Question

find the compounded ratio of (a-b):(a+b),(a+b)2:(a2+b2)and (a4-b4):(a2-b2)2

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Answer 1

Step-by-step explanation:

the answer is given in the attachment

Answer 2

Given that the three ratios are (a-b):(a+b), $(a+b)^2:(a^2+b^2)$ and $(a^4-b^4):(a^2-b^2)^2$

To find :

The compounded ratio for the given three ratios

Solution :

The formula for Compounded ratio is :

$Compounded ratio=\frac{product of antecedents}{product of consequents}$

$Compounded ratio=\frac{(a-b)\times (a+b)^2\times (a^4-b^4)}{(a+b)\times (a^2+b^2)\times (a^2-b^2)^2}$

By using the formula :

$\frac{1}{a^m}=a^{-m}$

$=\frac{(a-b)(a+b)^2.(a+b)^{-1}(a^4-b^4)}{(a^2+b^2)(a^2-b^2)^2}$

By using the formula :

$a^m.a^{-n}=a^{m-n}$

$=\frac{(a-b)(a+b)^{2-1}(a^4-b^4)}{(a^2+b^2)(a^2-b^2)^2}$

$=\frac{(a-b)(a+b)^{1}(a^2)^2-(b^2)^2}{(a^2+b^2)(a^2-b^2)^2}$

$=\frac{(a-b)(a+b)(a^2-b^2)(a^2+b^2)}{(a^2+b^2)(a^2-b^2)^2}$

$=\frac{(a^2-b^2)(a^2-b^2)}{(a^2+b^2)(a^2+b^2)^{-1}(a^2-b^2)^2}$

$=\frac{(a^2-b^2)^{1+1}}{(a^2+b^2)^{1-1}(a^2-b^2)^2}$

By using the identity :

$a^o=1$

$=\frac{(a^2-b^2)^{2}}{(a^2+b^2)^{0}(a^2-b^2)^2}$

$=\frac{(a^2-b^2)^{2}(a^2-b^2)^{-2}}{1}$

$=\frac{(a^2-b^2)^{2-2}}{1}$

$=\frac{(a^2-b^2)^{0}}{1}$

$=\frac{1}{1}$

Therefore $Compounded ratio=\frac{(a-b)\times (a+b)^2\times (a^4-b^4)}{(a+b)\times (a^2+b^2)\times (a^2-b^2)^2}=\frac{1}{1}$

The compounded ratio for the given three ratios (a-b):(a+b), $(a+b)^2:(a^2+b^2)$ and $(a^4-b^4):(a^2-b^2)^2$ is 1:1