Which statement is always true about the graph of an equation in the form y=kx? a)the point (0,k) is on the graph. b)the point (1,1) is on the graph. c)the point (k,0) is on the graph d)the point (0,0) is on the graph.

d)The point (0,0) is on the graph.

Step-by-step explanation:

Select statement is always true about the graph of an equation in the form y=kx

(0,k)  , x=0  and y=k

LEts plug in x=0 in y= kx

y=k(0) =0, so the point (0,k) on the graph of y=kx is not true

(1,1)  , x=1  and y=1

LEts plug in x=1 in y= kx

y=k(1) =k, so the point (1,1) on the graph of y=kx is not true

(k,0)  , x=k  and y=0

LEts plug in x=k  in y= kx

y=k(k) =k^2, so the point (k,0) on the graph of y=kx is not true

(0,0)  , x=0  and y=0

Lets plug in x=0 in y= kx

y=k(0) =0, so the point (0,0) on the graph of y=kx is true

answer:   rlly

step-by-step explanation:

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2