### 计算机代写|计算机图形学作业代写computer graphics代考| Hexadecimal Numbers

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等楖率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 计算机代写|计算机图形学作业代写computer graphics代考|Hexadecimal Numbers

The hexadecimal number system has $B=16$, and $a$ to $h$ can be 0 to 15 , which presents a slight problem, as we don’t have 15 different numerical characters. Consequently, we use 0 to 9 , and the letters $A, B, C, D, E, F$ to represent $10,11,12,13,14,15$ respectively:

$$\ldots a 16^{3}+b 16^{2}+c 16^{1}+d 16^{0}+e 16^{-1}+f 16^{-2}+g 16^{-3}+h 16^{-4} \ldots$$
and the first 17 hexadecimal numbers are:
$$1_{16}, 2_{16}, 3_{16}, 4_{16}, 5_{16}, 6_{16}, 7_{16}, 8_{16}, 9_{16}, A_{16}, B_{16}, C_{16}, D_{16}, E_{16}, F_{16}, 10_{16}, 11_{16}$$
Thus $1 E .8_{16}$ is converted to decimal as follows:
$$\begin{gathered} (1 \times 16)+(E \times 1)+\left(8 \times 16^{-1}\right) \ (16+14)+(8 / 16) \ 30.5 . \end{gathered}$$
Although it is not obvious, binary, octal and hexadecimal numbers are closely related, which is why they are part of a programmer’s toolkit. Even though computers work with binary, it’s the last thing a programmer wants to use. So to simplify the manmachine interface, binary is converted into octal or hexadecimal. To illustrate this, let’s convert the 16-bit binary code 1101011000110001 into octal.
Using the following general binary integer
$$a 2^{8}+b 2^{7}+c 2^{6}+d 2^{5}+e 2^{4}+f 2^{3}+g 2^{2}+h 2^{1}+i 2^{0}$$
we group the terms into threes, starting from the right, because $2^{3}=8$ :
$$\left(a 2^{8}+b 2^{7}+c 2^{6}\right)+\left(d 2^{5}+e 2^{4}+f 2^{3}\right)+\left(g 2^{2}+h 2^{1}+i 2^{0}\right)$$
Simplifying:
$$\begin{gathered} 2^{6}\left(a 2^{2}+b 2^{1}+c 2^{0}\right)+2^{3}\left(d 2^{2}+e 2^{1}+f 2^{0}\right)+2^{0}\left(g 2^{2}+h 2^{1}+i 2^{0}\right) \ 8^{2}\left(a 2^{2}+b 2^{1}+c 2^{1}\right)+8^{1}\left(d 2^{2}+e 2^{1}+f 2^{0}\right)+8^{0}\left(g 2^{2}+h 2^{1}+i 2^{0}\right) \ 8^{2} R+8^{1} S+8^{0} T \end{gathered}$$
where
\begin{aligned} &R=a 2^{2}+b 2^{1}+c \ &S=d 2^{2}+e 2^{1}+f \ &T=g 2^{2}+h 2^{1}+i \end{aligned}
and the values of $R, S, T$ vary between 0 and 7 . Therefore, given 1101011000 110001 , we divide the binary code into groups of three, starting at the right, and adding two leading zeros.

## 计算机代写|计算机图形学作业代写computer graphics代考|Subtracting Binary Numbers

Two’s complement is a technique for converting a binary number into a form such that when it is added to another binary number, it results in a subtraction. There are two stages to the conversion: inversion, followed by the addition of 1 . For example, 24 in binary is 0000000000110000 , and is inverted by switching every 1 to 0 , and vice versa: 1111111111100111 . Next, we add 1: 1111111111101000, which now represents $-24$. If this is added to binary $36: 0000000000100100$, we have
\begin{tabular}{l}
$0000000000100100=+36$ \
$1111111111101000=-24$ \
\hline $0000000000001100=+12$ \
\hline
\end{tabular}
Note that the last high-order addition creates a carry of 1, which is ignored. Here is another example, $100-30$ :
\begin{tabular}{rll}
& 0000000000011110 & $=+30$ \
inversion & 111111111100001 & \
add 1 & 0000000000000001 & \
\hline & 1111111111100010 & $=-30$ \
add 100 & 0000000001100100 & $=+100$ \
\hline & 0000000001000110 & $=+70$ \
\hline
\end{tabular}

## 计算机代写|计算机图形学作业代写computer graphics代考|Algebraic and Transcendental Numbers

Polynomial equations with rational coefficients have the form:
$$f(x)=a x^{n}+b x^{n-1}+c x^{n-2} \ldots+C$$
such as
$$y=3 x^{2}+2 x-1$$
and their roots belong to the set of algebraic numbers $A$. A consequence of this definition implies that all rational numbers are algebraic, since if
$$x=\frac{p}{q}$$
then
$$q x-p=0$$
which is a polynomial. Numbers that are not roots to polynomial equations are transcendental numbers and include most irrational numbers, but not $\sqrt{2}$, since if
$$x=\sqrt{2}$$
then

$$x^{2}-2=0$$
which is a polynomial.

## 计算机代写|计算机图形学作业代写computer graphics代考|Hexadecimal Numbers

116,216,316,416,516,616,716,816,916,一种16,乙16,C16,D16,和16,F16,1016,1116

(1×16)+(和×1)+(8×16−1) (16+14)+(8/16) 30.5.

(一种28+b27+C26)+(d25+和24+F23)+(G22+H21+一世20)

26(一种22+b21+C20)+23(d22+和21+F20)+20(G22+H21+一世20) 82(一种22+b21+C21)+81(d22+和21+F20)+80(G22+H21+一世20) 82R+81小号+80吨

R=一种22+b21+C 小号=d22+和21+F 吨=G22+H21+一世

## 计算机代写|计算机图形学作业代写computer graphics代考|Subtracting Binary Numbers

\begin{表格}{l} $0000000000100100=+36$ \ $1111111111101000=-24$ \ \hline $0000000000001100=+12$ \ \hline \end{表格}\begin{表格}{l} $0000000000100100=+36$ \ $1111111111101000=-24$ \ \hline $0000000000001100=+12$ \ \hline \end{表格}

\begin{tabular}{rll} & 0000000000011110 & $=+30$ \ 反转 & 111111111100001 & \ 添加 1 & 0000000000000001 & \ \hline & 111111111100010 & $=-30$ \ 添加 100 & 00000=0+1001 $\hline & 0000000001000110 &$=+70$\ \hline \end{表格}\begin{tabular}{rll} & 0000000000011110 &$=+30$\ 反转 & 111111111100001 & \ 添加 1 & 0000000000000001 & \ \hline & 111111111100010 &$=-30$\ 添加 100 & 00000=0+1001$ \hline & 0000000001000110 & $=+70$ \ \hline \end{表格}

## 计算机代写|计算机图形学作业代写computer graphics代考|Algebraic and Transcendental Numbers

F(X)=一种Xn+bXn−1+CXn−2…+C

X=pq

qX−p=0

X=2

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## MATLAB代写

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