find the perimeter of a rectangular object with the length of (square root 75) feet and the width (square root 48 feet)???????????????

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find the perimeter of a rectangular object with the length of (square root 75) feet and the width (square root 48 feet)???????????????

Mathematics
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perimeter is the total length of all the edges of the object so P=2*sqrt(75)+2*sqrt(48) P=2*sqrt(3*25)+2*sqrt(3*16) P=2*5sqrt(3)+2*4sqrt(3) P=10sqrt(3)+8sqrt(3) P=18sqrt(3)
is that how to do it or the answer?
that's how to do it. but if you struggle with changing sqrt(75) ans sqrt48 into simpler form then I can get you more detail

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ok, so i think you have problem with that. sqrt = square root sqrt(75)=sqrt(3*5*5)=sqrt(3*5^2)=sqrt(3)*sqrt(5^2)=5*sqrt(3) 48 is similar
i dont get what the peremiter would be tho of the two ?
rectangular has 2 longer edges and 2 shorter edges Perimeter = total lengths of 2 longer edges + total length of 2 shorter edges you know the longer edge is sqrt75 ft and shorter edge is sqrt48 ft, so parameter: P= sqrt75 + sqrt75 + sqrt48 +sqrt48 = 2sqrt75 + 2sqrt48
rectangle lol
so it would be 5 sqrt 3 plus 5 sqrt 3 plus 4 sqrt 3 or something
5 sqrt 3 plus 5 sqrt 3 plus 4 sqrt 3 plus 4 sqrt 3 since 2 edges each.
do i add them together for the answer
yeah, you will get 18 sqrt3
thanks so much
no problem, as long as you understand how it works, there shouldn't be any other problems
nope do you no what a house would look like if using those measures
what do you mean? the base should be a nice squared rectangle

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