How do you solve $_^{3} \sqrt{5 z - 4} + 7 = 10$ $_^3sqrt(5z-4) + 7 = 10$ ?

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Answer 1 · 2015-04-14T00:59:06

$\sqrt[3]{5 x - 4} + 7 = 10$ $root(3)(5x-4)+7 = 10$

isolate the cube root
$\sqrt[3]{5 z - 4} = 3$ $root(3)(5z-4) = 3$

cube both sides
$5 z - 4 = 27$ $5z-4 = 27$

isolate the $z$ $z$ term
$5 z = 31$ $5z = 31$

$\to z = \frac{31}{5}$ $rarr z = (31)/(5)$

How do you solve _^3sqrt(5z-4) + 7 = 10_3√5z−4+7=10_^3sqrt(5z-4) + 7 = 10?