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Category: Arithmetic

Question-48090

Question Number 48090 by peter frank last updated on 19/Nov/18 Answered by tanmay.chaudhury50@gmail.com last updated on 19/Nov/18 1)truevalueofproductoflength×width=2.29×1.29=(2.300.01)(1.300.01)=2.30×1.300.01(2.30+1.30)+0.0001$$=\mathrm{2}.\mathrm{30}×\mathrm{1}.\mathrm{30}−\mathrm{3}.\mathrm{60}×\mathrm{0}.\mathrm{01}+\mathrm{0}.\mathrm{0001} \

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}}…