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Answer :
(i)We have:\(3^0+4^{-1}\times 2^2\)
\(=3^0+4^{-1}\times 2^2=(1+\frac{1}4)\times4\)
\(=(\frac{4+1}4)\times4=\frac{5}4\times 4=5\)
(ii)We have:\(2^{-1}\times4^{-1}÷ 2^{-2}\)
\(=2^{-1}\times4^{-1}÷ 2^{-2}=(\frac{1}2\times\frac{1}4)÷\frac{1}{2^2}\)
\(=\frac{1}8÷\frac{1}4=\frac{1}8\times\frac{4}1\)
\(=\frac{1}2\)
(iii)We have:\((\frac{1}2)^{-2}+(\frac{1}3)^{-2}+(\frac{1}4)^{-2}\)
\(=(\frac{1}2)^{-2}+(\frac{1}3)^{-2}+(\frac{1}4)^{-2}=2^2+3^2+4^2\)
\(=4+9+16=29\)
(iv)We have:\((3^{-1}+4^{-1}+5^{-1})^0\)
\(=(3^{-1}+4^{-1}+5^{-1})^0=1\quad\)[∵\(a^0=1\)]
(v)We have:\([(\frac{-2}3)^{-2}]^2\)
\(=[(\frac{-2}3)^{-2}]^2=[(-\frac{3}2)^2]^2\)
\(=(\frac{9}4)^2=\frac{81}{16}\)