Mathematics, 17.03.2020 05:10 189652
Given: m∠AEB = 45° ∠AEC is a right angle. Prove: Ray E B bisects ∠AEC. 3 lines extend from point E. A horizontal line extends to the left to point A and a vertical line extends up to point C. Another line extends halfway between the other 2 lines to point B. Lines A E and E C form a right angle. Proof: We are given that m∠AEB = 45° and ∠AEC is a right angle. The measure of ∠AEC is 90° by the definition of a right angle. Applying the gives m∠AEB + m∠BEC = m∠AEC. Applying the substitution property gives 45° + m∠BEC = 90°. The subtraction property can be used to find m∠BEC = 45°, so ∠BEC ≅ ∠AEB because they have the same measure. Since Ray E B divides ∠AEC into two congruent angles, it is the angle bisector.
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Maria found the least common multiple of 6 and 15. her work is shown below. multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, . . multiples of 15: 15, 30, 45, 60, . . the least common multiple is 60. what is maria's error?
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Given: m∠AEB = 45° ∠AEC is a right angle. Prove: Ray E B bisects ∠AEC. 3 lines extend from point E....
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