From 1912 to 1971, the Milwaukee Road Railroad Company had a train stop every 15 days in Dubuque, Iowa. During this same period, the same train also stopped in Des Moines, Iowa, every 9 days. If on April 18, 1964, the train made a stop in Dubuque and a stop in Des oines, how many days wait it until the train again stooped in both cities on the same day?

46 days ; kindly check explanation for more

Step-by-step explanation:

Given that:

Train stops in Dubuque, Iowa every 15 days

The same train also stops in Des Moines, Iowa every 9 days

If train stops in both location on April 18, 1964

How many will it take the train to stop in the same location again

To obtain the time it will take, find the lowest common multiple of 15 and 9

Multiples of 15 : 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 175,

Multiples of 9 : 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108,

Hence, the lowest common multi of 15 and 9 is 45

Hence, train will stop in both locations again after 45 days;

April 18, 1964 + 45 days = 3 June, 1964

answer: b.36 feet

step-by-step explanation:

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.