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nika2105 [10]
2 years ago
7

Type the correct answer in the box.

Mathematics
1 answer:
Marysya12 [62]2 years ago
8 0

Answer:(-2, 2)

Step-by-step explanation:

7q² - 28 = 0

Add 28 to bothside

7q² - 28 + 28 = 0+ 28

7q² = 28

Divide bothside by 7

7q² / 7 = 28/7

q² = 4

Take the square root of bothside

√q² = ±√4

q = ± 2

q = -2 or q =2

q = (-2, 2)

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Max makes and sells posters. The function p(x)= -10x^2 +200x -250, graphed below, indicates how much profit he makes in a month
viktelen [127]
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>
7 0
3 years ago
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