A coffee merchant sells three blends of coffee. A bag of the house blend contains

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:

Abbreviation Used ( Colombian Bean = C , Kenyan Bean = K , French Roast Beans = F , House Blend = H , Special Blend = S , Gourmet Blend = G)

Now According to Question, We have 39kg of C , 15kg of K , 36kg of F

300gm in H + 200gm in S + 300gm in G = Total Colombian Beans(39000gm) ↔ Eq. 1

200gm in G + 200gm in S  = Total Kenyan Bean(15000gm) ↔ Eq. 2

300gm in H + 100gm in S + 300 in of G = Total FrenchRoast Beans(36000gm) ↔ Eq. 3

if we add all three Equations we get

600gm in H  + 500gm in S + 800gm in G = C+K+F(90000gm)↔Eq. 4

Now Subtract Eq. 1 - Eq. 3 We Get

100*S=3000

Total bag of Special Blend is 30

Now Put the value of S=30 in above equations(Eq. 1 & Eq. 4)

we get

3H+3G=330↔(eq. 5)  &   6H+8G=750↔(eq. 6)

solve these equation by subtracting (eq. 6 - 2*eq.5)

we get 2G=90  ⇔ Total bag of Gourmet Blend is 45

now By Putting G value in Eq. 6

We Get 6H=390 ⇔ Total bag of House Blend is 65

(5n²+10xn)(2n-3)

Step-by-step explanation:

10n³-15n²+20xn²-30xn

1. Factor by grouping

(10n³-15n²)+(20xn²-30xn)

2. Pull out Greatest Common Factor from each Pair

5n²(2n-3)+10xn(2n-3)

3. Rewrite Factored Expression Properly

(5n²+10xn)(2n-3)

=(5n²+10xn)(2n-3)

ik this doesnt make any sense but hope this helps