated on 07/May/17 Commented by ajfour last updated on 08/May/17 $${Touch}\:{and}\:{Draw} \\ $$ Commented by chux last updated on…
Question Number 12743 by tawa last updated on 30/Apr/17 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{eqilateral}\:\mathrm{triangle}\:\mathrm{whose}\:\mathrm{inscribed}\:\mathrm{circle}\:\mathrm{has}\:\mathrm{a}\:\mathrm{radius}\:\mathrm{2} \\ $$ Answered by mrW1 last updated on 30/Apr/17 $${OC}=\mathrm{2} \\ $$$${OB}=\mathrm{4}={DO} \\ $$$${CB}=\mathrm{2}\sqrt{\mathrm{3}} \\…