Daniel is constructing a fence that consists of parallel sides line AB and line EF. Complete the proof to explain how he can show that m∠GKB = 120° by filling in the missing justifications. Statement Justification
line AB ∥ line EF
m∠ELJ = 120° Given
m∠ELJ + m∠ELK = 180° Linear Pair Postulate
m∠BKL + m∠GKB = 180° Linear Pair Postulate
m∠ELJ + m∠ELK = m∠BKL + m∠GKB Transitive Property
∠ELK ≅ ∠BKL 1.
m∠ELK = m∠BKL 2.
m∠ELJ + m∠ELK = m∠ELK + m∠GKB Substitution Property
m∠ELJ = m∠GKB Subtraction Property
m∠GKB = m∠ELJ Symmetric Property
m∠GKB = 120° Substitution
1. Alternate Interior Angles Theorem; 2. Definition of Congruence
1. Alternate Exterior Angles Theorem; 2. Substitution Property
1. Definition of Congruence; 2. Alternate Interior Angles Theorem
1. Substitution Property; 2. Alternate Exterior Angles Theorem

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.

well, if it's triangular garden, the triangle could represent the garden.