Given: DE and DF are midsegments of ABC Prove: DE = 1/2 AC

1. midsegments DE and DF: given

2. DF ll BC and AC ll DE: Midsegments are || to the non-included side of the Δ.

3. AB is a transversal cutting DF and BC.

AB is a transversal cutting AC and DE: definition of transversal line

4. m∠FAD = m∠EDB
m∠ADF = m∠DBE: Corresponding Angles Theorem

5. AD = BD: definition of midpoint

6. DBE ≅ ADF: ASA

7. A F = 1/2 AC: definition of midpoint

8.

9. DE = 1/2 AC: Transitive Property of Equality

What is the missing step in this proof?

Draw —DE , which is a midsegment of △ABC. 0. 1. 2. 3. 4. 5. 6. −1. −2. 0. 1. 2. A. C ... 3. How are the midsegments of a triangle related to the sides of the triangle? 4. ... D(1.5, 4.5). E(5, 3). Segments. BC = 4. AC = 7.62. AB = 7.07. DE = ? DF = ? ... Given —. DE is a midsegment of △OBC. Prove —. DE —. OC and DE = 1— 2.

Step-by-step explanation:

since your problem is x^4 = 81, what you have to undeo is the "to the fourth power" part.

the inverse (opposite operation) of "to the fourth power" is "to the fourth root".

  (x^4) ^ ¼   = 81 ^ ¼  

  81 = 3^4     so (3 ^ 4) ^ ¼   =   3 ^ 1   = 3

by hand: (3 ^ 4) ^ ¼   =   3 ^ (4 x ¼) = 3 ^ 1   = 3   (when you take a power to a power, multiply the exponents.)

or, do it by calculator. one way of finding the fourth root is to use ^ (1/4)

step-by-step explanation: