Suppose ABCD is a rhombus such that the angle bisector of ∠ABD meets

AD at point K. Prove that m∠AKB = 3m∠ABK. Help meee!! What do I put in the box shown below??

Step-by-step explanation:

As we know that:

=> ∠ABK=∠KBD=x

=> ∠ABD=2x

=> ∠KDB=∠ABD=2x

∠AKB=∠KDB+∠KBD=2x+x=3x

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answer: a video game has three difficulty levels: easy, medium, and expert. the game's scoring works in such a way that a player has to score a fixed number of points to advance to the next level, and then the scoring starts over. the total score needed to clear all three difficulty levels is 7,500. two times the number of points needed to clear the easy level is 500 more than the number of points needed to clear the medium level. twice the number of points needed to clear the medium difficulty level is 1,500 more than the number of points needed to clear the expert level. if x, y, and z represent the easy, medium, and expert levels, respectively, and a is the coefficient matrix of the system of equations modeling this situation, identify the inverse matrix, a-1, and the solution matrix, x, of the system of equations.

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answers

rauschg

rauschg · hand

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step-by-step explanation: