She shall not reset her computer because she will have to load every thing again
Answer:
Nameplate
Explanation:
The banner on the front of a newsletter that identifies the publication is its nameplate.
The term "net radiation" refers to <span>the difference in amount of incoming and outgoing radiation the total amount of energy received by earth.
</span>
<span>Energy comes from the sunlight penetrates the top of the
atmosphere- that's the incoming energy. Then some of it is lost by reflection of clouds
or the Earth's surface, thermal radiation- that's the outgoing energy.</span>
Hi!
Well, this isn't exactly a question - but rather just a request. However, I'm going to attempt to try and <em>describe </em>to you how to approach this problem, instead of just writing the code for you and sending you on your way.
So, what's our general base goal here? We want to take a string into a function, and then print it out backwards. Seems simple enough!
Right away, we already have an idea how to set this code up. We need a main method which will call <em>PrintBackwards(), </em>which will have to take a parameter of type string.
This would look something like <em>PrintBackwards(string baseString). </em>Inside this method, we'd have to do something so we can see each character in this string and then store it in a new string.
I encourage you to try and tackle this on your own, but I can give you an idea. We can have a new valueless variable called reversedString, which will store our baseString but backwards.
We could try looping through the baseString for each character it possesses, and then keep adding onto our reversedString by doing something like +=. What I mean, is we'd access the very last index of baseString, and then keep appending characters into it.
So our loop would look something like <em>for(int i = baseString.length; i > 0; i--) {}.
</em>I haven't used C++ in awhile, so you'll have to find the specific syntax requirements. But with that loop, i represents the index of each character in baseString. It starts with the last index, and keeps going down in reverse.
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</em>Inside our loop, we could do something like reverseString += baseString.index(i); Again, I don't remember the specific syntax - so you'll have to do this on your own.
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</em>Hopefully, this helps! =)<em>
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