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MA_775_DIABLO [31]
3 years ago
12

Two boys, Shawn and Curtis, went for a walk. Shawn began walking 20 seconds earlier than Curtis. Shawn walked at a speed of 5 fe

et per second. Curtis walked at a speed of 6 feet per second. For how many seconds had Sean been walking at the moment on the two boys had walked exactly the same distance?
Mathematics
1 answer:
Alexeev081 [22]3 years ago
5 0
Distance = speed * time
for  curtis  d = st = 6t              (where t is the time that he walks)
 sean walks 20 seconds longer than curtis so we have the equation

d = s(t+20) = 5(t+20)

as the distances are the same we have

So sean has walked 100+20 = 2

6t = 5(t+20)
6t = 5t + 100
t = 100
so Sean walks 120 seconds when they had both walked the same distance.
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You and some friends are going to the fair. Small rides costs 4 tickets and while big rides costs 7 tickets. You and your friend
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Answer:

<em>4x+7y≤250</em>

Step-by-step explanation:

<u>Inequalities</u>

Let's call:

x = number of small rides

y = number of big rides

Since each small ride costs 4 tickets, x rides cost 4x tickets.

Since each big ride costs 7 tickets, y rides cost 7y tickets.

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The friends have 250 tickets, so the total tickets of the rides cannot be greater than the 250 tickets available, thus:

4x+7y≤250

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