The vertices of a trapezoid are shown below.

p(0, 0), q(3, 0), r(3, 6), s(0, 12)

this trapezoid is dilated about the origin by a scale factor of 4. what is the location of point s'?

-6 + 3 = 3 - 6

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18

1/3 + 1/5 = 8/15.

step-by-step explanation:

hope it