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Renee99
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Brian is creating a collage on a piece of cardboard that has an area of 110r3 square centimeters. The collage is covered entirely by pieces of paper that do not overlap. Each piece has an area of √r^5 square centimeters. Use the given information to determine an expression for the total number of pieces of paper used.
Can someone please tel me how I would solve this. Or could you just write this in equation form, I have trouble doing so. Word problems just aren't my best. Please and thank you to whoever is able to help me! :)
 2 years ago
 2 years ago
Renee99 Group Title
Brian is creating a collage on a piece of cardboard that has an area of 110r3 square centimeters. The collage is covered entirely by pieces of paper that do not overlap. Each piece has an area of √r^5 square centimeters. Use the given information to determine an expression for the total number of pieces of paper used. Can someone please tel me how I would solve this. Or could you just write this in equation form, I have trouble doing so. Word problems just aren't my best. Please and thank you to whoever is able to help me! :)
 2 years ago
 2 years ago

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Renee99 Group TitleBest ResponseYou've already chosen the best response.1
Does this sound good? Since each piece of paper has the same area ( √r^5 cm^2 ) You need to put √r^5 into exponential notation > √r^5 = r^5/2 This can be thought of in two ways: a) the square root of r taken to the fifth power or b) finding the square root of the fifth power or r. they are equivalent Multiply it by a variable and set it equal to 100r^3 cm^2. N(r^(5/2) ) =100 r^3 N = 100 r63 / (r^(5/2)) = 100 r^(3  (5/2)) = 100r^(1/2) N = 100 r^(1/2) = 100√r
 2 years ago

CHAD159753 Group TitleBest ResponseYou've already chosen the best response.1
Yeah thats correct
 2 years ago

mads4566 Group TitleBest ResponseYou've already chosen the best response.0
it sounds good...
 2 years ago

SchoolSlacker Group TitleBest ResponseYou've already chosen the best response.0
TRUE STORY
 2 years ago

Renee99 Group TitleBest ResponseYou've already chosen the best response.1
ur funny! :)
 2 years ago
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