Answer: 490 grams of the first alloy should be used.
30 grams of the second alloy should be used.
Step-by-step explanation:
Let x represent the weight of the first alloy in grams that should be used.
Let y represent the weight of the second alloy in grams that should be used.
A chemist has two alloys, one of which is 15% gold and 20% lead. This means that the amount of gold and lead in the first alloy is
0.15x and 0.2x
The second alloy contains 30% gold and 50% lead. This means that the amount of gold and lead in the second alloy is
0.3y and 0.5y
If the alloy to be made contains 82.5 g of gold, it means that
0.15x + 0.3y = 82.5 - - - - - - - - - - - -1
The second alloy would also contain 113 g of lead. This means that
0.2x + 0.5y = 113 - - - - - - - - - - - - -2
Multiplying equation 1 by 0.2 and equation 2 by 0.15, it becomes
0.03x + 0.06y = 16.5
0.03x + 0.075y = 16.95
Subtracting, it becomes
- 0.015y = - 0.45
y = - 0.45/- 0.015
y = 30
Substituting y = 30 into equation 1, it becomes
0.15x + 0.3 × 30 = 82.5
0.15x + 9 = 82.5
0.15x = 82.5 - 9 = 73.5
x = 73.5/0.15
x = 490
Answer:
-6
Step-by-step explanation:
-3 + (-4) 4 + (-3) is -6 I hope that helps because that what I got
Answer:
216
Step-by-step explanation:
put -3 where x is
3 ( 5 -3^2 - 9(-3) )
3 ( 45 + 27 )
3 (72)
216
Answer:
y= 10/3 + x/3
Step-by-step explanation:
not sure how to explain this
Step-by-step explanation:
h(x) = 3. g(x) + 5
x= -1 h(x) = 3×8 + 5= 29
x= 0h(x) = 3×5 + 5= 20
x= 2 h(x) = 3×1 + 5= 8
x= 5 h(x) = 3×-5 + 5= -10
