Let r and s be rings and define rx s = {(r, s) |ter, se s} with addition and multiplication defined componentwise on rx s which thus makes rx sa ring. (i) if r and s are both commutative, then prove that rx s is also commutative (ii) if r and s both have identity, then prove that rx s also has an identity

$4

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

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the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

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