Answer:Your left hand side evaluates to:
m+(−1)mn+(−1)m+(−1)mnp
and your right hand side evaluates to:
m+(−1)mn+(−1)m+np
After eliminating the common terms:
m+(−1)mn from both sides, we are left with showing:
(−1)m+(−1)mnp=(−1)m+np
If p=0, both sides are clearly equal, so assume p≠0, and we can (by cancellation) simply prove:
(−1)(−1)mn=(−1)n.
It should be clear that if m is even, we have equality (both sides are (−1)n), so we are down to the case where m is odd. In this case:
(−1)(−1)mn=(−1)−n=1(−1)n
Multiplying both sides by (−1)n then yields:
1=(−1)2n=[(−1)n]2 which is always true, no matter what n is
Answer:
122
Step-by-step explanation:
32+32+29+29=122
If we convert the given in its mathematical form, we have,
(30x⁶/14y⁵)(7y²/6x⁴)
It can be observed that the numerator of the first and the denominator of the second have a common factor of 6x⁴. Also, the denominator of the first and the numerator of the second expression have a common factor of 7y².
((6x⁴)(5x²)/(7y²)(2y³))(7y²/6x⁴)
Cancellation of the common terms will give us an answer of,
<em>5x²/2y³
</em><em />Therefore, the simplified version of the involved operation is 5x²/2y³. <em>
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The correct answer is: [D]: "no real solutions" .
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The only "answers" would be: " <span>± 6i " ;
</span> → <span>both of which are not "real solutions" .
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