Question

If f={(-2,4),(0,6), (2,8) and g={(-2,-1)(0,3)(2,5)} then [2f/3g+3g/2f](0) is

Answer 1

Answer:

Please see the attachment

Answer 2

Given :

$f=\{(-2,4),(0,6), (2,8)\}$

$g=\{(-2,-1)(0,3)(2,5)\}$

To find:

$(\dfrac{2f}{3g}+\dfrac{3g}{2f})(0)$

Solution:

$\dfrac{2f(0)}{3g(0)}+\dfrac{3g(0)}{2f(0)}\\\\=\dfrac{2\times 6}{3\times 3}+\dfrac{3\times 3}{2\times 6}\\\\=\dfrac{12}{9}+\dfrac{9}{12}\\\\=\dfrac{4}{3}+\dfrac{3}{4}\\\\=\dfrac{16+9}{12}\\\\=\dfrac{25}{12}$

Hence, the required value is $\dfrac{25}{12}$