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Find the positive solution to each equation. If the solution is irrational, write the solution using square root or cube root notation.
For each cube root, find the two whole numbers that it lies between.
Order the following values from least to greatest:
\(\displaystyle \sqrt[3]{530},\;\sqrt{48},\;\pi,\;\sqrt{121},\;\sqrt[3]{27},\;\frac{19}{2}\)
Select all the equations that have a solution of \(\frac{2}{7}\):
\(x^2=\frac27\)
\(x^2=\frac{4}{14}\)
\(x^2=\frac{4}{49}\)
\(x^3=\frac{6}{21}\)
\(x^3=\frac{8}{343}\)
\(x^3=\frac67\)
The equation \(x^2=25\) has two solutions. This is because both \(5 \boldcdot 5 = 25\), and also \(\text-5 \boldcdot \text-5 = 25\). So 5 is a solution, and \(\text-5\) is also a solution. But! The equation \(x^3=125\) only has one solution, which is 5. This is because \(5 \boldcdot 5 \boldcdot 5 = 125\), and there are no other numbers you can cube to make 125. (Think about why \(\text-5\) is not a solution!)
Find all the solutions to each equation.
Find the value of each variable, in units, to the nearest tenth.