Question Number 143077 by mathdanisur last updated on 09/Jun/21 $${sin}^{\mathrm{5}} {x}\:+\:{cos}^{\mathrm{5}} {x}\:=\:\mathrm{2}\:−\:{sin}^{\mathrm{4}} {x} \\ $$ Answered by MJS_new last updated on 10/Jun/21 $$\mathrm{sin}\:{x}\:={s} \\ $$$$\mathrm{cos}\:{x}\:={c}…
Question Number 77524 by BK last updated on 07/Jan/20 Commented by lémùst last updated on 07/Jan/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{question}\:? \\ $$ Commented by BK last updated on…
Question Number 77513 by BK last updated on 07/Jan/20 Commented by Kunal12588 last updated on 07/Jan/20 -99631646050232081463391082912554874214147786127212261300098296284926069581800964669753771227589064527729684298473399551387739150433334227370085293404740389579109830756667470298828368996394719909762120 Commented by MJS last updated on 07/Jan/20 $$\mathrm{let}'\mathrm{s}\:\mathrm{just}\:\mathrm{give}\:\mathrm{it}\:\mathrm{a}\:\mathrm{name}…
Question Number 143041 by bramlexs22 last updated on 09/Jun/21 $${If}\:{f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}\:{then}\:{the}\:{value} \\ $$$${of}\:\left(\mathrm{1}−\frac{\mathrm{2}}{{f}\left(\mathrm{1}\right)}\right)\left(\mathrm{1}−\frac{\mathrm{2}}{{f}\left(\mathrm{2}\right)}\right)\left(\mathrm{1}−\frac{\mathrm{2}}{{f}\left(\mathrm{3}\right)}\right)…\left(\mathrm{1}−\frac{\mathrm{2}}{{f}\left(\mathrm{2021}\right)}\right)=? \\ $$ Commented by SLVR last updated on 13/Jun/21 $${whats}\:{the}\:{answwer}???{and}\:{solution} \\ $$…
Question Number 77506 by BK last updated on 07/Jan/20 Commented by BK last updated on 07/Jan/20 $$\left.\mathrm{A}\left.\right)\left.\mathrm{2}\left.\mathrm{011}\:\:\:\:\:\:\:\:\:\:\:\mathrm{B}\right)\mathrm{2012}\:\:\:\:\:\:\:\:\:\:\:\mathrm{C}\right)\mathrm{4021}\:\:\:\:\:\:\:\:\:\mathrm{D}\right)\mathrm{4023} \\ $$ Commented by MJS last updated on…
Question Number 143036 by ajfour last updated on 09/Jun/21 $$\:{x}^{\mathrm{3}/\mathrm{2}} +{x}^{\mathrm{1}/\mathrm{2}} +\left({x}−{c}\right)\left(\frac{\mathrm{3}{x}+\mathrm{1}}{\mathrm{3}+{x}}\right)^{\mathrm{3}/\mathrm{2}} =\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 143024 by Feruzbek last updated on 09/Jun/21 Answered by bramlexs22 last updated on 09/Jun/21 $$\left(\mathrm{1}\right)\Leftrightarrow\:\mathrm{49}−{x}^{\mathrm{2}} \:\leqslant\:\mathrm{24}\: \\ $$$$\Leftrightarrow\:\mathrm{25}\:−{x}^{\mathrm{2}} \leqslant\mathrm{0} \\ $$$$\Leftrightarrow\:\left({x}+\mathrm{5}\right)\left({x}−\mathrm{5}\right)\geqslant\mathrm{0} \\ $$$$\Leftrightarrow\:{x}\leqslant−\mathrm{5}\:\cup\:{x}\geqslant\mathrm{5}…
Question Number 143018 by aupo14 last updated on 08/Jun/21 Answered by Dwaipayan Shikari last updated on 08/Jun/21 $$\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:\:\:\:\:\:\:\:{n}\:{starts}\:{from}\:\mathrm{1} \\ $$ Terms of Service Privacy…
Question Number 143011 by ajfour last updated on 08/Jun/21 $$\:\:{x}^{\mathrm{4}} +{bx}^{\mathrm{2}} +{cx}={s} \\ $$$${let}\:\:{x}^{\mathrm{2}} ={px}+{t} \\ $$$$\Rightarrow\:{p}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{ptx}+{t}^{\mathrm{2}} +{bx}^{\mathrm{2}} +{cx}={s} \\ $$$$\Rightarrow\:\left({p}^{\mathrm{2}} +{b}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{pt}+{c}\right){x}…
Question Number 143006 by mathdanisur last updated on 08/Jun/21 $$\mathrm{1}.\:{y}\frac{\partial{z}}{\partial{x}}\:+\:{z}\frac{\partial{z}}{\partial{y}}\:=\:\frac{{y}}{{x}} \\ $$$$\mathrm{2}.\:{x}^{\mathrm{2}} \frac{\partial{z}}{\partial{x}}\:−\:{xy}\frac{\partial{z}}{\partial{y}}\:+\:{y}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{3}.\:\begin{cases}{\frac{\partial{z}}{\partial{x}}\:=\:\frac{{z}}{{x}}}\\{\frac{\partial{z}}{\partial{y}}\:=\:\frac{\mathrm{2}{z}}{{y}}}\end{cases} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com