To find the answer we have to slowly analyze the command. For surely on the one side we want to have total cash spent, so 80.8 W e kn o wt ha tt h ere a re 2 t y p eso ff r am es 5 , 80 and 9.60 . W e kn o wt oo , t ha t f i s t h e n u mb ero ff r am es 5.80 bought. Simply, to find the cash spent on those frames we have to just mulitply 5.80 t im es f , so w eo b t ain [ t e x ] f ∗ 5.8 [ / t e x ] . W ec an s a y t ha t n u mb ero f bi gg er f r am es i s 20 − f . S o t o f in d c a s h s p e n t o nbi gg er f r am es w e ha v e t o m u lt i pl y 9.60 times (20-f), so 9.60 ∗ ( 20 − f ) Now we can just add cost of all frames together f ∗ 5.80 + 9.60 ( 20 − f ) And we know that is equal to 80.8, so f ∗ 5.80 + 9.60 ( 20 − f ) =80.8
As we can see, the answer is b.
The equation that models Lea's frame purchases is 5.8 f + 9.6 ( 20 − f ) = 80.8 from option B. This represents the total cost based on the number of each type of frame. Thus, option B is the correct choice.
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