Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 143072 by Feruzbek last updated on 09/Jun/21

Answered by mr W last updated on 09/Jun/21

$$sayx=n+f$$$$10000n={(n+f)}^2⩾n^2$$$$n^2-10000n⩽0$$$$\Rightarrow 0⩽n⩽10000...\left(i\right)$$$$$$$$10000n={(n+f)}^2<{(n+1)}^2$$$$n^2-9998n+1>0$$$$\Rightarrow n<1orn>9997...\left(ii\right)$$$$$$$$\Rightarrow n=0or9998or9999or10000$$$$\Rightarrow x=100\sqrt{n}$$$$i.e.x=0or$$$$=100\sqrt{9998}or$$$$=100\sqrt{9999}=300\sqrt{1111}or$$$$=10000$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com