PageSourceSearch

https://21ifm.com/static/js/9962.081792b3.chunk.js

js 21ifm.com collected 2026-10-07 11:02:23 UTC 3,746 bytes, 2 lines download raw bytes

1"use strict";(self.webpackChunkmagic_link_auth=self.webpackChunkmagic_link_auth||[]).push([[9962,7706,1257],{7706:(n,e,o)=>{o.r(e),o.d(e,{default:()=>r});o(82483);var t=o(56723);const r=n=>{let{url:e,label:o,color:r,openVideo:l}=n;return(0,t.jsx)("div",{style:{textAlign:"center",margin:"20px 0"},children:(0,t.jsxs)("button",{onClick:()=>l(e),style:{backgroundColor:r||"#007BFF",color:"white",padding:"10px 20px",fontSize:"16px",borderRadius:"8px",border:"none",cursor:"pointer"},children:["\ud83c\udfa5 ",o]})})}},31257:(n,e,o)=>{o.r(e),o.d(e,{default:()=>r});o(82483);var t=o(56723);const r=n=>{let{videoUrl:e,onClose:o}=n;return e?(0,t.jsx)("div",{style:{position:"fixed",top:0,left:0,width:"100%",height:"100%",backgroundColor:"rgba(0, 0, 0, 0.8)",display:"flex",justifyContent:"center",alignItems:"center",zIndex:1e3},children:(0,t.jsxs)("div",{style:{position:"relative",backgroundColor:"#fff",padding:"20px",borderRadius:"8px",maxWidth:"800px",width:"90%",textAlign:"center"},children:[(0,t.jsx)("button",{onClick:o,style:{position:"absolute",top:"10px",right:"10px",background:"red",color:"white",border:"none",fontSize:"16px",padding:"5px 10px",borderRadius:"5px",cursor:"pointer"},children:"\u2716 Close"}),(0,t.jsx)("iframe",{width:"100%",height:"400px",src:e,title:"Video Explanation",frameBorder:"0",allowFullScreen:!0})]})}):null}},29962:(n,e,o)=>{o.r(e),o.d(e,{default:()=>c});var t=o(82483),r=o(80736),l=o(82327),a=o(25413),i=(o(70119),o(31257)),s=o(7706),d=o(56723);const u={exponentialGrowthDecay:"\n# **\ud83d\udccc Exponential Growth & Decay**\n\ud83d\udca1 **Reminder for Students**  \n- **Growth Formula:**  \n$\ny = y_0 e^{kt}\n$\n- **Decay Formula (Half-Life):**  \n$\ny = y_0 e^{-kt}\n$\n- **Doubling Time Formula:**  \n$\nT = \\frac{\\ln(2)}{k}\n$\n\n---\n\n## **\ud83d\udd35 Problem 348: Doubling Time Formula**\n\u2714 **Step 1: Use the formula**\n$\nT = \\frac{\\ln(2)}{\\ln(c)}\n$\n\n\ud83d\udccc **True or False?** \u2705 **TRUE**.\n\n---\n\n## **\ud83d\udd35 Problem 349: Investment Growth Comparison**\n\u2714 **Step 1: Compare Compounded Interest**\n- **3% continuous interest formula:**\n$\nA = P e^{0.03t}\n$\n- **2.5% continuous interest formula:**\n$\nA = P e^{0.025t}\n$\n\u2714 **Step 2: Compare after 1 year**\n$\n500e^{0.03} > 500e^{0.025}\n$\n\n\ud83d\udccc **Final Answer:** \u2705 **TRUE**.\n\n---\n\n## **\ud83d\udd35 Problem 352: Bacteria Growth**\n\u2714 **Step 1: Use Doubling Formula**\n$\nT = \\frac{\\ln(2)}{k}, \\quad 3 = \\frac{\\ln(2)}{k}\n$\n\u2714 **Step 2: Solve for k**\n$\nk = \\frac{\\ln(2)}{3}\n$\n\u2714 **Step 3: Find time for factor of 10**\n$\nt = \\frac{\\ln(10)}{k} = \\frac{\\ln(10)}{\\ln(2)/3} = \\frac{3 \\ln(10)}{\\ln(2)}\n$\n\n\ud83d\udccc **Final Answer:** **About 10 hours**.\n\n---\n\n## **\ud83d\udd35 Problem 354: Skull Age Calculation**\n\u2714 **Step 1: Use Half-Life Formula**\n$\ny = y_0 e^{-kt}\n$\n\u2714 **Step 2: Use Half-Life of 5730 Years**\n$\n\\frac{1}{5} = e^{-\\frac{t}{5730} \\ln 2}\n$\n\u2714 **Step 3: Solve for t**\n$\nt = \\frac{\\ln(1/5)}{-\\ln 2} \\times 5730\n$\n\n\ud83d\udccc **Final Answer:** **Around 13,200 years**.\n\n---\n\n## \ud83c\udfa5 **Want a Video Explanation?**\nClick below to watch a **step-by-step video explanation**.\n\n"},c=()=>{const[n,e]=(0,t.useState)(null);return(0,d.jsxs)("div",{className:"p-6 max-w-4xl mx-auto bg-white shadow-lg rounded-xl border",children:[(0,d.jsx)(r.oz,{children:u.exponentialGrowthDecay,remarkPlugins:[l.A],rehypePlugins:[a.A]}),(0,d.jsx)(i.default,{videoUrl:n,onClose:()=>e(null)}),(0,d.jsx)(s.default,{url:"https://www.youtube.com/embed/exponential_growth",label:"Watch Video on Exponential Growth & Decay",color:"#007BFF",openVideo:e})]})}}}]);
2//# sourceMappingURL=9962.081792b3.chunk.js.map

Line numbers count LF bytes from the start of the resource, as the search results do. Vendor segments are library code the classifier recognised; they are stored but not indexed. Bytes are shown as Latin1 characters, one per byte.