Math 206 Homework
Forfatter
Jibri L Kea
Sidst opdateret
9 år siden
Licens
Creative Commons CC BY 4.0
Resumé
Math 206 homework
Math 206 homework
%description: Math 290 HW Template
%%%%% Beginning of preamble %%%%%
\documentclass[12pt]{article} %What kind of document (article) and what size
%Packages to load which give you useful commands
\usepackage{graphicx}
\usepackage{amssymb, amsmath, amsthm}
%Sets the margins
\textwidth = 7 in
\textheight = 9.5 in
\oddsidemargin = -0.3 in
\evensidemargin = -0.3 in
\topmargin = -0.4 in
\headheight = 0.0 in
\headsep = 0.0 in
\parskip = 0.2in
\parindent = 0.0in
%defines a few theorem-type environments
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}{Definition}
%%%%% End of preamble %%%%%
\begin{document}
%Identification, Change as necessary!
{\Large Jibri Kea} \hfill
{\large Math 206, Section 2,} %Delete one
\hfill \today
I worked with:
\begin{enumerate}
\item \textbf{Problem 6: Number Systems} Perform D3F02 (hex) + C34D (hex). State your answer in hexadecimal and decimal.
D3F02h + C34Dh
\[ D 3 F 0 2 h\]
\[1101 + 0011 + 1111 + 0000 + 0010 b\]
\[ 13 + 3 + 15 + 0 + 2 d\]
\[13*16^{4} + 3*16^{3} + 15*16^{2} + 0*16^{1} + 2*16^{0}\]
\[851,968 + 12,288 + 3,840 + 0 + 2d\]
\[868,098d\]
\[C34Dh\]
\[1100+0011+0100+1101b\]
\[12+3+4+13d\]
\[12*16^{3} + 3*16^{2} + 4*16^{1} + 13*16^{0}\]
\[49,152+768+64+13d\]
\[49,997d\]
\[D3F02h+C34Dh= 918,095d\]
\[11010011111100000010b\]
\[+1100001101001101b\]
\[11100000001001001111b\]
\[14+0+2+4+15h\]
\[E024Fh = 918,095d\]
\item \textbf{Problem 8: Number Systems} Convert 109 to base 7.
\[109 / 7 = 15 R4\]
\[15 / 7 = 2 R1\]
\[214_{7}\]
\[2*2^{2}*7^{1}+4*7^{0}\]
\[98+7+4 =109d=214_{7}\]
\item \textbf{Problem 19} Find a closed form.
\[1,4,7,10,13,16,19\]
\[a_{n} = c+dn+fn^{2}\]
\[a_{1} = c+d+f=1\]
\[a_{2} = c+2d+4f=4\]
\[a_{3} = c+3d+9f=7\]
\[c=-2\]
\[d=3\]
\[f=0\]
\[a_{n}=-2+3n\]
\item \textbf{Problem 19} Find a closed form.
\[-1,1,3,5,7,9,11\]
For n is greater than/equal to 0.
\[GUESS a_{n}=2_{n-1}\]
\[n=0: True -1=0-1=-1\]
Assume for some n >= 1. a(k) = 2(k) - 1
\[=4n-2-2n+3=2n+1=2(n+1)-1\]
\end{enumerate}
\end{document}