Answer:
the answer is 29
Step-by-step explanation:
what is the best and easiest method to use to find sides A and B?
B
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>
To solve this problem, we need to use the Pythagorean Theorem, which states that a^2 + b^2 must equal c^2. "a" is the length of one of the legs (shorter sides of the triangle) and "b" is the length of the other leg. "c" is the length of the hypotenuse.
a = y = 36.25cm
b = x = 20.93cm
a^2 + b^2 = c^2
(36.25)^2 + (20.93)^2 = c^2
1314.0625 + 438.0649 = c^2
√1752.1274 = √c^2
c = 41.8584209 ≈ 41.86
So your final answer is...
The length of the hypotenuse is about 41.86 cm.
You can figure out how many miles per hour it travels by dividing 18/3. This gives us the answer of 6. 3/4 = 0.75 so we can multiply 6 * 0.75 to get 4.5 as our final answer. It will travel 4.5 miles in 3/4 of an hour.