Pure copper was mixed with a 10% copper alloy to produce an alloy thát was 25% copper. How much of the pure copper and how much 10% alloy were used to produce 36 kg of 25% alloy?

6kg pure copper and 30 kg 10% copper was mixed to give 36kg of 25% alloy

Step-by-step explanation:

Here, we want to produce 36 kg of 25% alloy

Let the Pure copper be x kg while 10% alloy be y kg

Pure copper is simply 100% copper

Thus;

x + y = 36 •••••(i)

Then;

100% of x + 10% of y = 25% of 36

= x + 0.1y = 9 •••••• ii)

From i x = 36-y

from ii, x = 9-0.1y

Equate both x

36-y = 9-0.1y

36-9 = 0.1y + y

0.9y = 27

y = 27/0.9

y = 30

x = 36-y

x = 36-30

x = 6 kg

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: