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Answer:
Marine pollution is a major problem because it <em>endangers marine life that ingests the pollution which affects us when we eat fish or seafood, makes the ocean less safe for humans to swim in, and the ocean can't function the way it's meant to when unpolluted.</em>
Take the expression in chunks:
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
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


>>
Note that
. So

where the cyclic sum notation means

In other words, we take the sum over all possible cycles of the sequence of arguments to the summand. The second square-bracketed chunk reduces to

So to recap, we've reduced the starting expression to

>>

Finally, we have

Then distributing
and rewriting to decode the message, we have

We get the values of m as 1 and n as 0.
We are given a function:
f(x) = m / (x - 1) + n
We are also given that:
f (-2) = 1 and f (4) = 3
We have to find the value of m and n.
f (-2) = m (2 - 1) + n
1 = m + n
n = 1 - m
f (4) = m / (4 - 1) + n
3 = 3 m + n
3 = 3 m + 1 - m
3 = 2 m + 1
2 m = 3 - 1
2 m = 2
m = 2 / 2
m = 1
m + n = 1
1 + n = 1
n = 0.
Therefore, we get the values of m as 1 and n as 0.
Learn more about function here:
brainly.com/question/4025726
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Tony drank 1/3 of a cup of water in the morning.
Tony drank 3/5 of a cup of water in the afternoon.
You need to find out how much water did he drink it all so we need to add 3/5 with 1/3.
3/5 + 1/3 =

= 5/15 + 9/15
= 
The answer is 14/15
hope this helps!