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raketka [301]
3 years ago
8

In a board game, players lose 50 points every time they land on a red space. If there is no other way to lose points and Jamie h

as a score of -450 points, what is the minimum number of times he could have landed on a red space?
Mathematics
1 answer:
Simora [160]3 years ago
7 0
Since players are losing points every time they land on a red space, it is -50,
Here's our equation where c equals the amount of times.
-50C=-450
We need C by itself, we need to divide.
-450÷-50=9
Since there is no other way to lose points, we don't have to worry about the slope in this equation. But, there might be a way to gain points. We have it set at 0 assuming that Jamie did not score any points. This allows us to get the minimum times her could have landed.
The minimum number of times he could have landed on a red space is 9 times.
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Answer:

MOP=33

Step-by-step explanation:

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Kisachek [45]
So, I'm going to break it down to help you understand it a bit more.

If it starts at (0,-2) and crosses through (1,0) that means it moved to the right once and up twice. Which means, that the slope will be 2. If it were one it would be to the right 1 up one, if it were 4 it would be to the right 1 up 4, and finally if it were 1/2 it would be to the right 2 up 1. 

So, your answer is C. or 2. 
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