The width of a rectangle is 3 less than 2 times the length. If the perimeter is 51 what is the actual length and width of the re
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So, you had done everything right so far (other than squaring the 2), but that was only half of the question.
to find the least common multiple, you need to first figure out what the prime factors have in common.
each have two twos. both have one 5, so we know our answer will look something like
now to figure out the other stuff... we have to represent the greatest amount of everything that is left, and we have 3s and 7s left over, so we need to figure out how many of each we need.
one has one 3 and one has two, so we need two threes. now our equation is
what's the only number we have to deal with? 7...
how many sevens does 60 have? 0, and 630 has 1, so we know we need one 7. our answer becomes
to find the least common multiple, you need to first figure out what the prime factors have in common.
each have two twos. both have one 5, so we know our answer will look something like
now to figure out the other stuff... we have to represent the greatest amount of everything that is left, and we have 3s and 7s left over, so we need to figure out how many of each we need.
one has one 3 and one has two, so we need two threes. now our equation is
what's the only number we have to deal with? 7...
how many sevens does 60 have? 0, and 630 has 1, so we know we need one 7. our answer becomes