Which operation would you do third in the following problem?
8(10-4-6
addition
subtraction
multiplication
division
Answers
Answered by
2
BY USING BODMAS RULE
WE HAVE TO FIRST SUBTRACT THE NUMBERS WHICH ARE INSIDE THE BRACKET AND LASTLY WE HAVE TO MULTIPLY THE NUMBER BY THE OUTSIDE NUMBER.
i.e.
8×(10-4-6)
=8×(10-10)
=8×0
=0
SO, THE ANSWER = 0
WE HAVE TO FIRST SUBTRACT THE NUMBERS WHICH ARE INSIDE THE BRACKET AND LASTLY WE HAVE TO MULTIPLY THE NUMBER BY THE OUTSIDE NUMBER.
i.e.
8×(10-4-6)
=8×(10-10)
=8×0
=0
SO, THE ANSWER = 0
Answered by
5
By using the BODMAS rule we should first we should solve the brakets.Then, multiplication.
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