For #t = 1/2 a^(3/2)#, what is #t# when #a = 36#?

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Answer 1 · 2017-10-15T11:56:07

See a solution process below:

We can rewrite this equation as:

#t = 1/2a^(1/2 *3)#

#t = 1/2(a^(1/2))^3#

#t = 1/2(sqrt(a))^3#

Substituting #color(red)(36)# for #color(red)(a)# gives:

#t = 1/2(sqrt(color(red)(a)))^3# becomes:

#t = 1/2(sqrt(color(red)(36)))^3#

We must remember the square root of a number produces a positive and negative result:

#t = 1/2(-sqrt(color(red)(36)))^3# and #t = 1/2(sqrt(color(red)(36)))^3#

#t = 1/2(-6)^3# and #t = 1/2(6)^3#

#t = 1/2 * -216# and #t = 1/2 * 216#

#t = -108# and #t = 108#