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Sagot :
[tex]Let's\ use\ a\ simple\ math\ solution\ regarding\ with\ this\ situation... \\ \\ If\ the\ question\ states\ "evenly"\ therefore\ they\ are\ equal. \\ \\ Solution; \\ 18\ \cdot x=900\ meters \\ \\ x= \frac{900\ meters}{18} \\ \\ \boxed{\boxed{x=50\ meters}} \\ \\ Conclusion; \\ \\ Thus,\underline{50\ meters}\ are\ the\ gaped\ between\ a\ post\ to\ the\ other\ post. \\ \\ Hope\ it\ Helps:) \\ Domini[/tex]
Instead of 19x = 900, where x is the even spaces between each posts.
I will use 18x = 900. Why?
Well, to explain it clearly. Take a look at your hand (either left or right). If your fingers are the posts and the spaces in between are the x or the even spaces. It would make a 5 posts and 4 spaces in between. I hope my explanation is clear.
So, in this case. There would be 19 posts and 18 even spaces.
18x = 900
[tex] \frac{18x = 900}{x} [/tex]
x = 50 m
it would look like this:
| 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m |
| <- this sign is the post. I just draw it to emphasize more.
Now count the spaces and the posts. :) There would be 19 posts and 18 spaces. I hope you get my point. :)
I will use 18x = 900. Why?
Well, to explain it clearly. Take a look at your hand (either left or right). If your fingers are the posts and the spaces in between are the x or the even spaces. It would make a 5 posts and 4 spaces in between. I hope my explanation is clear.
So, in this case. There would be 19 posts and 18 even spaces.
18x = 900
[tex] \frac{18x = 900}{x} [/tex]
x = 50 m
it would look like this:
| 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m | 50m |
| <- this sign is the post. I just draw it to emphasize more.
Now count the spaces and the posts. :) There would be 19 posts and 18 spaces. I hope you get my point. :)
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