Ask
your own question, for FREE!
Mathematics
18 Online
Heeeeeeeeeeeeeeeelp I need Math Heeeeeeeeeeeeeeeeeeeeeeeeeeeeeelp
Still Need Help?
Join the QuestionCove community and study together with friends!
@mathmath333
$$ \text{area_of_rectangle} = (\sqrt[5]{16}) \cdot 2^\frac{1}{5} = \bigg(\sqrt[5]{2^4}\bigg) \cdot \sqrt[5]{2} = \\ = \big(\sqrt[5]{2}\big)^4 \cdot \sqrt[5]{2} = (\sqrt[5]{2})^5 = 2 $$
@pitamar could you help with one more?!?!?!?
sure
Still Need Help?
Join the QuestionCove community and study together with friends!
Ok. Well we have: $$ \bigg(5^\frac{3}{4}\bigg)^\frac{2}{3} $$ We can say $$ \bigg(5^\frac{3}{4}\bigg)^\frac{2}{3} = \large5^{\big(\frac{3}{4} \cdot \frac{2}{3}\big)} = 5^{\frac{2}{4}} = 5^\frac{1}{2} = \sqrt{5} $$
thank so very very much!!!!!!!!!!!!!!!
Sure np
Can't find your answer?
Make a FREE account and ask your own questions, OR help others and earn volunteer hours!
Join our real-time social learning platform and learn together with your friends!
Join our real-time social learning platform and learn together with your friends!
Latest Questions
VERNEISH:
Two objects are moving along separate linear paths where each path is described by position, d, and time, t.
Maci:
what are the best stock apps for people who want to start investing?
Twaylor:
genuienly how do i get better self control i keep watcing youtube and theres this
IzzyCookie:
|dw:1792074056138:dw| Uhm, I don't know where to ask this, but I'm in college welding at my community college, and I need help learning vertical welding for
thehearken:
not sure if this is considered spam, im sorry if it is, but does anyone have tips
12 hours ago
0 Replies
0 Medals
4 days ago
3 Replies
2 Medals
1 week ago
3 Replies
0 Medals
2 days ago
9 Replies
1 Medal
1 week ago
2 Replies
1 Medal