Question Number 66988 by Mr Jor last updated on 21/Aug/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 1453 by 112358 last updated on 06/Aug/15 $${Show}\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{{rx}} =\frac{{e}^{{x}} \left({e}^{{nx}} −\mathrm{1}\right)}{{e}^{{x}} −\mathrm{1}}\:\:\:\:\:\:\:\left(\ast\right) \\ $$$${if}\:\:\:{e}^{{rx}} ={cos}\left({irx}\right)−{isin}\left({irx}\right)\:{and}\:{x}\neq\mathrm{0}\:. \\ $$$$\left[{Do}\:{not}\:{treat}\:\left(\ast\right)\:{as}\:{a}\:{GP}\:{to}\right. \\ $$$$\left.{directly}\:{obtain}\:{the}\:{result}.\right]…
Question Number 66986 by Mr Jor last updated on 21/Aug/19 Commented by Mr Jor last updated on 21/Aug/19 $${The}\:{metal}\:{solid}\:{above}\:{is}\:{made}\:{up} \\ $$$${by}\:{joining}\:{a}\:{hemisphere}\:{of}\:{radius} \\ $$$$\mathrm{7}{cm}\:{to}\:{a}\:{cylinder}\:{of}\:{the}\:{same}\:{radius}. \\ $$$${The}\:{mass}\:{and}\:{density}\:{of}\:{the}\:{solid}…
Question Number 66985 by Mr Jor last updated on 21/Aug/19 $$\mathrm{The}\:\mathrm{external}\:\mathrm{length},\mathrm{width}\:\mathrm{and}\:\mathrm{height} \\ $$$$\mathrm{of}\:\mathrm{an}\:\mathrm{open}\:\mathrm{rectangular}\:\mathrm{container}\:\mathrm{are} \\ $$$$\mathrm{41cm},\mathrm{21cm}\:\mathrm{and}\:\mathrm{15}.\mathrm{5cm}\:\mathrm{respectively}. \\ $$$$\mathrm{The}\:\mathrm{thickness}\:\mathrm{of}\:\mathrm{the}\:\mathrm{material}\:\mathrm{making} \\ $$$$\mathrm{the}\:\mathrm{container}\:\mathrm{is}\:\mathrm{5mm}.\mathrm{If}\:\mathrm{the}\:\mathrm{container} \\ $$$$\mathrm{has}\:\mathrm{8litres}\:\mathrm{of}\:\mathrm{water},\mathrm{calculate}\:\mathrm{the} \\ $$$$\mathrm{internal}\:\mathrm{height}\:\mathrm{above}\:\mathrm{the}\:\mathrm{water}\:\mathrm{level}. \\ $$$$\:\:\:\:\:\:\:\:…
Question Number 1448 by 123456 last updated on 05/Aug/15 $$\boldsymbol{{x}}=\left({x}_{\mathrm{1}} ,{x}_{\mathrm{2}} \right),\boldsymbol{{y}}=\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right) \\ $$$$\eta:\left[\mathrm{0},\mathrm{1}\right)^{\mathrm{4}} \rightarrow\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\eta\left(\boldsymbol{{x}},\boldsymbol{{y}}\right):=\mathrm{med}\left[\frac{\left(\mathrm{1}−{x}_{\mathrm{1}} \right)^{{y}_{\mathrm{1}} } +\left(\mathrm{1}−{y}_{\mathrm{1}} \right)^{{x}_{\mathrm{1}} } }{\mathrm{2}},\frac{\left(\mathrm{1}−{x}_{\mathrm{2}}…
Question Number 132519 by mnjuly1970 last updated on 14/Feb/21 $$\:\:\:\:\:….\:\:{nice}\:\:{calculus}…. \\ $$$$\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {ln}\left({n}\right)}{{n}}=\gamma{ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$ \\ $$ Answered by mindispower…
Question Number 66980 by Mr Jor last updated on 21/Aug/19 Commented by Mr Jor last updated on 21/Aug/19 $${The}\:{figure}\:{above}\:{is}\:{a}\:{right}\:{pyramid} \\ $$$${with}\:{a}\:{rectangular}\:{base}\:{ABCD}\:{and} \\ $$$${VO}\:{as}\:{the}\:{height}. \\ $$$${The}\:{vectors}\:\boldsymbol{{AD}}=\boldsymbol{{a}},\boldsymbol{{AB}}=\boldsymbol{{b}}\:{and}\:\boldsymbol{{DV}}=\boldsymbol{{c}}…
Question Number 66981 by Mr Jor last updated on 21/Aug/19 Commented by Mr Jor last updated on 21/Aug/19 $${In}\:{the}\:{figure}\:{above},{ABCD}\:{is}\:{a} \\ $$$${parallelogram}.{AOC}\:{and}\:{BOD}\: \\ $$$${are}\:{diagonals}\:{of}\:{the}\:{parallologram}. \\ $$$${Show}\:{that}\:{the}\:{diagonals}\:{of}\:{the}\:…
Evaluate-the-following-integral-0-n-x-1-x-dx-n-N-Here-x-is-the-integer-part-of-x-e-g-0-12-0-5-896-5-
Question Number 1444 by 112358 last updated on 04/Aug/15 $${Evaluate}\:{the}\:{following}\:{integral}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:{n}} \lfloor{x}\rfloor^{\mathrm{1}/\lfloor{x}\rfloor!} {dx}\:\:\:\:\:\:\:\left({n}\in\mathbb{N}\right) \\ $$$${Here}\:\lfloor{x}\rfloor\:{is}\:{the}\:{integer}−{part}\:{of}\:{x} \\ $$$${e}.{g}\:\lfloor\mathrm{0}.\mathrm{12}\rfloor=\mathrm{0},\:\lfloor\mathrm{5}.\mathrm{896}\rfloor=\mathrm{5} \\ $$$$ \\ $$ Answered by…
Question Number 66979 by Mr Jor last updated on 21/Aug/19 Commented by Mr Jor last updated on 21/Aug/19 $${In}\:{the}\:{figure}\:{above},\:{OC}=\mathrm{3}{CA}\:{and} \\ $$$${OD}=\mathrm{3}{DB}.{By}\:{taking}\:\boldsymbol{{OA}}=\boldsymbol{{a}},\boldsymbol{{OB}} \\ $$$$=\boldsymbol{{b}},{show}\:{that}\:{CB}//{AB}. \\ $$…