Witch property of exponents must be used first to solve the expression (xy^2)1/3
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An aritmetic sequence is like this
where a1=first term and d=common difference
geometric is
where a1=first term and r=common ratio
can it be both aritmetic and geometric
hmm, that means that the starting terms should be the same
therfor we need to solve
what values of d and r make all natural numbers of n true?
are there values that make all natural numbers for n true?
when n=1, then d(1-1)=0 and r^(1-1)=1, so already they are not equal
the answer is no, a sequence cannot be both aritmetic and geometric
geometric is
can it be both aritmetic and geometric
hmm, that means that the starting terms should be the same
therfor we need to solve
what values of d and r make all natural numbers of n true?
are there values that make all natural numbers for n true?
when n=1, then d(1-1)=0 and r^(1-1)=1, so already they are not equal
the answer is no, a sequence cannot be both aritmetic and geometric