57
Since ( y ) varies directly as ( x ), we can express this relationship as ( y = kx ), where ( k ) is the constant of variation. Given the values ( y = 80 ) when ( x = 40 ), we can find ( k ) by substituting these values into the equation: ( 80 = k(40) ). Solving for ( k ) gives ( k = 2 ). Therefore, the equation of variation is ( y = 2x ).
direct variation: y = kx y = kx k = y/x = 0.8/0.4 = 2
Since ( y ) varies directly as ( x ), we can express this relationship as ( y = kx ), where ( k ) is the constant of variation. Given that ( y = 28 ) when ( x = 7 ), we can substitute these values into the equation to find ( k ): [ 28 = k(7) \implies k = 4. ] Thus, the equation of variation is ( y = 4x ).
y=k*x 535=k*140 k=535/140=107/28 y=(107/28)*x
What is y? Fx varies with x, but y could be anything.
Find an equation of variation where y varies directly as x. One pair of values is y = 80 when x = 40
y = 10x
4x= y
a varies directly as b and a = 12 when b = 4. What is the constant of variation?
40
y = -5
y = 8
Since ( y ) varies directly as ( x ), we can express this relationship as ( y = kx ), where ( k ) is the constant of variation. Given the values ( y = 80 ) when ( x = 40 ), we can find ( k ) by substituting these values into the equation: ( 80 = k(40) ). Solving for ( k ) gives ( k = 2 ). Therefore, the equation of variation is ( y = 2x ).
direct variation: y = kx y = kx k = y/x = 0.8/0.4 = 2
y = 3x y = 3*19 = 57
y = kx: 10 = 37k so k = 10/37 and y = 10x/37
2