Answer:
Let P be the external point. O be the origin. join O and P get OP and nearest point on the circle from P be A.
Let Q be the point onthe circle in which, tangent make 90° with radius at Q.
PQ = 8 and OQ = 6
we get a right angled triangle PQO right angled at Q.
so, OP^2 = OQ^2 + PQ^2= 8^2 + 6^2 = 64 + 36 =1==
therefore OP =10cm
we need nearest point from P, which is PA
PA = OP - OA= 10 -6=4cm
I really don’t know sorry I would help you but I can genesis hdbdbhs hebdbh
(a/b)(c/d)=(ac/bd)
Therefore:
(5/4).(2/5)=(5*2)/(4*5)=10/20=1/2
(3/8).(-1/2)=(3*(-1)) /(8*2)=-3/16
(2/5)(5/2)=(2*5)/(5*2)=10/10)=1
(1/2)(1/3)=1/6
(-1/4)(-2/3)=[(-1)*(-2)] / (4*3)=2/12=1/6
Find a common denominator.
5/6 and 3/4
10/12 and 9/12.
If she ran 10/12 of a mile away and turned around and ran 9/12's back, she is 1/12 of a mile away.
