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:
3A^2 dA/dt + 3B^2 dB/dt = 0
If A=2, then 8+B^3=9, B=1
12 dA/dt + 9 = 0
dA/dt = -3/4
Answer:
Lin lives 1 1/20 miles away from school
Step-by-step explanation:
1 3/10 becomes 1 6/20
1/4 becomes 5/20
1 6/20 - 5/20
= 1 1/20
165=15×11
15=5×3
Prime factorization of 165 is 3×5×11.
<u>Answer:</u>
The correct answer option is AAS.
<u>Step-by-step explanation:</u>
We are given two triangles: ΔABC and ΔCBD and we are two figure out by which postulate can their congruence be proved.
We can see from the given diagram that angle A and angle B are equal to that of angle C and angle B respectively. Also, the non-included side of these two angles, side AD and CD, are equal here.
According to the Angle Angle Side postulate, two triangles are said to be congruent if two angles and the non-included side of one triangle are congruent to the two angles and the non-included side of another triangle.