In reducing one's speed from 70 mph to 50 mph, how much of a percentage decrease in stopping distance is realized (to the neares
t percent)?
1 answer:
You might be interested in
A x = 0
using the law of exponents = 1
for (6² )^ x = 1 then x = 0
B note that = 1 ⇒ x = 1
2 → 2^8 × 3^(-5) × 1^(-2) × 3^(-8) × 2^(-12) × 2^(28)
= 2^(8 -12 + 28) × 1 × 3^(- 5 - 8)
= 2^24 × 3^(- 13) = 2^(24)/3^(13) = 10.523 ( 3 dec. places)