Acoffee merchant sells three blends of coffee. a bag of the house blend contains 100 grams of colombian beans and 100 grams of french roast beans. a bag of the special blend contains 100 grams of colombian beans, 300 grams of kenyan beans, and 200 grams of french roast beans. a bag of the gourmet blend contains 100 grams of colombian beans, 200 grams of kenyan beans, and 300 grams of french roast beans. the merchant has on hand 14 kilograms of colombian beans, 18 kilograms of kenyan beans, and 26 kilograms of french roast beans. if he wishes to use up all of the beans, how many bags of each type of blend can be made?

Step-by-step explanation:

Since we want to use all the beans, the constraints can be written as equations in h, s, and g, the numbers of bags of the House, Special, and Gourmet blends that can be made from existing stock. If units are 0.1 kg, then the equations are ...

  h + s + g = 140

  h +2s +3g = 260

       3s +2g = 180

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Subtracting the first equation from the second gives ...

  s + 2g = 120 . . . . . . . derived fourth equation

Subtracting that from the third equation gives ...

  2s = 60

  s = 30

Then filling that back into the derived fourth equation, we have ...

  30 + 2g = 120

  g = (120 -30)/2 = 45

And substituting into the first equation gives ...

  h + 30 + 45 = 140

  h = 65

The merchant can use all of his beans making 65 bags of house blend, 30 bags of special blend, and 45 bags of gourmet blend.

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

idk

step-by-step explanation:

i don't know