Answer:
Its the second choice.
Step-by-step explanation:
g(x) = (x - 3)^2 + 9
The -3 in the parentheses moves the graph of x^2 3 units to the right.
The + 9 moves it upwards 9 units.
Here is our profit as a function of # of posters
p(x) =-10x² + 200x - 250
Here is our price per poster, as a function of the # of posters:
pr(x) = 20 - x
Since we want to find the optimum price and # of posters, let's plug our price function into our profit function, to find the optimum x, and then use that to find the optimum price:
p(x) = -10 (20-x)² + 200 (20 - x) - 250
p(x) = -10 (400 -40x + x²) + 4000 - 200x - 250
Take a look at our profit function. It is a normal trinomial square, with a negative sign on the squared term. This means the curve is a downward facing parabola, so our profit maximum will be the top of the curve.
By taking the derivative, we can find where p'(x) = 0 (where the slope of p(x) equals 0), to see where the top of profit function is.
p(x) = -4000 +400x -10x² + 4000 -200x -250
p'(x) = 400 - 20x -200
0 = 200 - 20x
20x = 200
x = 10
p'(x) = 0 at x=10. This is the peak of our profit function. To find the price per poster, plug x=10 into our price function:
price = 20 - x
price = 10
Now plug x=10 into our original profit function in order to find our maximum profit:
<span>p(x)= -10x^2 +200x -250
p(x) = -10 (10)</span>² +200 (10) - 250
<span>p(x) = -1000 + 2000 - 250
p(x) = 750
Correct answer is C)</span>
Answer:
13.22
Step-by-step explanation:
were solving for t and we know:
a(t)=p(1+(r/n))^nt
5000=a-the total
2940=p-the starting amount
.041=r-the rate
1=n-compound (annual)
plug this into a graph :
5000=2940(1+(.041/1))^x
and you get : 13.22
