quick computing company produces calculators. they have found that the cost, c(x), of making x calculators…

Explanation:

Step1: Set up the quadratic - function

Let the quadratic cost function be $c(x)=ax^{2}+bx + c$. We know that $c(2)=4a + 2b + c=45$, $c(4)=16a+4b + c = 143$, and $c(10)=100a + 10b + c=869$.

Step2: Subtract equations to eliminate $c$

Subtract the first equation from the second: $(16a + 4b + c)-(4a + 2b + c)=143 - 45$, which simplifies to $12a+2b = 98$ (Equation 1). Subtract the second equation from the third: $(100a + 10b + c)-(16a+4b + c)=869 - 143$, which simplifies to $84a+6b = 726$ (Equation 2).

Step3: Solve the system of linear equations

Multiply Equation 1 by 3: $3(12a + 2b)=3\times98$, so $36a+6b = 294$. Subtract this new - equation from Equation 2: $(84a+6b)-(36a + 6b)=726 - 294$. $84a+6b - 36a - 6b=432$, $48a=432$, $a = 9$.

Step4: Find $b$

Substitute $a = 9$ into $12a+2b = 98$. $12\times9+2b=98$, $108+2b=98$, $2b=98 - 108=-10$, $b=-5$.

Step5: Find $c$

Substitute $a = 9$ and $b=-5$ into $4a + 2b + c=45$. $4\times9+2\times(-5)+c=45$, $36-10 + c=45$, $26 + c=45$, $c = 19$. So the cost function is $c(x)=9x^{2}-5x + 19$.

Step6: Calculate $c(7)$

Substitute $x = 7$ into $c(x)=9x^{2}-5x + 19$. $c(7)=9\times7^{2}-5\times7 + 19=9\times49-35 + 19=441-35 + 19=425$.

Answer:

425