The point which would lie on the graph of f(x) = log2x is; (1, 0.301).
<h3>Which point would lie on the graph of f(x) = log2x?</h3>
Since it follows from the task content that the given point (–1, 0.5) lies on the graph of f –1(x) = 2x.
The subsequent graph of f(x) = log2x would result from taking logarithms in which case, we have the graph passing through points determined as follows;
f(x) = log2^x
f(x) = x log2
Hence, when x = 1;
f(1) = 1log 2
Hence, we have point; (1, 0.301).
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Answer:
2
Step-by-step explanation:
Please see attached picture for full solution.
Answer:
x^3(x-4)
Step-by-step explanation:
x^4-4x^3
x^3(x-4)
It's very important to use the symbol " ^ " here to denote exponentiation.
You probably meant "-10x^(-10), which would be "-10 times x to the minus tenth power."
If you want the value, you could do the following:
-10 1
------------------ = - ---------- = - 10^(-9).
10^10 10^9
Answer:
ionevebknokg00glek
Step-by-step explanation:
useg00glemayne