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What is the product of 138 multiplied by 570 but showing all work in long multiplication from start to finish using Roman numerals?
Wiki User
∙ 7y ago
Using long multiplication in Roman numerals:-
D*CXXXVIII = M(LXX)
L*CXXXVIII = C(V)MM
XX*CXXXVIII = MMDCCLX
So: M(LXX)+C(V)MM+MMDCCLX = (LXXVIII)DCLX
The above in English numbers are equivalent to the following:-
500*138 = -1000+70,000
50*138 = -100+7,000
20*138 = 2,760
So: -1,000+70,000-100+7,000+2,760 = 78,660
Wiki User
∙ 7y ago
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Just tackle the problem in the same way as you would work out a problem in long multiplication and then add up the products as follows:- C times DLXX = (LVII) => 100*570 = 57,000 XXX times DLXX = (XVII)C => 333*570 = 17,100 V times DLXX = (II)DCCCL => 5*570 = 2,850 III times DLXX = (I)DCCX => 3*570 = 1,710 (LVII)+(XVII)C+(II)DCCCL+(I)DCCX = (LXXVIII)DCLX 57,000+17,100+2,850+1,710 = 78,660 Note that numerals within brackets indicate multiplication by 1,000.
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The exponent shows how many times the number is being multiplied by itself. So if it's 10 to the power of 3 (which is 1000) ur are showing a shorter way of showing 10x10x10=1000.
What is 318 multiplied by 720 showing work?
318 multiplied by 720 is 228,960.
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What is the product of 138 and 570 but showing all work from start to finish in Roman numerals?
Just tackle the problem in the same way as you would work out a problem in long multiplication and then add up the products as follows:- C times DLXX = (LVII) => 100*570 = 57,000 XXX times DLXX = (XVII)C => 333*570 = 17,100 V times DLXX = (II)DCCCL => 5*570 = 2,850 III times DLXX = (I)DCCX => 3*570 = 1,710 (LVII)+(XVII)C+(II)DCCCL+(I)DCCX = (LXXVIII)DCLX 57,000+17,100+2,850+1,710 = 78,660 Note that numerals within brackets indicate multiplication by 1,000.
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