Question Number 217064 by ArshadS last updated on 28/Feb/25 $$\mathrm{Two}\:\mathrm{numbers}\:\mathrm{differ}\:\mathrm{by}\:\mathrm{6}.\:\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{reciprocals}\:\mathrm{is}\:\frac{\mathrm{2}}{\mathrm{15}}\:. \\ $$$$\mathcal{D}{etermine}\:{the}\:{numbers}. \\ $$ Answered by A5T last updated on 28/Feb/25 $$\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}=\frac{\mathrm{2}}{\mathrm{15}}\Rightarrow\mathrm{15}\left(\mathrm{a}+\mathrm{b}\right)=\mathrm{2ab} \\ $$$$\mathrm{a}−\mathrm{b}=\underset{−} {+}\mathrm{6}\Rightarrow\mathrm{a}=\mathrm{b}\underset{−}…
Question Number 217065 by ArshadS last updated on 28/Feb/25 $${To}\:{Tinku}\:{Tara}\:{sir}, \\ $$$${How}\:{can}\:{I}\:{import}\:{something}\:{in}\:{Latex}\:{form}. \\ $$$$\bullet{There}\:{is}\:{an}\:{option}\:\mathrm{Get}\:\mathrm{Latex}\:{but}\:{there}\:{is}\:{no} \\ $$$$“\mathrm{Paste}\:\mathrm{Latex}''\:{option} \\ $$$$\bullet{When}\:{I}\:{tried}\:{to}\:{import}\:{something}\:{using}\:{option} \\ $$$$“\mathrm{Paste}\:\mathrm{plain}\:\mathrm{text}''\:{all}\:{the}\:{whitespaces}/{new}\:{line}\: \\ $$$${were}\:{lost}\:{in}\:{imported}\:{material}. \\ $$$$\boldsymbol{{Please}}\:{solve}\:{these}\:{issues}\:{as}\:{soon}\:{as}\:{possible} \\…
Question Number 217066 by ArshadS last updated on 28/Feb/25 $${Find}\:{all}\:{integer}\:{x},{y}\:{such}\:{that} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{100} \\ $$ Answered by mehdee7396 last updated on 28/Feb/25 $$\left({x}−{y}\right)\left({x}+{y}\right)=\mathrm{2}×\mathrm{50}=\mathrm{10}×\mathrm{10} \\…
Question Number 217088 by alcohol last updated on 28/Feb/25 $${show}\:{that}\:\int_{\:{n}} ^{\:{n}\:+\:\mathrm{1}} {ln}\left({t}\right)\:{dt}\:\leqslant\:{ln}\left({n}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${Given}\:{u}_{{n}} \:=\:\frac{\left(\mathrm{4}{n}\right)^{{n}} {n}!{e}^{−{n}} }{\left(\mathrm{2}{n}\right)!},\:\forall{n}\:\geqslant\:\mathrm{1} \\ $$$${prove},\:{using}\:{the}\:{preceding}\:{question}\:{that} \\ $$$${u}_{{n}} \:{is}\:{decreasing}\:{and}\:{convergent} \\ $$ Answered…
Question Number 217058 by ArshadS last updated on 28/Feb/25 $$\mathrm{The}\:\mathrm{quadratic}\:\mathrm{equation}\:\:\mathrm{has}\:\mathrm{two}\:\mathrm{equal}\:\mathrm{roots}: \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +\left({k}−\mathrm{3}\right){x}+{k}=\mathrm{0} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:{k}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{For}\:\mathrm{this}\:\mathrm{value}\:\mathrm{of}\:\:{k},\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{for}\:{x} \\ $$$$\left(\mathrm{c}\right)\mathrm{If}\:\:{x}\:\mathrm{is}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rectangle}\:\mathrm{and}\:\mathrm{its}\:\mathrm{width}\:\mathrm{is}\:{x}−\mathrm{2},\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{area}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle}. \\ $$ Answered by…
Question Number 217032 by Tawa11 last updated on 27/Feb/25 Answered by issac last updated on 27/Feb/25 $$\mathrm{true} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 217049 by nECxx2 last updated on 27/Feb/25 $${Hello}\:{here}\:{its}\:{been}\:{a}\:{while}.\:{I}'{ve} \\ $$$${been}\:{watching}\:{Ajfour}\:{sir}\:{dancing} \\ $$$${lately}\:{on}\:{Youtube}.\:{I}\:{believe}\:{he}\:{has} \\ $$$${gotten}\:{to}\:{saturation}\:{point}. \\ $$$${Is}\:{there}\:{anyone}\:{here}\:{into}\:{AI}\:{and} \\ $$$${Machine}\:{learning}? \\ $$$$ \\ $$ Commented…
Question Number 217050 by nECxx2 last updated on 27/Feb/25 Commented by nECxx2 last updated on 27/Feb/25 $${Please}\:{help}\:{me}\:{out}. \\ $$ Answered by mr W last updated…
Question Number 217046 by OmoloyeMichael last updated on 27/Feb/25 $${form}\:{the}\:{differential}\:{equation}\:{by}\: \\ $$$${eliminating}\:{the}\:{arbritrary}\:{constant} \\ $$$${y}^{\mathrm{2}} ={Ax}^{\mathrm{2}} +{Bx}+{C} \\ $$ Answered by mr W last updated on…
Question Number 217030 by ArshadS last updated on 27/Feb/25 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:\:\mathrm{n}\:\:\mathrm{such}\:\mathrm{that}\:\: \\ $$$$\:\mathrm{n}\:+\:\mathrm{1}\:\:\mathrm{divides}\:\:\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$ Answered by issac last updated on 27/Feb/25 $$\frac{{n}+\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}=\frac{{n}+\mathrm{1}}{\left(\mathrm{1}+{n}\boldsymbol{{i}}\right)\left(\mathrm{1}−{n}\boldsymbol{{i}}\right)} \\…