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

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Answer 1 · 2016-10-30T02:34:38

The answer is #-2#.

Evaluate #root(3)(2(color(blue)t-color(red)r))# given #r=color(red)(-3), s=4, t=color(blue)(-7)#

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

#root(3)(2(color(blue)(-7)-(color(red)(-3)))#

#root(3)(2(-7+3))#

#root(3)(2(-4)#

#root(3)(-8)=-2#

Note that the cube root of #-8# is #-2# because #-2 * -2 * -2=-8#