Question Number 61521 by necx1 last updated on 03/Jun/19 Answered by ajfour last updated on 04/Jun/19 Commented by ajfour last updated on 04/Jun/19 $${FB}\:{line}=\frac{{b}}{\mathrm{4}}+\lambda\left({a}−\frac{{b}}{\mathrm{4}}\right) \\…
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Question Number 192560 by Ari last updated on 20/May/23 Commented by dubylee last updated on 22/May/23 $${what}\:{is}\:{the}\:{question}\:{here}? \\ $$ Commented by Skabetix last updated on…
Question Number 127018 by ‘-E/9 last updated on 26/Dec/20 $${Vf}×\mathrm{2}={Vi}×\mathrm{2}+\mathrm{2}{a}\Delta{d} \\ $$$$\mathrm{0}=\mathrm{16}.\mathrm{5} \\ $$ Answered by physicstutes last updated on 26/Dec/20 $$\mathrm{you}\:\mathrm{should}\:\mathrm{write}\:\mathrm{it}\:\mathrm{this}\:\mathrm{way}. \\ $$$$\:\:{v}_{{f}} ^{\mathrm{2}}…
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Question Number 61449 by Tawa1 last updated on 02/Jun/19 Commented by Tawa1 last updated on 02/Jun/19 $$\frac{\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} }{\mathrm{c}^{\mathrm{2}} }\:\:=\:\:? \\ $$ Answered by perlman…
Question Number 61424 by Tawa1 last updated on 02/Jun/19 Answered by mr W last updated on 02/Jun/19 Commented by mr W last updated on 02/Jun/19…
Question Number 192469 by Mingma last updated on 18/May/23 Answered by a.lgnaoui last updated on 19/May/23 $$\bigtriangleup\mathrm{ABC}\:\:\mathrm{triangle}\:\mathrm{isocele}\:\:\mathrm{AB}=\mathrm{AC} \\ $$$$\mathrm{donc}\:\:\measuredangle\mathrm{B}=\measuredangle\mathrm{C}=\mathrm{90}−\frac{\mathrm{78}}{\mathrm{2}}=\mathrm{51}° \\ $$$$\:\:\:\:\:\:\frac{\mathrm{sin}\:\mathrm{78}}{\boldsymbol{\mathrm{BC}}}=\frac{\mathrm{sin}\:\mathrm{51}}{\boldsymbol{\mathrm{AC}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\bigtriangleup\mathrm{ABD}\:\:\:\:\mathrm{AB}=\mathrm{BD}\:\:\:\measuredangle\mathrm{CAD}=\mathrm{A}_{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:…
Question Number 61347 by Tawa1 last updated on 01/Jun/19 Commented by mr W last updated on 02/Jun/19 $${x}=\mathrm{20}°,\:{see}\:{also}\:{Q}\mathrm{43728}. \\ $$ Answered by peter frank last…
Question Number 61335 by necx1 last updated on 01/Jun/19 Answered by mr W last updated on 02/Jun/19 $$\mathrm{cos}\:\angle{A}=\frac{\mathrm{5}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} }{\mathrm{2}×\mathrm{5}×\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{7}} \\ $$$$\Rightarrow\mathrm{sin}\:\angle{A}=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{7}} \\ $$$$\mathrm{cos}\:\angle{B}=\frac{\mathrm{5}^{\mathrm{2}}…