+231 One ordered pair (a,b) satisfies the two equations ab^4 = 384 and a^2 b^5 = 4608. What is the value of a in this ordered pair?
+658 Hello peepeepuupuu, your username is bad.
I'm not sure if this is even possible, but:
(1) \(ab^4=384\)
(2) \(a^2b^5=4608\)
(2) divided by (1) is \(\frac{a^2b^5}{ab^4}=\frac{4608}{384}\)
\(ab=12\)
Plugging back in:
\(b=\frac{12}{a}\)
\(a(\frac{12}{a}^4)=384\)
\(12(\frac{12}{a}^3)=384\)
\((\frac{12}{a})^3=32\)
\(32a^3=12^3\)
\(a^3=54\)
\(\boxed{a=3\sqrt[3]{2}}\) I'm not sure if this is correct because I've never divided two equations before.
+658 Hello peepeepuupuu, your username is bad.
I'm not sure if this is even possible, but:
(1) \(ab^4=384\)
(2) \(a^2b^5=4608\)
(2) divided by (1) is \(\frac{a^2b^5}{ab^4}=\frac{4608}{384}\)
\(ab=12\)
Plugging back in:
\(b=\frac{12}{a}\)
\(a(\frac{12}{a}^4)=384\)
\(12(\frac{12}{a}^3)=384\)
\((\frac{12}{a})^3=32\)
\(32a^3=12^3\)
\(a^3=54\)
\(\boxed{a=3\sqrt[3]{2}}\) I'm not sure if this is correct because I've never divided two equations before.
+231 Thanks!! It was correct though, so yay! also my sister made the username so yee
+105 i'm his sister! my name is actually charis but my user is matthew macdell based on a tv show character. thank you :))
