Respuesta :
You use ax^2+bx+c=c(x)
Use (3,61), (7,165), and (12, 385)= those are your a and b (a,b)...
a= 2
b= 6
c= 25
c(x)=2x^2+6x+25
Since you're looking for the price of six calculators, substitute 6 in for x
c(6)=2*36 + 6*6 + 25
Calculate...
Answer: $169
Use (3,61), (7,165), and (12, 385)= those are your a and b (a,b)...
a= 2
b= 6
c= 25
c(x)=2x^2+6x+25
Since you're looking for the price of six calculators, substitute 6 in for x
c(6)=2*36 + 6*6 + 25
Calculate...
Answer: $169
Answer:
Hence, the cost for 6 calculators is:
$ 133
Step-by-step explanation:
Let it follows the quadratic model.
i.e let
[tex]p(x)=ax^2+bx+c[/tex]
where a,b and c are real numbers.
where p(x) denotes the cost of x calculators.
Now, it is given that:
The company discovered that it costs $61 to produce 3 calculators.
i.e. when x=3 p(x)=61
Hence, we get:
[tex]9a+3b+c=61-------------(1)[/tex]
and cost of 7 calculators is: $ 165
i.e. [tex]49a+7b+c=165--------------(2)[/tex]
and cost of 12 calculators is: $385
i.e. [tex]144a+12b+c=385--------------(3)[/tex]
on subtracting equation (2) from (1) we get:
[tex]40a+4b=104[/tex]
on dividing both side by 4 we get:
[tex]10a+b=26---------(4)[/tex]
on subtracting equation (3) from equation (2) and dividing by 5 we get:
[tex]19a+b=44-------------(5)[/tex]
Now on subtracting equation (4) from (5) we get:
[tex]9a=18\\\\i.e.\\\\\\a=2[/tex]
and by putting the value of a in equation (4) we get:
[tex]b=6[/tex]
and now putting the value of a and b in equation (1) we get:
[tex]c=25[/tex]
Hence, the function that models the cost is:
[tex]p(x)=2x^2+6x+25[/tex]
Now, we are asked to find the cost of 6 calculators
when x=6 we have:
[tex]p(x)=2\times 6^2+6\times 6+25\\\\\\i.e.\\\\\\p(x)=133[/tex]
The cost of 6 calculators is:
$ 133
