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
6sqrt2 for both since there equivalent
Answer:
31st term = 84
Step-by-step explanation:
Given:
204, 200,196,...
Find:
31st term
Computation:
204, 200,196,...
First term a = 204
Difference d = a3 - a2
Difference d = 196 - 200
Difference d = -4
An = a + (n-1)d
A31 = 204 + (31-1)(-4)
A31 = 204 - 120
A31 = 84
Answer:
-3/7 > b
Step-by-step explanation:
2/7> b+5/7
Subtract 5/7 from each side
2/7-5/7> b+5/7-5/7
-3/7 > b
