100+0=100
x + 50 = 2x + 100 50 = 2x+100-x 50 = x + 100 50-100=x -50=x -50+50=0 -100+100=0 it is zero
113+0
When we add a positive number to a negative number, we must consider their signs. In this case, when we add +100 and -100, they will cancel each other out. The positive 100 will offset the negative 100, resulting in a sum of zero. Mathematically, 100 + (-100) = 0.
108 remainder 3 :D
The discriminant in the quadratic equation x2 + 11x + 121 = x + 96 is 0.Simplify the equation to the form Ax2 + Bx + C = 0 and you get x2 + 10x + 25 = 0. The discriminant is B2 - 4AC or 100 - (4)(1)(25) or 100 - 100 or 0.Not asked, but answered for completeness - since the discriminant is 0, there is one real solution, namely x = (-10 +/- 00.5) / 2 = -5
No.
75 - 100 + 0 = -25
If: y = x2+20x+100 and x2-20x+100 Then: x2+20x+100 = x2-20x+100 So: 40x = 0 => x = 0 When x = 0 then y = 100 Therefore point of intersection: (0, 100)
x + 50 = 2x + 100 50 = 2x+100-x 50 = x + 100 50-100=x -50=x -50+50=0 -100+100=0 it is zero
Yes. 25x4=100 or 143-43=100
#include <iostream> using namespace std; int main() { cout << "Even from 0-100: \n\n"; for(int i = 0; i <= 100; i++) { if(i % 2 == 0 && i != 0) { cout << i << endl; } } cout << "Odd from 0-100: \n\n"; for(int i = 0; i < 100; i++) { if(i % 2 != 0) { cout << i << endl; } } char wait; cin >> wait; return 0; }
Interesting. Assuming "times" is a variable: You're question is what is 0/times + times * (0/+0*100) That would be 0 + times * (0/0) 0/0 = infinity(Anything over 0 = infinity) So then, you can figure out that it is times * infinity which is infinity.
(1000 × 100 × 0) + 1 = 1
113+0
0 plus 0 plus 0, repeated multiple times, equals 0. Adding zero to zero does not change the value, so regardless of how many times you add it, the sum remains 0. Therefore, the final result is simply 0.
x2+2x=0 4x=0 x=0 if you meant soemthing times 2 plus 2 times x =0 solve for x then y2+2x=0 y+x=0 x=-y so if x is 2 y is -2, if x is 100 y is -100
It is B, 0.