Answer:
$77
Step-by-step explanation:
Let Friday be the variable f.
Let Saturday be the variable s.
Substitute 11 for f in the second equation.
Therefore, Kaitlin collected $77 on Saturday
Help me with this plz Choose: Same Or opposite
1 answer:
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I think you're asking if it's possible to have a cube root, fifth root, 7th root, etc of a number as a solution to f(x). The answer is yes it's possible.
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Example:
f(x) = x^3 - 29
This function has one real-number root of (cube root of 29) and the other two roots are complex or imaginary roots.