-100(negative 100)+100(positive 100)=0. If you have 100 of something(the +100) and then take away 100 (the -100), you're left with nothing (the 0). So it's 0.
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
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)
-100(negative 100)+100(positive 100)=0. If you have 100 of something(the +100) and then take away 100 (the -100), you're left with nothing (the 0). So it's 0.
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
Answer: 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