Since n=5, take the 6th row of Pascal Triangle: (a+b)⁵:
(a+b)⁵ = a⁵+5a⁴b+10a³b²+10a²b³+5ab⁴+b⁵.
But a= 2x and b=y, plug in:
(2x)⁵ + 5(2x)⁴y + 10(2x)³y² + 10(2x)²y³ + 5(2x)y⁴ +y⁵
32x⁵ + 80x⁴y +80x³y² + 40x²y³ + 10xy⁴ + y⁵
The 3rd term has a:
coefficient = 80
power of x= 3
power of y =2
Answer:
They will do rounds together again in 11 minutes and 40 seconds.
Step-by-step explanation:
Security A rounds every 2 minutes and 20 seconds in the gate 1.
That is, after 2*60 + 20 = 120 + 20 = 140 seconds.
Security B rounds every 1 minute and 40 seconds in the gate 2.
That is, after 1*60 + 40 = 60 + 40 = 100 seconds.
After how many minutes they will rounds together again?
First, we find this is seconds, which is the least common multiple between 140 and 100. Factoring them by prime factors:
140 - 100|2
70 - 50|2
35 - 25|5
7 - 5|5
7 - 1|7
1 - 1
lcm(140,100) = 2*2*5*5*7 = 700 seconds
They will do rounds together again in 700 seconds, which is 11 minutes and 40 seconds.
To simplify this all you need to do is distribute the 4x across each term in the parenthesis
700 cal out of the 2000
2000*.35
