Answer:
20kg of 25% copper alloy
30kg of 60% copper alloy
Step-by-step explanation:
There are 2 kinds of metal, first metal(A) have 25% copper while the second metal(B) has 60% copper. The metalworker want to create 50kg of metal, which means the total weight of both metals is 50kg (A+B = 50). The metalworker also want the metal made of a 46% which is 23kg(0.25A + 0.6B = 0.46*50). From these sentences, we can derive 2 equations. We can solve this with substitution.
A+B = 50
A= 50-B
Let's put the first equation into the second.
0.25A + 0.6B = 0.46 *50
0.25(50-B) + 0.6B =23
12.5 - 0.25B + 0.6B =23
0.35B=23 -12.5
B= 10.5/ 0.35= 30
Then we can solve A
A= 50-B
A= 50-30
A=20
Answer:
I'm sorry I can help you maybe a little bit if that's okay. Btw what grade are you in?
Step-by-step explanation:
Answer:
I don't understand the question
Step-by-step explanation:
Graph C. The equation of the line is y=2x+5. 5 is the y-intercept so the line has to go through that point.
Answer:
The values of x for which the model is 0 ≤ x ≤ 3
Step-by-step explanation:
The given function for the volume of the shipping box is given as follows;
V = 2·x³ - 19·x² + 39·x
The function will make sense when V ≥ 0, which is given as follows
When V = 0, x = 0
Which gives;
0 = 2·x³ - 19·x² + 39·x
0 = 2·x² - 19·x + 39
0 = x² - 9.5·x + 19.5
From an hint obtained by plotting the function, we have;
0 = (x - 3)·(x - 6.5)
We check for the local maximum as follows;
dV/dx = d(2·x³ - 19·x² + 39·x)/dx = 0
6·x² - 38·x + 39 = 0
x² - 19/3·x + 6.5 = 0
x = (19/3 ±√((19/3)² - 4 × 1 × 6.5))/2
∴ x = 1.288, or 5.045
At x = 1.288, we have;
V = 2·1.288³ - 19·1.288² + 39·1.288 ≈ 22.99
V ≈ 22.99 in.³
When x = 5.045, we have;
V = 2·5.045³ - 19·5.045² + 39·5.045≈ -30.023
Therefore;
V > 0 for 0 < x < 3 and V < 0 for 3 < x < 6.5
The values of x for which the model makes sense and V ≥ 0 is 0 ≤ x ≤ 3.
