Answer:
60 miles
Step-by-step explanation:
Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?
Statement 1. Brian’s walking speed is twice the difference between Ashok’s walking speed and his own
Statement 2. If Ashok’s walking speed were five times as great, it would be three times the sum of his and Brian’s actual walking speeds
Solution
A. Brian’s walking speed is twice the difference between Ashok’s walking speed and his own.
Let Brian speed=b
Ashok speed=a
Brian's walking speed=2(a-b)
b=2(a-b)
Divide both sides by 2
b/2=a-b
Ashok catches up in (time)= distance /( relative rate
=30/(a-b)
=30/(b/2)
=30÷b/2
=30*2/b
=60/b.
By that time Brian will cover a distance of
distance=rate*time
=b*60/b
=2(a-b)*60/2(a-b)
=60 miles
(2) If Ashok’s walking speed were five times as great, it would be three times the sum of his and Brian’s actual walking speeds.
5a=3(a+b)
5a=3a+3b
5a-3a=3b
2a=3b
Here is the answer for the problem.
X^3-(2y)^3
( x-2y)(x^2+x * 2y+4y^2)
final answer ( x-2y)(x^2+2xy+4y^2)
Answer:
If it is 
then it is the solution 
because the square root of an expression multiplied by itself gives that expression.
Answer:
i need a pic to answer
Step-by-step explanation:
sorry
