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:
they are equal because while solving 2b+5+1b it will come 3b+5 so they are equal
First box: 7 to the power of 2
second box: 7
third box: 7 to the power of four
first box: m to the power of 2
second box: m to the power of 5
Answer:
Step-by-step explanation:
Convert 2y=-3x+2 into y=mx+b form
Y intercept is 1, slope is -3/2
Graph looks like this:
Answer:
please write the full questions
