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Q: What math property is used in 0x9 equals 0?
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What is 0 x 9?

0x9=0


Which statement is NOT true about the number 0x9 equals 0 A. 0 is a factor B. 9 is a factor C 0 is the product D. 9 is the only factor?

D is the choice that is not true.


Which number is not divisible by 9?

The multiples of 9 are:0x9 = 01x9 = 92x9 = 183x9 = 27etc. Those numbers are divisible by 9. Any other number is not.The multiples of 9 are:0x9 = 01x9 = 92x9 = 183x9 = 27etc. Those numbers are divisible by 9. Any other number is not.The multiples of 9 are:0x9 = 01x9 = 92x9 = 183x9 = 27etc. Those numbers are divisible by 9. Any other number is not.The multiples of 9 are:0x9 = 01x9 = 92x9 = 183x9 = 27etc. Those numbers are divisible by 9. Any other number is not.


If there were only 9 digits -0 through 8- how much would the number 100 represent?

The number 100, using digits 0 thru 8, would equal the number 81 using digits 0 thru 9. 1x92 + 0x91 + 0x90 which is 1x9x9 + 0x9 + 0x1


What is procedure given in class to convert binary to hex?

Split the binary value into groups of 4 bits (half-a-byte). Translate each nybble to its corresponding hex digit. Use the following table to translate each nybble: 0000 = 0x0 0001 = 0x1 0010 = 0x2 0011 = 0x3 0100 = 0x4 0101 = 0x5 0110 = 0x6 0111 = 0x7 1000 = 0x8 1001 = 0x9 1010 = 0xA 1011 = 0xB 1100 = 0xC 1101 = 0xD 1110 = 0xE 1111 = 0xF


What is the 4-binary number assoiated with the hexadecimal symbol C?

0xc = 1100 Hexadecimal digits use exactly 4 binary digits (bits). The 0x0 to 0xf of hexadecimal map to 0000 to 1111 of binary. Thinking of the hexadecimal digits as decimal numbers, ie 0x0 to 0x9 are 0 to 9 and 0xa to 0xf are 10 to 15, helps with the conversion to binary: 0xc is 12 decimal which is 8 + 4 → 1100 in [4 bit] binary.


What is the times nine table?

That means a table of multiples of 9: 0 x 9 = 0 1 x 9 = 9 2 x 9 = 18 etc. Usually such a table would multiply together only one-digit numbers (from 0x9 to 9x9). BUT the result of each multiplication produces two-digit numbers (as far as 99), and then 108 for 12 x 9. We need to think of 0 x 9 = 00 and 1 x 9 = 09, but we dont usually write out the zeroes which are not necessary. CAREFUL; that doesn't mean that all zeroes aren't necessary! Wouldn't you rather have £100 than £001?


How do you add hexadecimal?

Hexadecimal numbers can be manipulated in exactly the same was as decimal. The digits we can use are: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F. Let us take two numbers, say, 7 and 6. In decimal, we can count from 7 upward by 6... 8, 9, 10, 11, 12, 13. In Hexadecimal, we can do the same... 8, 9, A, B, C, D. Let's take two more numbers, D and 4. We count upwards from D by 4... E, F, 10, 11. Notice that after F, there is no other digit we can use, to we carry 1 to the 16's column and carry on counting to reach the final figure of 11Hex. This is just the same as the decimal procedure of carrying to the tens column after we pass 9. Once you have got your head around the idea of extra digits, the rest of it is as easy as decimal.


How do you convert exe to binary?

An exe is machine code and machine code is written entirely in binary. No conversion is necessary. A hex-editor is the simplest way to view the binary code. The code will be shown in hexadecimal rather than binary, however this actually makes it much easier for humans to interpret the binary code because the conversion from hex to binary is so simple. Each hex digit represents a unique 4-bit binary pattern: 0x0 = 0000 0x1 = 0001 0x2 = 0010 0x3 = 0011 0x4 = 0100 0x5 = 0101 0x6 = 0110 0x7 = 0111 0x8 = 1000 0x9 = 1001 0xA = 1010 0xB = 1011 0xC = 1100 0xD = 1101 0xE= 1110 0xF = 1111 Thus the hex value 0x9A translates directly to the 8-bit binary value 10011010. That is, 8 binary digits reduce to just 2 hex digits and therefore makes it much easier for humans to interpret the binary value.


How do you convert hex decimal number in to octal?

First convert the hexadecimal to binary. Every hexadecimal digit corresponds to 4 binary digits: 0x0 = 0000 0x1 = 0001 0x2 = 0010 0x3 = 0011 0x4 = 0100 0x5 = 0101 0x6 = 0110 0x7 = 0111 0x8 = 1000 0x9 = 1001 0xA = 1010 0xB = 1011 0xC = 1100 0xD = 1101 0xE = 1110 0xF = 1111 Next, after you have converted each hexadecimal digit into binary digits, convert the binary digits to Octal. Each octal digit corresponds to 3 binary digits: 00 = 000 01 = 001 02 = 010 03 = 011 04 = 100 05 = 101 06 = 110 07 = 111 Make sure that you don't accidentally mix up the digits. If you are computer scientist, then you may also have to worry about endianess (whether the most significant digit comes first or the least significant digit comes first).


How do you say daddy in binary code?

It depends on the encoding but if we assume standard ASCII encodings, the representation is the same for all systems, the only difference being the number of leading 0 bits per character. 7-bit ASCII (ISO/IEC 646): 1100100 1100001 1100100 1100100 1111001 0000000 8-bit ASCII (ISO/IEC 8859, Windows-1252 and UTF8): 01100100 01100001 01100100 01100100 01111001 00000000 UTF16: 00000000 01100100 00000000 01100001 00000000 01100100 00000000 01100100 00000000 01111001 00000000 00000000 To perform these conversions, convert each character to its ASCII representation (in decimal): d = 100 a = 97 d = 100 d = 100 y = 121 For completeness, we should also include the null-terminator, character code 0. null = 0 Now convert each decimal value to its 8-bit representation in hexadecimal: 100 = 0x64 97 = 0x61 100 = 0x64 100 = 0x64 121 = 0x79 0 = 0x00 Convert each hexadecimal digit to its 4-bit binary representation: 0x6 = 0110 0x4 = 0100 0x6 = 0110 0x1 = 0001 0x6 = 0110 0x4 = 0100 0x6 = 0110 0x4 = 0100 0x7 = 0111 0x9 = 1001 0x0 = 0000 0x0 = 0000 Place the binary codes in sequence. "daddy" = 01100100 01100001 01100100 01100100 01111001 00000000 Finally, add or remove leading zero bits to suit the actual encoding.


How do you convert a decimal into a BCD in C plus plus?

The following program shows one way to convert to BCD using a variation of packed BCD 8421 encoding. Each digit is represented by 4 bits in the range 0x0 to 0x9. 0xa denotes a minus symbol (for negative values) while 0xb denotes a decimal point for floating point values. The unused value 0xf is used to pad a BCD to a full byte when the number of symbols is odd. Values 0xc, 0xd and 0xe are ignored, but could be used for other purposes such as when representing exponents or fractions/ratios. Values may be any length, the BCD being represented by a vector of type unsigned char. Since the value of a BCD could easily exceed the range of a long double or a 64-bit integer, conversion to these types is disabled. BCD values are constructed from an input string and converted back to a string for output. Arithmetic operations can be overloaded to handle a BCD (using decimal notation arithmetic), however this has been left as an exercise for the reader. #include<iostream> #include<sstream> #include<vector> #include<exception> using uchar = unsigned char; class BCD_8421 { private: std::vector<uchar> v; static const uchar lo_mask {0x0f}; static const uchar hi_mask {0xf0}; public: BCD_8421 (void); BCD_8421 (const BCD_8421&); BCD_8421 (BCD_8421&&); BCD_8421 (const std::string&); BCD_8421& operator= (const BCD_8421&); BCD_8421& operator= (BCD_8421&&); BCD_8421& operator= (const std::string&); static bool is_valid (const std::string& s); operator std::string (void) const; friend std::ostream& operator<< (std::ostream&, const BCD_8421&); }; BCD_8421::BCD_8421 (void): v(1, 0) {} BCD_8421::BCD_8421 (const BCD_8421& bcd): v (bcd.v) {} BCD_8421::BCD_8421 (BCD_8421&& bcd): v (std::move(bcd.v)) {} BCD_8421::BCD_8421 (const std::string& s): v{} { *this = s; } BCD_8421& BCD_8421::operator= (const BCD_8421& bcd) { v=bcd.v; return *this; } BCD_8421& BCD_8421::operator= (BCD_8421&& bcd) { v=std::move(bcd.v); return *this; } BCD_8421& BCD_8421::operator= (const std::string& s) { v.clear(); uchar c {}; bool msb {false}; bool minus {false}; bool point {false}; bool error {false}; bool digit {false}; for (auto i=s.begin(); i!=s.end(); ++i) { if (!msb) c = hi_mask; uchar x; switch (*i) { case ('0'): if (!digit) continue; case ('1'): case ('2'): case ('3'): case ('4'): case ('5'): case ('6'): case ('7'): case ('8'): case ('9'): digit = true; x = *i - '0'; break; case ('.'): if (!point) { point = !point; x = 10; break; } else error = !error; case ('-'): if (!minus && i==s.begin()) { minus = !minus; x = 11; break; } default: error = !error; } if (error) { v.clear(); v.push_back (hi_mask); std::stringstream ss; ss << "Error: invalid argument in BCD_8421.operator= ("" << s << "")"; throw std::invalid_argument (ss.str()); } if (msb) v.push_back ((c & lo_mask) | (x << 4)); else c |= x; msb = !msb; } if (msb && c) v.push_back (c); else if (!digit) { v.clear(); v.push_back (hi_mask); } return *this; } bool BCD_8421::is_valid (const std::string& s) { const std::string valid_chars {"0123456789-."}; if (s.find_first_not_of (valid_chars) != valid_chars.npos) return false; auto f = s.find ('-'); if (f != s.rfind('-') (f && f!=s.npos)) return false; if (s.find ('.') != s.rfind('.')) return false; return true; } BCD_8421::operator std::string (void) const { std::stringstream ss; bool digit {false}; bool point {false}; for (auto i : v) { bool msb {false}; do { uchar c {}; uchar x = msb ? i >> 4 : i & lo_mask; switch (x) { case (0x0): case (0x1): case (0x2): case (0x3): case (0x4): case (0x5): case (0x6): case (0x7): case (0x8): case (0x9): c = '0' + x; digit = true; break; case (0xa): c = '.'; point = true; break; case (0xb): c = '-'; break; default: break; } if (point && !digit) ss << '0'; if (c) ss << c; msb = !msb; } while (msb); } if (ss.str().back()=='.') ss << '0'; return ss.str(); } std::ostream& operator<< (std::ostream& os, const BCD_8421& bcd) { return os << static_cast<std::string> (bcd); } int main() { for (unsigned loop=0; loop<10; ++loop) { std::string input; while (true) { std::cout << "Enter a numeric value: "; std::cin >> input; if (BCD_8421::is_valid (input)) break; std::cerr << "Invalid input.\n"; } try { BCD_8421 a = input; std::cout << "You entered the value: " << a << std::endl; } catch (std::range_error& e) { std::cerr << e.what() << std::endl; } } }