Answer: a = 3∛2
<u>Step-by-step explanation:</u>
ab⁴ = 384 --> a = 384/b⁴
Substitute a = 384/b⁴ into the second equation to solve for "b".
a²b⁵ = 4608
![\bigg(\dfrac{384}{b^4}\bigg)^2\cdot b^5=4608\\\\\\\dfrac{147,456b^5}{b^8}=4608\\\\\\\dfrac{147,456}{b^3}=4608\\\\\\\dfrac{147,456}{4608}=b^3\\\\\\32=b^3\\\\\\\sqrt[3]{32} =b\\\\\\2\sqrt[3]{4} =b](https://tex.z-dn.net/?f=%5Cbigg%28%5Cdfrac%7B384%7D%7Bb%5E4%7D%5Cbigg%29%5E2%5Ccdot%20b%5E5%3D4608%5C%5C%5C%5C%5C%5C%5Cdfrac%7B147%2C456b%5E5%7D%7Bb%5E8%7D%3D4608%5C%5C%5C%5C%5C%5C%5Cdfrac%7B147%2C456%7D%7Bb%5E3%7D%3D4608%5C%5C%5C%5C%5C%5C%5Cdfrac%7B147%2C456%7D%7B4608%7D%3Db%5E3%5C%5C%5C%5C%5C%5C32%3Db%5E3%5C%5C%5C%5C%5C%5C%5Csqrt%5B3%5D%7B32%7D%20%3Db%5C%5C%5C%5C%5C%5C2%5Csqrt%5B3%5D%7B4%7D%20%3Db)
Substitute b = 2∛4 into the first equation to solve for "a".
ab⁴ = 384
a(2∛4)⁴ = 384
a = 384/(2∛4)⁴
a = 24/4∛4
= 6/∛4
= 6(∛2)/2
= 3∛2
Let x be the page number then x+1 would be the other page number (because the pages are two facing pages which means they follow on consecutively)
Then
(x)(x+1) = 156
x^2 + x - 156 = 0
x = 12 or x = -13( but x can't be a negative number)
so x = 12
and the next page is 13
I have nooo clue the problem doesn’t make sense
G=11 because the math and the numbers all add one to one round number
