Question Number 136375 by nadovic last updated on 22/Mar/21 $$\mathrm{A}\:\mathrm{bowl}\:\mathrm{contains}\:\mathrm{carefully}\:\mathrm{shredded} \\ $$$$\mathrm{confetti},\:\mathrm{6}\:\mathrm{of}\:\mathrm{which}\:\mathrm{are}\:\mathrm{blue}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{remaining}\:\mathrm{12}\:\mathrm{are}\:\mathrm{red}. \\ $$$$\mathrm{Each}\:\mathrm{piece}\:\mathrm{of}\:\mathrm{confetti}\:\mathrm{is}\:\mathrm{either}\:\mathrm{circular} \\ $$$$\mathrm{triangular}\:\mathrm{or}\:\mathrm{rectangular}\:\mathrm{in}\:\mathrm{shape}\:\mathrm{and} \\ $$$$\mathrm{are}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{1}:\mathrm{3}:\mathrm{2}\:\mathrm{for}\:\mathrm{each}\:\mathrm{colour} \\ $$$$\mathrm{respectively}.\:\mathrm{If}\:\mathrm{they}\:\mathrm{are}\:\mathrm{thrown}\:\mathrm{up}\: \\ $$$$\boldsymbol{\mathrm{thrice}}\:\mathrm{in}\:\mathrm{a}\:\mathrm{certain}\:\mathrm{competition}\:\mathrm{and}\:\mathrm{a} \\…
Question Number 70834 by Maclaurin Stickker last updated on 08/Oct/19 Answered by mind is power last updated on 08/Oct/19 $$\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \frac{\mathrm{1}}{{log}_{{x}} \left(\mathrm{2}^{{k}} \right){log}_{{x}} \left(\mathrm{2}^{\left.{k}+\mathrm{1}\right)}…
Question Number 5297 by sanusihammed last updated on 05/May/16 Commented by sanusihammed last updated on 05/May/16 $${Where}\:{F}\:{is}\:\mathrm{917}\:{and}\:{L}\:{is}\:\mathrm{6} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 136370 by mathmax by abdo last updated on 21/Mar/21 $$\mathrm{letf}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} \:\mathrm{arctan}\left(\frac{\pi}{\mathrm{x}}\right) \\ $$$$\left.\mathrm{1}\right)\:\mathrm{calculate}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right) \\ $$$$\left.\mathrm{2}\right)\mathrm{calculate}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{1}\right) \\ $$$$\left.\mathrm{3}\right)\:\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{integer}\:\mathrm{serie} \\ $$ Answered by…
Question Number 136365 by liberty last updated on 21/Mar/21 $$\int\:\frac{{dx}}{\mathrm{sin}\:{x}\:\sqrt{\mathrm{cos}\:{x}}}\:=? \\ $$ Answered by mathmax by abdo last updated on 21/Mar/21 $$\mathrm{I}\:=\int\:\frac{\mathrm{dx}}{\mathrm{sinx}\sqrt{\mathrm{cosx}}}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{cosx}=\mathrm{t}^{\mathrm{2}} \:\Rightarrow\mathrm{x}=\mathrm{arcos}\left(\mathrm{t}^{\mathrm{2}} \right) \\…
Question Number 70831 by ajfour last updated on 08/Oct/19 Commented by ajfour last updated on 08/Oct/19 $${Stick}\:{of}\:{mass}\:{m},\:{length}\:{b}\:{is} \\ $$$${released}\:{as}\:{shown}.\:{As}\:{it}\:{slides} \\ $$$${down}\:{and}\:{breaks}\:{contact}\:{with} \\ $$$${hemisphere},\:{the}\:{top}\:{end}\:{is}\:{at} \\ $$$${angular}\:{position}\:\beta.\:{Assume}…
Question Number 70826 by naka3546 last updated on 08/Oct/19 Commented by naka3546 last updated on 08/Oct/19 $${Find}\:\:{x}\:. \\ $$ Answered by mind is power last…
Question Number 5291 by Rasheed Soomro last updated on 05/May/16 $$\mathrm{A}\:\mathrm{circle}\:\mathrm{has}\:\mathrm{been}\:\mathrm{drawn}\:\mathrm{with}\:\mathrm{one} \\ $$$$\mathrm{unit}\:\mathrm{opened}\:\mathrm{compass}\:\mathrm{on}\:\:\mathrm{surface} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{sphere}\:\mathrm{of}\:\mathrm{one}\:\mathrm{unit}\:\mathrm{radius}.\: \\ $$$$\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}? \\ $$ Commented by Yozzii last updated on…
Question Number 5290 by sanusihammed last updated on 05/May/16 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 136356 by SOMEDAVONG last updated on 21/Mar/21 $$\mathrm{L}=\underset{\mathrm{n}\rightarrow\propto} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{2}}\:+\:…+\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{n}}\right] \\ $$ Answered by mindispower last updated on 21/Mar/21 $$\underset{{k}\leqslant{n}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}}…