Answer:
6/25 km in 1 min; or 0.24 km/min
Step-by-step explanation:
The info about Angel is not necessary.
Jayden runs 3 laps in 5 mins
Each lap is 2/5 km, so Jayden runs 3 x (2/5 km) in 5 min = 6/5 km in 5 min.
Distance = rate x time
6/5 km = rate (5 min)
We want to isolate r (rate), so divide both sides by 5 min
6/5 km ÷ 5 min = r
6/5 km (1/5 min) = r
6/25 km/min = r (notice how the units worked out correctly to km/min)
So, Jayden runs 6/25 km in 1 min.
9a + (9t+27)=x
To write it another way you could flip the numbers around:
(9t+27)+9a=x
Or flip the variable to the other side.
x=9a + (9t+27)
Any way you put the numbers as long as x is on the either side, you will get the same answer.
Hope this helps!
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>
I think it’s 55.6 square inches
Sorry if I’m wrong! :)
Evaluate f(9) + f(5)= 14f If I didn't answer Your Question fully than tell me please!
