anonymous
  • anonymous
SR (600y^23)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\sqrt{600y ^{23}}\] is the actual equation
bahrom7893
  • bahrom7893
\[(600y^{23})^{1/2}\] = \[10 (y^{22/2})\sqrt{6*y} \]
bahrom7893
  • bahrom7893
\[10y^{11}\sqrt{68y}\]

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bahrom7893
  • bahrom7893
woops sorry: \[10y^{11}\sqrt{6*y}\]
bahrom7893
  • bahrom7893
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anonymous
  • anonymous
any questions?
anonymous
  • anonymous
yeah I am still lost as can be...
anonymous
  • anonymous
your in alegebra 2 right? im might be in the same chapter as you.
anonymous
  • anonymous
essentially it is algebra 2
anonymous
  • anonymous
okay soo first see in you can find a sqrt of 600.
anonymous
  • anonymous
it comes out as a decimal so I know that I have to simplify first and to me it would make the most sense to have it be \[\sqrt{100}\sqrt{6}\]
anonymous
  • anonymous
being as 100 is a perfect square
anonymous
  • anonymous
what my teacher taught me was to write it like Sqrt(100x6) then break those down to more simple numbers. Sqrt(50x2x2x3) [50x2=100] and 2x3=6
anonymous
  • anonymous
Do you guys want to go through the problem
anonymous
  • anonymous
?
anonymous
  • anonymous
i am going throught the problem...
anonymous
  • anonymous
the sqrt will eventually look like this (5x5x2x2x2x3)
anonymous
  • anonymous
you have 2 5's and 2 2's. take both and multiple 5 and 2 and put their product on the outside of the squareroot.
anonymous
  • anonymous
your equation should look like this. 10sqrt(6y^23) [6=product of remianing numbers in root]
anonymous
  • anonymous
for just figure out how many times 2 can go into 23 any remaing numbers remain in the root. answer = \[10y^{11}\sqrt{6y}\]

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