anonymous
  • anonymous
4squ2-squ32+squ16 simply @phi
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[4 \sqrt2-\sqrt{32}+\sqrt{16}=4 \sqrt2-\sqrt{4^2 \times 2}+\sqrt{4^2}\] \[=4 \sqrt2-4\sqrt{ 2}+4=4\]
anonymous
  • anonymous
@jdoe0001
jdoe0001
  • jdoe0001
yeap

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phi
  • phi
looks like you are getting the hang of it.
anonymous
  • anonymous
I got 4squ2+8squ2-4squ2 but its not a choice
phi
  • phi
4squ2-squ32+squ16 I got 4squ2+8squ2-4squ2 but its not a choice 4 sqr(2) is ok - sqr(32) is - 4 sqr(2). which matches your -4squ2 how did you get 8 squ2 ?
anonymous
  • anonymous
16*2
phi
  • phi
which term did you start with ?
anonymous
  • anonymous
the first one
phi
  • phi
I am lost. If we start at the beginning 4squ2-squ32+squ16 simplify if we write this nicely \[ 4 \sqrt{2} - \sqrt{32} + \sqrt{16} \] the first term \( 4 \sqrt{2}\) is already simplified (you can not factor the 2 inside the square root) now the second term \[ - \sqrt{32} \] first, can you factor the 32 ?
phi
  • phi
32= 2*16 (which if we know 16 is a perfect square is good enough) but to continue = 2* 2* 8= 2*2*2 *4 = 2*2*2*2*2 we have 2 multiplied by itself 5 times. How many pairs are there ? group into pairs (like noah's ark): (2*2) * (2*2) * 2 so \[ \sqrt{32} = \sqrt{ (2 \cdot 2)(2 \cdot 2)2}\]

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