Ametalworker has a metal alloy that is 30? % copper and another alloy that is 55? % copper. how many kilograms of each alloy should the metalworker combine to create 90 kg of a 50? % copper? alloy?

18 kilograms of 30% copper alloy is mixed with 72 kilograms of 55% copper alloy to makes  90 kg of a 50% copper alloy .

Step-by-step explanation:

Let us assume that the kilogram of 30% copper used to make alloy be x .

Let us assume that the kilogram of 55% copper used to make alloy be y .

As given

A metalworker has a metal alloy that is 30% copper and another alloy that is 55% copper.

The metalworker combine to create 90 kg of a 50% copper alloy .

Equation becomes

x  + y = 90

30% is written in the decimal form .

= 0.30

55% is written in the decimal form .

= 0.55

50% is written in the decimal form .

= 0.50

Equation becomes

Concentration of 30% copper alloy × Number of kilograms of 30% copper alloy used + Concentration of 55% copper alloy × Number of kilograms of 55% copper alloy used = Concentration of 50% copper alloy × Number of kilograms of 50% copper alloy .

0.30x + 0.55y = 0.50 × 90

Simplify the above equation

30x + 55y = 4500

Two equation becomes

x  + y = 90

30x + 55y = 4500

Multiply x  + y = 90 by 30 and subtracted form 30x + 55y = 4500 .

30x - 30x + 55y - 30y = 4500 - 2700

25y = 1800

y = 72 kilograms

Putting the value of y in the equation .

x  + 72 = 90

x = 90 - 72

x = 18 kilograms

Therefore 18 kilograms of 30% copper alloy is mixed with 72 kilograms of 55% copper alloy to makes  90 kg of a 50% copper alloy .

i think 1: 2

step-by-step explanation:

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb