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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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4 years ago
How do I solve this
Sladkaya [172]

Step-by-step explanation:

\bf \underline{Solution-} \\

\textsf{Given,}\\

\sf{ \frac{ {a}^{6} {b}^{4}  }{ {a}^{2} {b}^{3}  } } \\

\textsf{Using law of intergal expo.. aᵐ ÷ aⁿ = aᵐ - ⁿ}\\

\sf{ \Rightarrow \:  {a}^{(6 - 2)}  {b}^{(4 - 3)} } \\

\sf{ \Rightarrow \:  {a}^{4}  {b}^{1} } \\

\sf{ \Rightarrow \:  {a}^{4}  {b}^{} } \\

\bf \underline{Hope\: this\: helps.} \\

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