How to solve #(3^-2+4)/(5^0-9^-1)#?

#(3^-2+4)/(5^0-9^-1)#

2 Answers
Bhavin B.
May 9, 2018

#37/8#

Explanation:

we need to use exponential law here:
#a^-1 = 1/a#
#a^0 = 1# ---> a nything to power #0#, is equal to #1#

So from:
#(3^-2 +4)/(5^0 - 9^-1)#

We get:
#(1/3^2 + 4)/(1-(1/9))# = #(1/9+4)/(1-(1/9)#

**multiply by 9 both in numerator and denumerator and we get:

#((9xx1/9) +(4xx9))/((9xx1) - (9xx1/9))# = #(1+36)/(9-1)# = #37/8#

Jim G.
May 9, 2018

#37/8#

Explanation:

#"using the "color(blue)"laws of exponents"#

#•color(white)(x)a^-mhArr1/a^m" and "a^0=1#

#rArr(3^-2+4)/(5^0-9^-1)#

#=(1/3^2+4)/(1-1/9^1)#

#=(1/9+36/9)/(9/9-1/9)#

#=(37/9)/(8/9)=37/cancel(9)xxcancel(9)/8=37/8#

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