Mindblast3r
  • Mindblast3r
help pls
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
with?
Mindblast3r
  • Mindblast3r
\[r^2=\frac{ 9 }{ 2 }\]
Mindblast3r
  • Mindblast3r
\[r=\frac{ 3\sqrt{2} }{ 2 }\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Mindblast3r
  • Mindblast3r
how does the \[\frac{ 9 }{ 2 }\]
Mindblast3r
  • Mindblast3r
turn into \[\frac{ 3\sqrt{2} }{ 2 }\]
anonymous
  • anonymous
because 3 goes into 9 2 times so 2 is the square root of 3 and 9 and then the denominator stays the same
ParthKohli
  • ParthKohli
\[r^2 = \frac{9}{2}\]\[\Rightarrow r^2 = 9 \cdot \frac{1}{2}\]\[\Rightarrow r = \sqrt 9 \cdot \sqrt{\frac{1}{2}}\]\[\Rightarrow r = 3 \cdot \frac{1}{\sqrt2}= 3 \cdot \frac{\sqrt 2}{\sqrt 2 \cdot \sqrt 2} = \frac{3\sqrt 2}{2}\]
anonymous
  • anonymous
at least thats what i get out of it thats what it looks like to me
Mindblast3r
  • Mindblast3r
oh!!! I get it now! Thank you soo much!! I've been stuck on that forever!

Looking for something else?

Not the answer you are looking for? Search for more explanations.