R=5
B=4+G
Y=G-3
R+B+Y+G=24
(5)+(4+G)+(G-3)+G=24
5+4+G+G-3+G=24
3G+6=24
3G=18
G=6
B=4+(6)
B=10
Y=(6)-3
y=3
answer
R+B+Y+G=24
5+10+3+6=24
Answer:
Theres no picture?
Step-by-step explanation:
The first statement is false. A square DOES have 4 right angles.
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: