Question Number 223685 by fantastic last updated on 02/Aug/25 $$\mathrm{25}^{{x}} −\mathrm{8}.\mathrm{5}^{{x}} =−\mathrm{16} \\ $$ Answered by Rasheed.Sindhi last updated on 02/Aug/25 $$\left(\mathrm{5}^{{x}} \right)^{\mathrm{2}} −\mathrm{8}\left(\mathrm{5}^{{x}} \right)=−\mathrm{16}…
Question Number 223703 by Rojarani last updated on 02/Aug/25 Answered by Rasheed.Sindhi last updated on 02/Aug/25 $$\left(\frac{{x}−\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} =\frac{\mathrm{4}}{\mathrm{9}}+\sqrt[{\mathrm{4}}]{\frac{\mathrm{3}}{\mathrm{27}}}\:+\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}}{\mathrm{27}}}\: \\ $$$$\frac{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1}}{\mathrm{27}}=\frac{\mathrm{4}}{\mathrm{9}}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}}}{\mathrm{3}}\:+\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}}}{\mathrm{3}}\: \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}}…
Question Number 223626 by fantastic last updated on 01/Aug/25 $$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{25}}\\{{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{yz}=\mathrm{49}}\\{{z}^{\mathrm{2}} +{x}^{\mathrm{2}} +{zx}=\mathrm{64}}\end{cases} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} −\mathrm{100}=?? \\ $$ Commented by Frix last…
Question Number 223636 by yerlow last updated on 01/Aug/25 $$\mathrm{Factor}\:\mathrm{the}\:\mathrm{following}\:\mathrm{expression}: \\ $$$$\left(\sqrt[{\mathrm{5}}]{\mathrm{arctan}\left({x}^{\mathrm{5}} +\mathrm{1}\right)}\right)^{{x}^{−{x}^{\mathrm{2}} } } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 223655 by Rojarani last updated on 01/Aug/25 Commented by MathematicalUser2357 last updated on 01/Aug/25 $${is}\:{this}\:{like}: \\ $$ Commented by MathematicalUser2357 last updated on…
Question Number 223615 by fantastic last updated on 31/Jul/25 $$\mathrm{40}^{{x}−\mathrm{1}} =\mathrm{2}^{\mathrm{2}{x}+\mathrm{1}} \\ $$ Answered by mr W last updated on 31/Jul/25 $$\frac{\mathrm{40}^{{x}} }{\mathrm{40}}=\mathrm{2}×\mathrm{2}^{\mathrm{2}{x}} =\mathrm{2}×\mathrm{4}^{{x}} \\…
Question Number 223585 by hardmath last updated on 30/Jul/25 Commented by hardmath last updated on 30/Jul/25 $$\mathrm{Radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:=\:\mathrm{R} \\ $$$$\mathrm{AB}\:\bot\:\mathrm{CD} \\ $$$$\mathrm{AC}\:=\:\mathrm{a} \\ $$$$\mathrm{BD}\:=\:\mathrm{b} \\ $$$$\mathrm{Prove}\:\mathrm{that}:\:\mathrm{R}\:=\:\frac{\sqrt{\mathrm{a}^{\mathrm{2}}…
Question Number 223571 by hardmath last updated on 30/Jul/25 $$\mathrm{S}_{\mathrm{1}} \:=\:\mathrm{1}\centerdot\mathrm{1}!\:+\:\mathrm{2}\centerdot\mathrm{2}!\:+\:\mathrm{3}\centerdot\mathrm{3}!\:+…+\:\mathrm{16}\centerdot\mathrm{16}! \\ $$$$\mathrm{S}_{\mathrm{2}} \:=\:\mathrm{1}\centerdot\mathrm{1}!\:+\:\mathrm{2}\centerdot\mathrm{2}!\:+\:\mathrm{3}\centerdot\mathrm{3}!\:+…+\:\mathrm{14}\centerdot\mathrm{14}! \\ $$$$\mathrm{Find}:\:\:\:\frac{\mathrm{S}_{\mathrm{1}} }{\mathrm{S}_{\mathrm{2}} }\:=\:? \\ $$ Answered by parthasc last updated…
Question Number 223569 by Atomusmaths last updated on 29/Jul/25 Answered by Raphael254 last updated on 30/Jul/25 $$ \\ $$$$\mathrm{5}^{\mathrm{50}} \:=\:\left(\mathrm{5}^{{x}} \right)^{\frac{\mathrm{50}}{{x}}} \\ $$$$\mathrm{5}^{{x}} \:=\:\frac{\mathrm{50}}{{x}} \\…
Question Number 223508 by Rojarani last updated on 27/Jul/25 Terms of Service Privacy Policy Contact: info@tinkutara.com