A metallurgist needs to make 12 oz. Of an alloy containing 45% copper. He is going to melt and combine one metal that is 30% copper with another metal that is 50% copper. How much of each should he use?

The metallurgist should use 3 ounces of the 30 % copper alloy and 9 ounces of the 50 % copper alloy to make 12 ounces of 45 % copper alloy.

Step-by-step explanation:

The ounce is a mass unit, as we notice that the metallurgist wants to make 12 ounces of an alloy containing 45 % copper by mixing two metal with different copper proportions. We can use the following two equations:

Alloys

(Eq. 1)

Copper

(Eq. 2)

Where:

- Mass of the 30 % copper alloy, measured in ounces.

- Mass of the 50 % copper alloy, measured in ounces.

- Mass of the 45 % copper alloy, measured in ounces.

- Proportion of copper in the 30 % copper alloy, dimensionless.

- Proportion of copper in the 50 % copper alloy, dimensionless.

- Proportion of copper in the 45 % copper alloy, dimensionless.

Now, the mass of the 50 % copper alloy is cleared in Eq. 1 and eliminated in Eq. 2:

And we clear and calculate the mass of the 30 % copper alloy:

If we know that , , and , the mass of the 30 % copper alloy:

And the mass of the 50 % copper alloy is:

The metallurgist should use 3 ounces of the 30 % copper alloy and 9 ounces of the 50 % copper alloy to make 12 ounces of 45 % copper alloy.

2/3x + 7 = 29

step-by-step explanation:

two thirds = 2/3

plus a number = x

at least 29 = 29

there for you should get 2/3x + 7 = 29

step-by-step :