How many pounds of a 15% copper alloy must be mixed with 700lb of a 30% copper alloy to maybe a 25.5% copper alloy

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Step-by-step explanation:

Let represent the mass (in pounds) of that copper alloy required, such that the final mixture would contain copper by mass.

Consider: if pounds of that copper alloy is mixed with pounds that copper alloy, what would be the mass of copper in the mixture?

Therefore, the mixture would contain of copper.

The mass of that mixture would be . The mass fraction of copper in that mixture would be:

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This ratio is supposed to be equal to . These two pieces of equations combine to give an equation about :

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Simplify and solve for :

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Therefore, of that alloy would be required.

zero

step-by-step explanation:

f(2)= x-2

plug in the 2 in the "x"

f(2)=2-2

f(2)=0

use a calculator or a graph