anonymous
  • anonymous
Find the smallest value of k when a)280k is a perfect square b)882k is a cube It seems so simple, don't know why I can't solve it
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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baru
  • baru
clue: \[280k=2^2\times 7 \times 5\times 2\times k\]
Coolguy34
  • Coolguy34
The answer is a Because look at baru's clue
anonymous
  • anonymous
Baru, that's exactly where I get stuck. I don't know what to do next.

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denonakavro
  • denonakavro
Let me explain for a perfect square is a number you can square root without any decimals, so with the previous persons explanation, the first 2 is squared but the other 7,5,2 are not squared. thats why you have to multiply them 7x5x2 gives us 70, so the answer for a is 70.
anonymous
  • anonymous
Thanks! For b it's 882k = 2*3*3*7*7 No number seems to multiply itself thrice, what do I do?
baru
  • baru
if you factorize any perfect cube, it has to look like \((~)^3\times (~)^3\times (~)^3...\)
baru
  • baru
in this case, \( k=2^2 \times 3 \times 7 \)
hartnn
  • hartnn
think like this: i have 2 '7's, i need 3, so i need to multiply one 7 i have 2 '3's, i need 3, so i need to multiply one 3 i have 1 '2', i need 3, so i need to multiply two 2s 7*3*2*2
mathmate
  • mathmate
An example for finding the smallest value of k to make 12k a perfect cube: \(12=2^2\times 3\) To make it a perfect cube, we need the pattern: \(2^3\times 3^3\) so we need \(2^2×3k=2^3\times 3^3\) giving \(k=2\times 3^2\) or perfect cube = \(2^2 \times 3 \times (2\times 3^2) = 2^3\times 3^3=6^3\)
anonymous
  • anonymous
Excellent!

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