What is the value of the expression #a^2+b^3# when #a=1/3# and #b=1/2#?

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Answer 1 · 2016-09-14T16:28:35

In support of Brian's solution:

#17/72#

#a^2+b^2 -> (1/3)^2+(1/2)^3#

#=(1^2)/(3^2)+(1^3)/(2^3) = 1/9+1/8#

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#color(brown)("explanation about fractions")#

A fraction consists of:

#("count")/("size indicator of what you are counting") ->("numerator")/("denominator")#

You can only add the counts if the 'size indicators' are the same.

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Multiply by 1 and you do not change the vale but 1 comes in many forms:

#(1/9color(red)(xx1))+(1/8color(blue)(xx1))#

#(1/9color(red)(xx8/8))+(1/8color(blue)(xx9/9)) #

This makes the size indicators the same, giving:

#" "(1xx8)/(9xx8)color(white)(...)+color(white)(..)(1xx9)/(8xx9) ##color(white)(v)#

#" "8/72" "+" "9/72" "=" "(8+9)/72" "=" "17/72#