This phrase means that x = kyz where k is some constant.
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.
Since ( x ) varies directly with ( y ) and inversely with ( z ), we can express this relationship as ( x = k \frac{y}{z} ), where ( k ) is a constant. Given that ( x = 5 ) when ( y = 10 ) and ( z = 5 ), we can find ( k ): [ 5 = k \frac{10}{5} \implies k = 2. ] Now, to find ( x ) when ( y = 20 ) and ( z = 10 ): [ x = 2 \frac{20}{10} = 4. ] Thus, ( x ) equals 4.
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
Since ( x ) varies directly with ( y ) and ( z ), we can express this relationship as ( x = k \cdot y \cdot z ) for some constant ( k ). Given that ( x = 120 ) when ( y = 5 ) and ( z = 8 ), we can find ( k ) as follows: [ 120 = k \cdot 5 \cdot 8 \implies k = \frac{120}{40} = 3. ] Now, to find ( x ) when ( y = 6 ) and ( z = 2 ): [ x = 3 \cdot 6 \cdot 2 = 36. ] Thus, ( x = 36 ).
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