How do you evaluate $\sqrt[3]{2 (t - r)}$ $root3(2(t-r))$ given r=-3, s=4, t=-7?

Question

1581 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-30T02:34:38

The answer is $- 2$ $-2$ .

Evaluate $\sqrt[3]{2 (t - r)}$ $root(3)(2(color(blue)t-color(red)r))$ given $r = - 3, s = 4, t = - 7$ $r=color(red)(-3), s=4, t=color(blue)(-7)$

Substitute the given values for $r and t$ $color(red)r and color(blue)t$

$\sqrt[3]{2 (- 7 - (- 3))}$ $root(3)(2(color(blue)(-7)-(color(red)(-3)))$

$\sqrt[3]{2 (- 7 + 3)}$ $root(3)(2(-7+3))$

$\sqrt[3]{2 (- 4)}$ $root(3)(2(-4)$

$\sqrt[3]{- 8} = - 2$ $root(3)(-8)=-2$

Note that the cube root of $- 8$ $-8$ is $- 2$ $-2$ because $- 2 \cdot - 2 \cdot - 2 = - 8$ $-2 * -2 * -2=-8$

How do you evaluate 3√2(t−r)root3(2(t-r)) given r=-3, s=4, t=-7?