A horizontal line has a slope of 0. If you're using the slope formula, then when the numerator is equal to 0 then the slope is 0.
The length of ab can be found by using the Pythagorean theorem. The length of ab is equal to the square root of (0-8)^2 + (0-2)^2 which is equal to the square root of 68. Therefore, the length of ab is equal to 8.24.
Thing is, 3 doesn't equal 0.03... But... using false logic, it could be proved: Do we agree: 0/3 = 0/0.03 = 0 We know if: n/a = n/b then, a =b. Therefore, 3 = 0.03 However, you can't do that with a 0, so 3 doesn't equal 0.03
No; they are not equal.
0 + 0 = 0
Its truth table is: input output 0 1 1 0
Input Output 0 1 1 0
I don't really know what this is supposed to mean, if you want to print the truth-table of the NAND-gate that will be something like this: for (a=0; a<=1; ++a) for (b=0; b<=1; ++b) printf ("%d %d %d\n", a, b, !(a&&b))
. p . . . . . q. 0 . . . . . 1. 1 . . . . . 0
.....0 10 | 0 01 | 0 1.....0 10 | 0 11 | 1 10 | 11 | 0
apparently whenever you can swap the 0's for 1's and 1's for 0's in the truth table and the truth result remains unchanged.
Truth table of 'NAND' is 0 0 - 1 0 1 - 1 1 0 - 1 1 1 - 0 NAND is just opposite of AND as the name itself suggest NAND is the not of AND Truth table of "NOR" is 0 0 - 1 0 1 - 0 1 0 - 0 1 1 - 0 NOR is just opposite of OR as the name itself suggest NOR is the not of OR.
#include<iostream> int main() { std::cout << "Truth table for AND gate\n\n"; std::cout << " |0 1\n"; std::cout << "-+---\n"; for (unsigned a=0; a<2; ++a) { std::cout << a << '|'; for (unsigned b=0; b<2; ++b) { std::cout << (a & b) << ' '; } std::cout << '\n'; } std::cout << std::endl; }
It is the very same in every programming language. For example: AND: 0 && 0 = 0 0 && 1 = 0 1 && 0 = 0 1 && 1 = 1
Because if input A *and* input B is true, then the output is true! Truth table of AND gate: ┌─┬─╥───────┐ │A│B║Q (Output)│ ├─┼─╫───────┤ │0│0║0..............│ ├─┼─╫───────┤ │0│1║0............. │ ├─┼─╫───────┤ │1│0║0............. │ ├─┼─╫───────┤ │1│1║1............. │ └─┴─╨───────┘
To DeMorganize a Boolean expression, you apply De Morgan's laws, which involve changing the operation and flipping the logic of the expression. Specifically, the AND operation becomes an OR operation, and vice versa, while the NOT operation is distributed across the variables within parentheses. For example, DeMorganizing the expression ¬(A AND B) would result in ¬A OR ¬B. This process is crucial for simplifying and manipulating Boolean expressions in logic and computer science.
A horizontal line has a slope of 0. If you're using the slope formula, then when the numerator is equal to 0 then the slope is 0.