Answer:
Note that the tangents to the circles at A and B intersect at a point Z on XY by radical center. Then, since∠ZAB=∠ZQA and ∠ZBA=∠ZQB. ∠AZB+∠AQB=∠AZB+∠ZAB+∠ZBA=180°. ∴ZAQB is cyclic.But if O is the center of w, clearly ZAOB is cyclic with diameter ZO, so ∠ZQO is 90° ⇒Q is the mid-point of XY.Then, by Power of a Point, PY· PX = PA · PB = 15 and it is given that PY+PX = 11. Thus,PX=(11±
)/2. So, PQ=
, PQ²=
. Thus, the answer is 61+4=65.
Step-by-step explanation:
Answer:
Step-by-step explanation:
1) translate the given informations in a operation
8x^2 -7x - 2 - (6x^2-x+3)
2) change the sign of all the terms in the bracket
8x^2 -7x -2 -6x^2 + x -3
3) simplify the expression
2x^2 - 6x -5
Check the schedule and do the jobs as they relate to the schedule
9514 1404 393
Answer:
x = 154
Step-by-step explanation:
Angle x° is supplementary to the external angle:
x° = 180° -26°
x° = 154°
Answer:
15
Step-by-step explanation:
=
=15
