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# halllllpppp!! quicc as a cat meow

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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?

Apr 2, 2020

#1
+626
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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.

Apr 2, 2020

#1
+626
+3

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.

AnExtremelyLongName Apr 2, 2020
#2
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Thanks!! It was correct though, so yay! also my sister made the username so yee

Apr 2, 2020
#3
+626
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Nice sister

#4
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i'm his sister! my name is actually charis but my user is matthew macdell based on a tv show character. thank you :))

matthewmacdell  Apr 2, 2020