Given J(1, 1), K(3, 1), L(3, -4), and M(1, -4) and that J'(-1, 5), K'(1, 5), L'(1, 0), and M'(-1, 0). What is the rule that tran
anastassius [24]
(x; y) -> (x - 2; y + 4)
J(1; 1) ⇒ J'(1 - 2; 1 + 4) = (-1; 5)
K(3; 1) ⇒ K'(3 - 2; 1 + 4) = (1; 5)
L(3;-4) ⇒ L'(3 - 2; -4 + 4) = (1; 0)
M(1;-4) ⇒ M'(1 - 2;-4 + 4) = (-1; 0)
Yea you did jsixidjdjiejs
Answer:
y = 18
Step-by-step explanation:
<h3><u>i</u><u>)</u><u> </u><u>redu</u><u>ce</u><u> the</u><u> fraction</u><u> </u><u>with</u><u> </u><u>4</u></h3>



<h3><u>ii</u><u>)</u><u> </u><u>simpl</u><u>ify</u><u> the</u><u> </u><u>eq</u><u>uation</u><u> </u><u>with</u><u> </u><u>cross</u><u> </u><u>multiplication</u></h3>



<h3>iii) <u>divi</u><u>de</u><u> both</u><u> sides</u><u> of</u><u> the</u><u> equation</u><u> </u><u>by</u><u> </u><u>7</u></h3>
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Formula: n-2(180)
n = number of sides
6-2(180) = 720 degrees