You have one type of chocolate that sells for $2.40/lb and another type of chocolate that sells for $9.90/lb. You would like to have 30 lbs of a chocolate mixture that sells for $4.20/lb. How much of each chocolate will you need to obtain the desired mixture

We need 22.8 lbs of the first type of chocolate and 7.2 lbs of the second type of chocolate

Step-by-step explanation:

Step 1 :

Let x represent the quantity of the first type of chocolate

and y represent the quantity of the second type of chocolate

The quantity of the final mixture required = 30 lbs

Hence

x + y = 30

Step 2 :

Price of the first type of chocolate = $ 2.40 / lb

Price of the second type of chocolate = $9.90/lb

The total price of the final mixture = $4.20 / lb

Hence the price for 30 lbs = 4.2 × 30 = $126

Hence the equation representing this would be

2.4 x + 9.9 y = 126

Step 3 :

Solving the equations obtained in step 1 and 2 for x and y we get,

2.4 x + 2.4 y = 72

2.4 x + 9.9 y = 126

Subtracting equation 2 from 1 we get,

-7.5 y = -54  = > y = 7.2

x + y = 30 = > x = 30 - y = 30 - 7.2 = 22.8

Step 4 :

Answer :

We need 22.8 lbs of the first type of chocolate and 7.2 lbs of the second type of chocolate

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: