z = 8/3.
If x = y and y = z then x = z
#include <iostream> using namespace std; int main() { int x, y, z; cout << "Enter 3 numbers: \n"; cin >> x; cin >> y; cin >> z; if(x < y && x < z) { cout << x << " "; if(y < z) { cout << y << " " << z; } else if(z < y) { cout << z << " " << y; } } else if(y < x && y < z) { cout << y << " "; if(x < z) { cout << x << " " << z; } else if(z < x) { cout << z << " " << x; } } else if(z < y && z < x) { cout << z << " "; if(y < x) { cout << y << " " << x; } else if(x < y) { cout << x << " " << y; } } char wait; cin >> wait; return 0; }
3 out of 4. 8 possiableaties------------------ coins 1--- 2--- 3--- 4--- 5--- 6--- 7--- 8 x y--- x--- x--- x--- x--- y--- y---- y--- y x z--- x--- x--- z--- z--- x--- x----z--- z y z--- y----z---y--- z--- y--- z----y-----z There are 8 possiabilities for the three coins to land, you count the matches, there 6 out of 8 that match.
(x + y)/z = x/z + y/z where z is non-zero.
Y = K (z/x)
z = 8/3.
The answer is x=10. If: x=20 y=8 z=4 then: y=8/2=4 z=4*2=8 since x varies directly with y, meaning whatever happens to y, happens to x, so if y was divided by 2, then x should be divided by 2. After all, the inverse of division is multiplication.
1
Z = K Y / X 'K' can be any constant number.
y varies jointly with x and z if: when x is held fixed, y varies with z and when z is held fixed, y varies with x. Bothe x and z may vary together.
x = 24
mensos yo no me la seee
x = 5.y.z so 5.10.12 = 600
k = 2 Explanation: when z varies directly proportional to x means a division c = z x where c is a constant of proportionality when z varies inversely proportional to x it means a product z y = c now with both k = z y x z = 8 ; x = 12 and y = 3 k = 8.3 12 k = 2
80=4x, z=6x. find Z. x=80/4, x=20. z=6x20 z=120 answer= 120
Given x=k1y and x=k2/z x=125 ,y=5 then k1=25 x=125 , z=4 then k2=125(4)=500 If y=4 ,z=5 then x=25y = 100