Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 5771 by Rasheed Soomro last updated on 27/May/16

Find to the nearest hundredth the positive cube-root of 29 .

$$\mathrm{Find}\:\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\:\mathrm{hundredth}\:\mathrm{the}\:\mathrm{positive} \\ $$$$\mathrm{cube}-\mathrm{root}\:\mathrm{of}\:\mathrm{29}\:. \\ $$

Commented by Yozzii last updated on 27/May/16

(29)^(1/3) =(2+27)^(1/3) =3(1+(2/(27)))^(1/3) For ∣x∣<1 and n∈R, we can write (1+x)^n =1+nx+((n(n−1))/(2!))x^2 +((n(n−1)(n−2))/(3!))x^3 +...+((n(n−1)(n−2)...(n−r+1))/(r!))x^r +... ∴ (1+(2/(27)))^(1/3) =1+((1/3))((2/(27)))+(((1/3)(−2/3))/(2!))((2/(27)))^2 +(((1/3)(−2/3)(−5/3))/(3!))((2/(27)))^3 +... (1+(2/(27)))^(1/3) ≈1+(2/(81))−(4/(6561))+((40)/(1594323))=1.0241 (4d.p) ∴29^(1/3) ≈3×1.0241=3.07 (2d.p)

$$\left(\mathrm{29}\right)^{\mathrm{1}/\mathrm{3}} =\left(\mathrm{2}+\mathrm{27}\right)^{\mathrm{1}/\mathrm{3}} =\mathrm{3}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{27}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${For}\:\mid{x}\mid<\mathrm{1}\:{and}\:{n}\in\mathbb{R},\:{we}\:{can}\:{write} \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}} =\mathrm{1}+{nx}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}{x}^{\mathrm{3}} +...+\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)...\left({n}−{r}+\mathrm{1}\right)}{{r}!}{x}^{{r}} +... \\ $$$$\therefore\:\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{27}}\right)^{\mathrm{1}/\mathrm{3}} =\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\frac{\mathrm{2}}{\mathrm{27}}\right)+\frac{\left(\mathrm{1}/\mathrm{3}\right)\left(−\mathrm{2}/\mathrm{3}\right)}{\mathrm{2}!}\left(\frac{\mathrm{2}}{\mathrm{27}}\right)^{\mathrm{2}} +\frac{\left(\mathrm{1}/\mathrm{3}\right)\left(−\mathrm{2}/\mathrm{3}\right)\left(−\mathrm{5}/\mathrm{3}\right)}{\mathrm{3}!}\left(\frac{\mathrm{2}}{\mathrm{27}}\right)^{\mathrm{3}} +... \\ $$$$\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{27}}\right)^{\mathrm{1}/\mathrm{3}} \approx\mathrm{1}+\frac{\mathrm{2}}{\mathrm{81}}−\frac{\mathrm{4}}{\mathrm{6561}}+\frac{\mathrm{40}}{\mathrm{1594323}}=\mathrm{1}.\mathrm{0241}\:\left(\mathrm{4}{d}.{p}\right) \\ $$$$\therefore\mathrm{29}^{\mathrm{1}/\mathrm{3}} \approx\mathrm{3}×\mathrm{1}.\mathrm{0241}=\mathrm{3}.\mathrm{07}\:\:\left(\mathrm{2}{d}.{p}\right) \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com