Respuesta :

dalendrk
[tex]\dfrac{1}{3}x^2+5.2y\\\\for\ x=3\ and\ y=2\\\\\dfrac{1}{3}\cdot3^2+5.2\cdot2=\dfrac{1}{3}\cdot9+10.4=\dfrac{9}{3}+10.4=3+10.4=13.4[/tex]

Answer:

B. 13.4

Step-by-step explanation:

We have the fraction, [tex]L=\frac{1}{3}x^2+5.2y[/tex].

It is required to find the value of the fraction when x = 3 and y = 2.

So, on substituting these values, we get,

[tex]L=\frac{1}{3}(3)^2+5.2\times 2[/tex]

i.e. [tex]L=\frac{9}{3}+10.4[/tex]

i.e. [tex]L=3+10.4[/tex]

i.e. [tex]L=13.4[/tex]

So, the value of the fraction is 13.4

Hence, option B is correct.

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