Answer:
-279.63
Step-by-step explanation:
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Answer:
-4 sqrt(7)/7
Step-by-step explanation:
csc theta = -4/3
csc theta = hypotenuse / opposite side
hypotenuse = 4
opposite = 3
Using the pythagorean theorem
a^2 + b^2 = c^2
3^2 + b^2 = 4^2
9+b^2 = 16
b^2 = 16-9
b^2 = 7
Taking the square root
sqrt(b^2) = sqrt(7)
b = sqrt(7)
We are in the third quadrant so only tan and cot are positive
that means the x and y values are "negative" so a = -3 and b = - sqrt(7)
sec theta = hypotenuse / adjacent
= 4/ - sqrt(7)
rationalizing
-4 sqrt(7)/ sqrt(7)* sqrt(7)
= -4 sqrt(7)/7
Answer:
c) l x - 5 l - 4
Step-by-step explanation:
when you're graphing functions whatever is inside the paranthesis, square root or absolute value determines whether you go left or right, if its a negative you go to the right, if its positive you go to the left. So the opposite of what you would expect it to go
in this instance its - 5, so you go 5 units to the right
and the number outside determines if you go up or down, in this question its - 4 so you go down 4 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>
