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:
(10a+5)²= 100 a( a+1) + 25
Step-by-step explanation:
<u><em>Explanation:-</em></u>
Given
(10a+5)²
By using (a+b)² = a² +2ab +b²
= (10a)²+ 2 × 10a× 5 + (5)²
= 100a² + 100a + 25
= 100 a( a+1) + 25
After the distribution of x, you move 6 to the other side of the equation and then divide by three to get x=2. Therefore, the answer is c.
Hope this helps you!
Answer:
b=-3
Step-by-step explanation:
Answer:
D) 2/4
Step-by-step explanation:
2/4 is equivalent to 4/8 which is less than 5/8
