I'm doing an algebra equation and I'm having trouble with what seems like a simple step. Can someone please explain how you get 3(3/4)^(-4)+4 . The answer came out as 3 but I'm not sure the steps to it. Can someone please help, thank you.
You must have written the expression incorrectly as it certainly doesn't come out as 3.
$${\mathtt{3}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}^{\left(-{\mathtt{4}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}} = {\frac{{\mathtt{364}}}{{\mathtt{27}}}} = {\mathtt{13.481\: \!481\: \!481\: \!481\: \!481\: \!5}}$$
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Thanks for getting back to me. Here is the equation I'm working on. It says 3 on there.
p(-4) = 3*(3/4)(-4+4) = 3*(3/4)0 = 3*1 = 3 (because any number, including 3/4, to the power 0 is 1).
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