- #1
Albert1
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$n=\dfrac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)}$
$find\,\,\, n$
$find\,\,\, n$
Albert said:$n=\dfrac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)}$
$find\,\,\, n$
kaliprasad said:we have $x^4+ 324 = (x^2)^2 + 18^2 = (x^2+18)^2 - 36x^2 = (x^2 + 6x + 18)(x^2-6x+18$
using this we get
$10^4 + 324 = (10 * 16 + 18)(10 * 4 + 18)$
$22^4 + 324 = (22 * 28 + 18)(22 * 16 + 18)$
$34^4 + 324 = (34 * 40 + 18)(34 * 28 + 18)$
$46^4 + 324 = (46 * 52 + 18)(46 * 40 + 18)$
so numerator $= (10^4+324)(22^4+324)(34^4+324)(46^4+324)$
$= ( 4 * 10 + 18)(10 * 16 + 18) (16 * 22 + 18) ( 22 * 28 + 18)(28 * 34 + 18) (34 * 40 + 18) (40 * 46 + 18) (46 * 52 + 18)$
now for denominator
$4^4 + 324 = (4 * 10 + 18)( 4 * (-2) + 18)$
$16^4 + 324 = (16 * 22 + 18)( 16 * 10 + 18)$
$28^4 + 324 = (28 * 34 + 18)(28 * 22 + 18)$
$40^4 + 324 = (40 * 46 + 18)(40 * 34 + 18)$
so denominator $= (4^4+324)(16+324)(28+324)(40+324)$
$= ( 4 * (-2) + 18)(4 * 10 +18) (10 * 16 + 18) ( 16 * 22 + 18)(22 * 28 + 18) (28 * 34 + 18) (34 * 40 + 18) (40 *46 + 18)$
so ratio = $\frac{46* 52 +18}{4 * (-2) + 18} = 240$
or n = 240
Albert said:check your answer
The formula for finding n for a given ratio is n = a/b, where n is the unknown number, a is the first number in the ratio, and b is the second number in the ratio.
To solve for n in a given ratio, you can use the formula n = a/b. Plug in the values for a and b and then solve for n using basic algebraic principles.
Yes, you can use decimals or fractions in the ratio when finding n. Just make sure to use the same format for both numbers in the ratio.
Finding n for a given ratio is significant because it allows you to determine the unknown quantity in a proportional relationship. This can be useful in various mathematical and scientific applications.
One limitation of using the formula to find n for a given ratio is that it assumes a linear relationship between the two numbers in the ratio. This may not always be the case in real-world situations.