Ametalworker has a metal alloy that is 25% copper and another alloy that is 70% copper. how many kilograms of each alloy should the metalworker combine to create 50 kg of a 61% copper alloy? the metalworker should use nothing kilograms of the metal alloy that is 25% copper and nothing kilograms of the metal alloy that is 70% copper

10 kgs of 25% copper alloy and 40 kgs of 70% copper alloy

Step-by-step explanation:

Let a be kg of 25% copper alloy, and

b be kg of 70% copper alloy

We can write two equations:

1. a + b = 50

2. 0.25a+0.7b=0.61(50)

We can write #1 as b = 50 - a, and then plug it into #2. We have:

0.25a+0.7b=0.61(50)

0.25a+0.7(50 -  a) = 0.61(50)

0.25a + 35 - 0.7a = 30.5

-0.45a = 30.5 - 35

-0.45 a = -4.5

a = -4.5 / - 0.45

a = 10

Also, b = 50 - a, so b = 50 - 10 = 40

The metalworker should use 10 kgs of 25% copper alloy and 40 kgs of 70% copper alloy to make it.

idk

step-by-step explanation:

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places