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: 8 subsets----------------------------------------------------------------------
----------------------------------------------------------------------
There are n = 3 elements in the given set, so there are 2^n = 2^3 = 2*2*2 = 8 subsets. Those 8 subsets are listed below
{a,b,c}
{a,b}, {a,c}, {b, c}
{a}, {b}, {c}
{ }
The first row is the original set. Any set is a subset of itself.
The second row represents subsets with exactly 2 elements.
The third row represents subsets with exactly 1 element
The fourth row is the empty set which can be written as
72 ÷ o should be the algebraic expression.
Let
The origin of coordinates the tree
r1 = vector position of the child 1.
r2 = vector position of the child 2
Child 1:
r1 = (12i + 12j)
Child 2:
r2 = (-18i + 11j)
The scalar product will be given by:
r1.r2 = ((12) * (- 18)) + ((12) * (11)) = - 84
The scalar product of their net displacements from the tree is -84m ^ 2
You'll want to use the quadratic formula:
-b (+/-) sqrt(b^2 - 4ac), all divided by 2a.
Under the square root you'll get:
-11
remember that the square root of -1 is i.
sqrt(-11) can be factored to sqrt(11*-1) and then sqrt(-1) * sqrt(11)
which becomes i*sqrt(11)
so your complex solution is:
-3 (+/-) (i*sqrt(11)), all over 4