Lines ma and mb tangent circle k(o) at a and b. point c is symmetric to point o with respect to point b . prove: m∠amc=3m∠bmc.

See explanation

Step-by-step explanation:

If MA is tangent to the circle k(O), then radius OA is perpendicular to segment MA.

If MB is tangent to the circle k(O), then radius OB is perpendicular to segment MB.

Consider two right triangles MOA and MOB. In these triangles:

Thus, triangles MOA and MOB are congruent by HL theorem. So

∠AMO=∠BMO.

If point C is symmetric to point O with respect to point B, then OC⊥MB. Consider two right triangles MOB and MCB. In these triangles:

Thus, triangles MOB and MCB are congruent by HL theorem. So

∠BMO=∠BMC.

Hence,

∠AMC=∠AMO+∠BMO+∠BMC=3∠BMC.

he has now walked a total of what?

if i know what this is maybe i can

step-by-step explanation:

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560