It could be add 15, or it could be multiply by 3 plus 1, for example.
Subtract two.
Multiply by 5 and add 20 (or equivalently, add 4 and then multiply by 5)
It is not. Suppose the function is "add 7".Then an input of 1 gives an output of 1+7 = 8.Double the input to 2 and the output is 2+7 = 9Whereas simply halving the output gives 9/2 = 4.5So the question is based on false premises.
A VHDL program for an 8-to-3 priority encoder using data flow style can be implemented using the when-else construct. The encoder outputs a 3-bit binary representation of the highest-priority active input (from 7 to 0), while also providing an output for invalid conditions. Here’s a simple example: library IEEE; use IEEE.STD_LOGIC_1164.ALL; use IEEE.STD_LOGIC_ARITH.ALL; use IEEE.STD_LOGIC_UNSIGNED.ALL; entity priority_encoder is Port ( input : in STD_LOGIC_VECTOR(7 downto 0); output : out STD_LOGIC_VECTOR(2 downto 0); valid : out STD_LOGIC); end priority_encoder; architecture dataflow of priority_encoder is begin process(input) begin case input is when "00000000" => output <= "000"; valid <= '0'; when others => output <= "111"; -- Default output for higher priority valid <= '1'; if input(7) = '1' then output <= "111"; elsif input(6) = '1' then output <= "110"; elsif input(5) = '1' then output <= "101"; elsif input(4) = '1' then output <= "100"; elsif input(3) = '1' then output <= "011"; elsif input(2) = '1' then output <= "010"; elsif input(1) = '1' then output <= "001"; elsif input(0) = '1' then output <= "000"; end if; end case; end process; end dataflow; This code checks the input vector and determines the highest active bit, setting the output accordingly.
The operation appears to involve subtracting 1 from the quotient of each input number divided by 5. Specifically, for each input number ( x ), the output can be calculated as ( \text{output} = \frac{x}{5} + 3 ). For example, for the input 5, the output is ( \frac{5}{5} + 3 = 4 ). This pattern holds for all given input numbers.
Subtract two.
Multiply by 5 and add 20 (or equivalently, add 4 and then multiply by 5)
if one answer is 6 and the other answer is 7, how do the output numbers from the input/output machines compare
It is not. Suppose the function is "add 7".Then an input of 1 gives an output of 1+7 = 8.Double the input to 2 and the output is 2+7 = 9Whereas simply halving the output gives 9/2 = 4.5So the question is based on false premises.
the rule is add 7 because 4 plus 7 = 11
you really have been waiting around 7 years I believe I got the answer I think its a input device I'm sorry you had to wait around 7 years
There are many rules for that pair. Without further information about the kind of operators expected to be used it is impossible to give an answer. Let me give you a very simple example: What is the rule for input 2 output 4: rule 1: x → x + 2 rule 2: x → 2x rule 3: x → x^2 rule 4: x → x^3 ÷ 2 rule 5: x → x^3 - x^2 rule 6: x → x^3 - 2x rule 7: x → x^3 - 4 rule 6: x → x^4 - x^3 - x^2 etc Even with a list of permissible operators, there is no one definite answer given only one pair of input/output - it would require further examples of inputs and their corresponding outputs to be able to narow the possible answers: only slightly as it is still possible to find [infinitely] many polynomials that provide the given input/output pairs.
Both the CD4026 and CD4033 are BCD to 7 Segment counter/decoders. The 4026 has a display enable input/output, while the 4033 has a ripple blanking input/output.
3 + 4 = 7 7 x 2 = 14
you really have been waiting around 7 years I believe I got the answer I think its a input device I'm sorry you had to wait around 7 years
L293D is having 20 pin IC and also 16 pin IC. description of 20 pin is: 1-enable 1 2- input 1 3- output 1 4,5,6,7,14,15,16,17- ground 8- output 2 9- input 2 10,20-vs 11-enable 2 12- input 3 13-output 3 18-output 4 19-input 4 description for 18 pin: 1-enable 1 2- input 1 3- output 1 4,5,12,13- ground 6-output 2 7- input 2 8,18-vs 9-enable 2 10-input 3 11-output 3 14-output 4 15-input 4
An input/output table works like this:You input something, and through a function, it outputs something else!Say I Had a function that is: input+2If I were to input 5, It would output 7All an input/output table does is displays a couple examples of multiple inputs with their outputs! Put tables only operate on one function....Example:Function: Input x 5 + 3INPUTS - OUTPUTS----------------------1 - 82 - 133 - 186 - 3310 - 53