A community for students.
Here's the question you clicked on:
 0 viewing
Tia_Sweetheart17
 4 years ago
square root of 512m^3. show work
Tia_Sweetheart17
 4 years ago
square root of 512m^3. show work

This Question is Closed

Rogue
 4 years ago
Best ResponseYou've already chosen the best response.2Radicals... In order to find the square root, we need to find a perfect square factor of 512. 64 is a perfect square factor, which can be multiplied by 8 to give 512. So we can write the square of 512 this way. \[\sqrt{512 m^3} = \sqrt{64 * 8 m^3}\]

Rogue
 4 years ago
Best ResponseYou've already chosen the best response.2\[\sqrt{64}*\sqrt{8 m^3} = 8 * \sqrt{4 * 2 m^3}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0256 is also a perfect square factor :)

Rogue
 4 years ago
Best ResponseYou've already chosen the best response.2\[8*\sqrt{4}*\sqrt{2 m^3} = 8*2*\sqrt{2 m^3} = 16\sqrt{2m^3}\]

Rogue
 4 years ago
Best ResponseYou've already chosen the best response.2Yeah, we could've skipped all that using 256, thanks Rick :)

Rogue
 4 years ago
Best ResponseYou've already chosen the best response.2The cube root of meters cubed is meters to the 1.5 power (yeah, I know, its weird). So you can write your final answer as:\[\sqrt{512 m^3} = 16\sqrt{2} m^{1.5}\]

Rogue
 4 years ago
Best ResponseYou've already chosen the best response.2Well, is that meters cubed, or is it a variable?

Rogue
 4 years ago
Best ResponseYou've already chosen the best response.2Either way, the answer is still the same.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I assume it's a variable

Rogue
 4 years ago
Best ResponseYou've already chosen the best response.2Yeah, makes more sense, meters to the 1.5 power, lol.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.