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Category: Arithmetic

6-5-14-6-14-23-6-23-32-6-9n-4-9n-5-

Question Number 113593 by bemath last updated on 14/Sep/20 $$\frac{\mathrm{6}}{\mathrm{5}×\mathrm{14}}+\frac{\mathrm{6}}{\mathrm{14}×\mathrm{23}}+\frac{\mathrm{6}}{\mathrm{23}×\mathrm{32}}+…+\frac{\mathrm{6}}{\left(\mathrm{9}{n}−\mathrm{4}\right)\left(\mathrm{9}{n}+\mathrm{5}\right)}=? \\ $$ Commented by bemath last updated on 14/Sep/20 $${thank}\:{you}\:{both}\:{sir} \\ $$ Answered by bobhans…

Question-47976

Question Number 47976 by peter frank last updated on 17/Nov/18 Commented by maxmathsup by imad last updated on 17/Nov/18 $${let}\:{A}\:=\int_{\frac{\mathrm{1}}{\mathrm{4}}} ^{\frac{\mathrm{3}}{\mathrm{4}}} \:\:\:\frac{\left(\sqrt{{x}−{x}^{\mathrm{2}} }\right)^{−\mathrm{1}} }{\left({dx}\right)^{−\mathrm{2}} }\:\left({dx}\right)^{−\mathrm{1}}…

Question-178932

Question Number 178932 by cherokeesay last updated on 22/Oct/22 Answered by MJS_new last updated on 23/Oct/22 $${S}_{{n}} =\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{8}^{{j}} =\frac{\mathrm{8}}{\mathrm{7}}\left(\mathrm{8}^{{n}} −\mathrm{1}\right) \\ $$$${S}_{\mathrm{1}} =\mathrm{8}=\mathrm{0}×\mathrm{73}+\mathrm{8}…

Question-178869

Question Number 178869 by ARUNG_Brandon_MBU last updated on 22/Oct/22 Answered by Rasheed.Sindhi last updated on 22/Oct/22 $$\:\:\:\:\:\:\:\:\:\begin{array}{|c|c|c|c|c|}{\:\mathrm{2}\:}&\hline{\:\:\:\:\:\:\:\:\mathrm{4}}&\hline{\:\mathrm{8}}\\{\:}&\hline{\:}&\hline{\:}\\{\:\mathrm{7}}&\hline{\:}&\hline{\:\mathrm{3}}\\{\:}&\hline{\:\begin{array}{|c|}{?=\mathrm{1}}\\\hline\end{array}\:}&\hline{\:}\\{\:\mathrm{5}}&\hline{\:}&\hline{\:\mathrm{6}\:}\\\hline\end{array}\:\: \\ $$ Commented by ARUNG_Brandon_MBU last updated on…

Question-178870

Question Number 178870 by ARUNG_Brandon_MBU last updated on 22/Oct/22 Answered by Rasheed.Sindhi last updated on 22/Oct/22 $$\mathrm{Rule}: \\ $$$$\begin{array}{|c|c|c|}{\begin{array}{|c|}{{a}}\\\hline\end{array}}&\hline{\:\:\:\:\:}&\hline{\begin{array}{|c|}{{b}}\\\hline\end{array}}\\{\:\:\:}&\hline{\begin{array}{|c|}{\frac{{ad}+{bc}}{\mathrm{2}}}\\\hline\end{array}}&\hline{\:\:}\\{\begin{array}{|c|}{{c}}\\\hline\end{array}}&\hline{\:\:\:}&\hline{\begin{array}{|c|}{{d}}\\\hline\end{array}}\\\hline\end{array}\:\:\: \\ $$$$\mathrm{Applying}\:\mathrm{rule}: \\ $$$$\begin{array}{|c|c|c|}{\begin{array}{|c|}{{x}+\mathrm{2}}\\\hline\end{array}}&\hline{\:\:\:\:\:}&\hline{\begin{array}{|c|}{{x}+\mathrm{1}}\\\hline\end{array}}\\{\:\:\:}&\hline{\begin{array}{|c|}{\frac{\left({x}+\mathrm{2}\right)\left(\mathrm{3}\right)+\left({x}\right)\left({x}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{1}}\\\hline\end{array}}&\hline{\:\:}\\{\begin{array}{|c|}{\:\:\:\:{x}\:\:\:}\\\hline\end{array}}&\hline{\:\:\:}&\hline{\begin{array}{|c|}{\:\:\:\:\mathrm{3}\:\:}\\\hline\end{array}}\\\hline\end{array} \\ $$$$\:\:\:\frac{\left({x}+\mathrm{2}\right)\left(\mathrm{3}\right)+\left({x}\right)\left({x}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{1}…