ariana004 ariana004
  • 07-05-2020
  • Mathematics
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PLEASE HELP!! S is the center of the circle. Suppose that JK=12, LK=12, NS=8, and PS= 2x - 6. Fine the following

PLEASE HELP S is the center of the circle Suppose that JK12 LK12 NS8 and PS 2x 6 Fine the following class=

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mysticchacha
mysticchacha mysticchacha
  • 07-05-2020

Answer:

x = 7

LP = 6

Step-by-step explanation:

JK = 12

LK = 12

NS = 8

PS = 2x - 6

radius bisects the chords JK and LK  because it is perpendicular to them.

JK = JN + NK = 6 + 6 = 12

LK = LP + PK = 6 + 6 = 12

LP = 6

radius= root(6^2 + 8^2) = 10  then  PS = 8 = 2x -6,

8 = 2x - 6

14 = 2x

x = 7

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