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

im pretty sure it would be 5 am, hope that , and give brainliest and guys if this is right then press the button and give 5 stars

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length