How do you evaluate #b=sqrt(a^2-c^2)# given a=29 and c=20?

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Answer 1 · 2016-11-24T11:56:10

#b = 35.23# rounded to the nearest 100th.

You substitute #29# for #a# and #20# for #c# in the equation and then do the math to solve for #b#:

#b = sqrt(29^2 - 20^2)#

#b = sqrt(29*29 - 20*20)#

#b = sqrt(841 - 400)#

#b = sqrt(1241)#

#b = 35.23#

Answer 2 · 2016-11-24T11:57:59

#b=21#

To evaluate b, substitute the values given for a and c into the radical.

#rArrb=sqrt(29^2-20^2)#

#=sqrt(841-400)=sqrt441=21#