This is a mixture problem.
Let the 9 karat gold to be mixed be x. The 18 karat would weight (200 - x). Since the total = 200g.
mass1 * karat1 + mass2*karat = total mass * total carat
Let mass1 be x g. mass 2 would be = (200 -x)
karat 1 = 9 karat 2 = 18
Total mass = 200g.
Total karat = 14.
9*x + 18*(200 -x) = 14*200
9x + 18*200 - 18*x = 14*200
18*200 + 9x - 18x = 14*200
3600 - 9x = 2800
3600 - 2800 = 9x
800 = 9x
9x = 800
x = 800/9
x ≈ 88.89
Therefore 88.89 grams of 9 karat gold must be mixed.
Answer:
Step-by-step explanation:
Bấm máy tính hệ phương trình là ra
f(x) = 3 - 2sin(x)
0 = 3 - 2sin(x)
- 3 - 3
-3 = -2sin(x)
-2 -2
1¹/₂ = sin(x)
sin⁻¹(1¹/₂) = sin⁻¹[sin(x)]
sin⁻¹(1¹/₂) = x
10x+20y=540
x+y=34
y=34-x
Plug in (34-x) for y:
10x+20(34-x)=540
10x+680-20x=540
-10x=-140
10x=140
x=14
There were 14 ten cent coins and 20 20 cent coins
