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Question-159744

Question Number 159744 by 0731619 last updated on 20/Nov/21 Answered by gsk2684 last updated on 20/Nov/21 $$\frac{\mathrm{0}}{\mathrm{0}}\:{form},\:{apply}\:{L}'{hospital}\:{rule} \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{−\left(\mathrm{cos}\:\frac{\pi}{{x}}\right)\left(−\frac{\pi}{{x}^{\mathrm{2}} }\right)}{\left(\mathrm{sin}\:\pi{x}\right)\pi}=\underset{{x}\rightarrow\mathrm{2}} {\frac{\mathrm{1}}{\mathrm{4}}\mathrm{lim}\frac{\left(\mathrm{cos}\:\frac{\pi}{{x}}\right)}{\left(\mathrm{sin}\:\pi{x}\right)}\:} \\ $$$${again}\:{apply}\:{the}\:{same}\:{rule} \\…

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} \\…