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The number of kilowatt/hour consumption you will consume if you use 3000 w twelve volts dc to 110 v inverter in full load for ten hours a day in one month is achieved by doing the correct calculations.

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Q: How much kilowatt hour consumption will you consume if you use 3000 w twelve volts dc to 110 v inverter in full load for ten hours a day in one month?
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Multiply What is 1B7twelve x 29twelve?

1B7 twelve x 29 twelve = 1B7 twelve x 9 twelve plus 1B7 twelve x 20 twelve First 1B7 twelve x 9 twelve is sum of 100 twelve x 9 twelve + B0 twelve x 9 twelve + 7 twelve x 9 twelve in base 10 7 x 9 = 63 = and that equals 53 twelve Next B0 twelve x 9 twelve is 11x12 in base 10 + 0 base 10 and that's 132 since 132 x 9 = 1188 in base ten. . 1188 = 8 x 144 + 3 x 12 + 0 x 1 so 1188 = 830 twelve Now 100 twelve times 9 twelve = 900 twelve so we have to add 53 twelve + 830 twelve + 900 twelve the ones column adds to 3, then twelves column is 5+3 =8 and the hundredfortyfours column is 8+9 = 17 base 10 which is 15 twelve so B0 twelve times 9 twelve equals 1583 twelve Now we have to determine 1B7 twelve x 20 twelve 1B7 twelve times 2 twelve is 200 twelve plus (22 base 10) x 10 twelve + 14 base 10 and that is 200 twelve + 1A0 twelve + 12 twelve = 3B2 twelve. we need to put a zero after that so it is 3B20 The final product is equal to 1583 twelve + 3B20 twelve and that is 4 [16]A3 twelve. we must get rid of that base 10 sixteen. Sixteen is 14 twelve so we put the 4 and carry the 1 and result is 54A3 twelve


Multiply 1b7 Twelve x 29 twelve?

1B7 twelve x 29 twelve