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Algebra
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OpenStudy (anonymous):
Simplify. ∜400/∜5
12 years ago
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OpenStudy (mathstudent55):
This rule may be helpful to get started:
\( \dfrac{\sqrt[n]{a}} {\sqrt[n]{b}} = \sqrt[n]{\dfrac{a}{b} } \)
12 years ago
OpenStudy (anonymous):
yea but i got these A. 5∜2
B. 2∜5
C. 2√5
D. ∜80
12 years ago
OpenStudy (anonymous):
@mathstudent55
12 years ago
OpenStudy (mathstudent55):
Try my suggestion first. It won't give you the fianl answer, but it'll get you started. Then we can work on the next step.
12 years ago
OpenStudy (mathstudent55):
\( \dfrac{\sqrt[4]{400}} {\sqrt[4]{8}} = \sqrt[4]{\dfrac{400}{8} } \)
Now divide 400 by 8, what do you get?
12 years ago
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OpenStudy (anonymous):
i gott 50 @mathstudent55
12 years ago
OpenStudy (mathstudent55):
Sorry. It's 400 divided by 5, not 8.
Let me do it again.
12 years ago
OpenStudy (anonymous):
its 80
12 years ago
OpenStudy (mathstudent55):
Right.
\(\dfrac{\sqrt[4]{400}} {\sqrt[4]{5}} = \sqrt[4]{\dfrac{400}{5}} = \sqrt[4]{80} \)
12 years ago
OpenStudy (anonymous):
THANKS
12 years ago
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OpenStudy (mathstudent55):
Now we need to simplify
\( \sqrt[4]{80} \)
We need to find the prime factors of 80.
\(80 = 2 \times 2 \times 2 \times 2 \times 5 = 2^4 \times 5\)
\(\sqrt[4]{80} =\sqrt[4]{2^4 \times 5} \)
\(= \sqrt[4]{2^4} \times \sqrt[4]{5}\)
\(= 2\sqrt[4]{5} \)
12 years ago
OpenStudy (anonymous):
GOTT THANKS
12 years ago
OpenStudy (mathstudent55):
wlcm
12 years ago
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