Given D, E, and F are collinear points, E is between D and F, DE is six times the length of EF, and DF=84 units, find the length of segment DE.

Step-by-step explanation:

Given D,E, and F are collinear points and E is between D and F, then the expression DE + EF = DF 1 is true

If DE is six times the length of EF, then;

DE = 6DF

DF = DE/6 2

Substitute equation 2 into 1;

DE + EF = DF

DE + DE/6 = DF

find the LCM

(6DE+DE)/6 = DF

cross multiply

6DE+DE = 6DF

7DE = 6DF

Given DF = 84units

7DE = 6(84)

7DE = 504

Divide both sides by 7

7DE/7 = 504/7

DE = 72

Hence  the length of segment DE is 72units

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)