complexity theory – Is the multiplicative constant in the Big O notation are ignored because of Linear Speed-Up theorem?

I just want to know if Big O notation was designed to be used as a consequences of the Linear speed up theorem or not. For me I guess the answer is yes.

For example, if we didn’t have a linear speed-up theorem, then does it mean that we would have a different measure of time/space complexity? i.e. multiplicative constants does makes different. For example, $f(n) = 100 n$ isn’t the same as $g(n)=10^{82}n$. Therefore, in this regard, Big O notation is not useful. So, probably we have another way to measure algorithms.

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