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Question-94191

Question Number 94191 by oustmuchiya@gmail.com last updated on 17/May/20 Answered by mr W last updated on 17/May/20 $$\overset{\rightarrow} {{AD}}=\boldsymbol{{c}}+\frac{\boldsymbol{{a}}}{\mathrm{2}} \\ $$$$\overset{\rightarrow} {{BE}}=\boldsymbol{{a}}+\frac{\boldsymbol{{b}}}{\mathrm{2}} \\ $$$$\overset{\rightarrow} {{CF}}=\boldsymbol{{b}}+\frac{\boldsymbol{{c}}}{\mathrm{2}}…

Question-94158

Question Number 94158 by seedhamaieng@gmail.com last updated on 17/May/20 Commented by prakash jain last updated on 17/May/20 $$\mathrm{mass}\:{m} \\ $$$$\mathrm{acceleration}\:{a}=−\frac{{R}}{{m}} \\ $$$$\:\:\:\:\:−{ve}\:{acclerarion}\:{as}\:{mention}\:{in} \\ $$$$\:\:\:\:\:\:{question} \\…

Question-159690

Question Number 159690 by 0731619 last updated on 20/Nov/21 Answered by MJS_new last updated on 21/Nov/21 $$\mathrm{very}\:\mathrm{obviously}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{is}\:+\infty \\ $$$$\mathrm{check}\:\mathrm{for}\:{x}=\mathrm{5} \\ $$$$\frac{\mathrm{5}^{\mathrm{5}!} }{\left(\mathrm{5}!\right)^{\mathrm{5}} }=\frac{\mathrm{5}^{\mathrm{120}} }{\mathrm{120}^{\mathrm{5}} }=\frac{\approx\mathrm{10}^{\mathrm{84}}…

lim-x-1-1-x-1-x-ln-x-Without-L-Hospital-

Question Number 94134 by naka3546 last updated on 17/May/20 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\left(\:\frac{\mathrm{1}}{{x}−\mathrm{1}}\:−\:\frac{{x}}{\mathrm{ln}\:{x}}\:\right)\:=\:… \\ $$$$\left(\:{Without}\:\:{L}'\:{Hospital}\:\right) \\ $$ Commented by $@ty@m123 last updated on 17/May/20 $${Go}\:{To}\:\:{Q}.\:{No}.\:\mathrm{87105}\: \\ $$…