The production cost is an illustration of a quadratic function
<h3>How to determine the equation</h3>
We have the following parameters:
- Computing components (a)
- Number of devices (x),
- Labor (b)
- Savings on reusable material (c)
Using the instructions in the question, the equation of the production cost would be:
![y =ax^2 + bx -c](https://tex.z-dn.net/?f=y%20%3Dax%5E2%20%2B%20bx%20-c)
Hence, the production cost is ![y =ax^2 + bx -c](https://tex.z-dn.net/?f=y%20%3Dax%5E2%20%2B%20bx%20-c)
<h3>Change the subject of the equation</h3>
In (a), we have:
![y =ax^2 + bx -c](https://tex.z-dn.net/?f=y%20%3Dax%5E2%20%2B%20bx%20-c)
Subtract bx - c from both sides
![y - bx + c = ax^2](https://tex.z-dn.net/?f=y%20-%20bx%20%2B%20c%20%3D%20ax%5E2)
Divide through by x^2
![a = \frac{y - bx + c}{x^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7By%20-%20bx%20%2B%20c%7D%7Bx%5E2%7D)
Hence, the equation that represents the computing component is ![a = \frac{y - bx + c}{x^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7By%20-%20bx%20%2B%20c%7D%7Bx%5E2%7D)
Read more about equations and functions at:
brainly.com/question/1214333
The exact calculations:
Principal, P=3000
interest rate = 33% per year
Period, t=10 years
Future value
![F=Pe^(it)](https://tex.z-dn.net/?f=F%3DPe%5E%28it%29)
![=3000e^(0.33*10)](https://tex.z-dn.net/?f=%3D3000e%5E%280.33%2A10%29)
![=81337.92](https://tex.z-dn.net/?f=%3D81337.92)
Interest
= future value - principal
= 81337.92-3000
=
78337.92 (to the nearest cent)
Answer:
please upload the figure so i could help you :)
Step-by-step explanation:
Answer:
1) 1/10
2)16/100
3)6..../10
4) 3/5
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Step-by-step explanation: