<?xml version="1.0" encoding="UTF-8"?><rss xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:content="http://purl.org/rss/1.0/modules/content/" xmlns:atom="http://www.w3.org/2005/Atom" version="2.0"><channel><title><![CDATA[RSS Feed]]></title><description><![CDATA[RSS Feed]]></description><link>https://ecency.com</link><image><url>https://ecency.com/logo512.png</url><title>RSS Feed</title><link>https://ecency.com</link></image><generator>RSS for Node</generator><lastBuildDate>Tue, 15 Sep 2026 22:56:27 GMT</lastBuildDate><atom:link href="https://ecency.com/@winko2019/rss" rel="self" type="application/rss+xml"/><item><title><![CDATA[Strrejohn say]]></title><description><![CDATA[Thanks bytebsll 4 your free adrp]]></description><link>https://ecency.com/@winko2019/strrejohn-say</link><guid isPermaLink="true">https://ecency.com/@winko2019/strrejohn-say</guid><category><![CDATA[mse]]></category><dc:creator><![CDATA[winko2019]]></dc:creator><pubDate>Mon, 23 Jul 2018 14:02:09 GMT</pubDate></item><item><title><![CDATA[Tommow cet]]></title><description><![CDATA[logp-logq , then p=………. A. p=q=1 B. p=q1-q C. p=q21-q D. p= q1+q E. p=q21+q 25.If loga=5, logb =3, then the value of ab is A. 53 B. 2 C. 8 D. log53 E.100 Given that loga2 =0.301 and loga3 =0.477, then]]></description><link>https://ecency.com/@winko2019/tommow-cet</link><guid isPermaLink="true">https://ecency.com/@winko2019/tommow-cet</guid><category><![CDATA[intro]]></category><dc:creator><![CDATA[winko2019]]></dc:creator><pubDate>Mon, 23 Jul 2018 13:49:33 GMT</pubDate></item><item><title><![CDATA[Lostof Ml]]></title><description><![CDATA[Lost ml fking server]]></description><link>https://ecency.com/@winko2019/lostof-ml</link><guid isPermaLink="true">https://ecency.com/@winko2019/lostof-ml</guid><category><![CDATA[ml]]></category><dc:creator><![CDATA[winko2019]]></dc:creator><pubDate>Mon, 23 Jul 2018 13:40:42 GMT</pubDate></item><item><title><![CDATA[Intro post]]></title><description><![CDATA[I m winko and from mawlamyaing.]]></description><link>https://ecency.com/@winko2019/intro-post</link><guid isPermaLink="true">https://ecency.com/@winko2019/intro-post</guid><category><![CDATA[intro]]></category><dc:creator><![CDATA[winko2019]]></dc:creator><pubDate>Mon, 23 Jul 2018 10:40:06 GMT</pubDate></item></channel></rss>