First we can simplify f(x) to make it a little easier. We can write it as
4x^3+2x^2-7x+4 by combining like terms
The we just subtract g(x) from f(x) and simplify
4x^3+2x^2-7x+4-(5x^3-7x+4) and remember to distribute the negative to g(x)
4x^3+2x^2-7x+4-5x^3+7x-4 (combine like terms)
-x^3+2x^2
In factored form, this becomes
-x^2(x-2)
Hope this helps
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:
105
Step-by-step explanation:
l is parallel m so angle 3+angle 6=180
