Question Number 225282 by fantastic last updated on 20/Oct/25 $${Q}\mathrm{225146} \\ $$$${here}\:{if}\:{the}\:{wedge}\:{and}\:{the} \\ $$$${ground}\:\:{had}\:{no}\:{friction} \\ $$$${it}\:{would}\:{start}\:{going}\:\rightarrow \\ $$$${what}\:{is}\:{acc}.\:{at}\:{time}\:{t}? \\ $$$$\mu={kx} \\ $$ Commented by mr…
Question Number 225241 by Spillover last updated on 18/Oct/25 Answered by som(math1967) last updated on 19/Oct/25 $$\:{tan}\theta=\frac{{nsin}\alpha{cos}\alpha.{sec}^{\mathrm{2}} \alpha}{\left(\mathrm{1}−{nsin}^{\mathrm{2}} \alpha\right){sec}^{\mathrm{2}} \alpha} \\ $$$$\Rightarrow{tan}\theta=\frac{{ntan}\alpha}{{sec}^{\mathrm{2}} \alpha−{ntan}^{\mathrm{2}} \alpha} \\…
Question Number 225240 by Spillover last updated on 18/Oct/25 Answered by mr W last updated on 19/Oct/25 $$\left({a}\right) \\ $$$${R}=\frac{\mid\mathrm{3}×\mathrm{5}−\mathrm{4}×\mathrm{4}+\mathrm{6}\mid}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }}=\mathrm{1} \\ $$$$\Rightarrow\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}}…
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Question Number 225230 by Abdulazim last updated on 18/Oct/25 $$\:\:\:{tg}^{\mathrm{4}} \mathrm{10}°+{tg}^{\mathrm{4}} \mathrm{50}°+{tg}^{\mathrm{4}} \mathrm{70}°=? \\ $$ Commented by Frix last updated on 18/Oct/25 $$\mathrm{Claim}: \\ $$$${x}^{\mathrm{3}}…
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Question Number 225098 by Ark55 last updated on 17/Oct/25 Answered by Rasheed.Sindhi last updated on 17/Oct/25 $$\mathrm{5}{x}+\frac{\mathrm{1}}{{x}}=\mathrm{6} \\ $$$$\left(\mathrm{5}{x}+\frac{\mathrm{1}}{{x}}=\mathrm{6}\right)^{\mathrm{2}} \\ $$$$\mathrm{25}{x}^{\mathrm{2}} +\mathrm{10}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{36} \\ $$$$\mathrm{25}{x}^{\mathrm{2}}…
Question Number 225152 by Spillover last updated on 18/Oct/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 225153 by Spillover last updated on 18/Oct/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 225075 by Ark55 last updated on 17/Oct/25 Answered by Frix last updated on 17/Oct/25 $$\mathrm{Obviously}\:{x}=\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{because}\:\mathrm{5}+\mathrm{1}=\mathrm{6}\:\mathrm{and}\:\mathrm{25}+\mathrm{1}=\mathrm{26} \\ $$$$\mathrm{or}\:{x}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\:\:\:\:\:\mathrm{because}\:\mathrm{1}+\mathrm{5}=\mathrm{6}\:\mathrm{and}\:\mathrm{1}+\mathrm{25}=\mathrm{26} \\ $$…