Xy = -150
x + y = 5
x + y = 5
x - x + y = -x + 5
y = -x + 5
xy = -150
x(-x + 5) = -150
x(-x) + x(5) = -150
-x² + 5x = -150
-x² + 5x + 150 = 0
-1(x²) - 1(-5x) - 1(-150) = 0
-1(x² - 5x - 150) = 0
-1 -1
x² - 5x - 150 = 0
x = -(-5) ± √((-5)² - 4(1)(-150))
2(1)
x = 5 ± √(25 + 600)
2
x = 5 ± √(625)
2
x = 5 ± 25
2
x = 2.5 ± 12.5
x = 2.5 + 12.5 or x = 2.5 - 12.5
x = 15 or x = -10
x + y = 5
15 + y = 5
- 15 - 15
y = -10
(x, y) = (15, -10)
or
x + y = 5
-10 + y = 5
+ 10 + 10
y = 15
(x, y) = (-10, 15)
The two numbers that add up to 5 and multiply to -150 are 15 and -10.
Answer:
2.4 hours
Step-by-step explanation:
David's time = 6 hours
Anthony's time = 4 hours
the combined time is given by the following formula
1/ (david's time) + 1/(anthony's time) = 1/total time
1/6 + 1/4 = 1/Total
4/24 + 6/24 = 1/Total
10/24 = 1/ total (taking reciprocal of both sides)
24/10 = total
total time = 24/10 = 2.4 hours
To check the decay rate, we need to check the variation in y-axis.
Since our interval is
![-2We need to evaluate both function at those limits.At x = -2, we have a value of 4 for both of them, at x = 0 we have 1 for the exponential function and 0 to the quadratic function. Let's call the exponential f(x), and the quadratic g(x).[tex]\begin{gathered} f(-2)=g(-2)=4 \\ f(0)=1 \\ g(0)=0 \end{gathered}](https://tex.z-dn.net/?f=-2We%20need%20to%20evaluate%20both%20function%20at%20those%20limits.%3Cp%3E%3C%2Fp%3E%3Cp%3EAt%20x%20%3D%20-2%2C%20we%20have%20a%20value%20of%204%20for%20both%20of%20them%2C%20at%20x%20%3D%200%20we%20have%201%20for%20the%20exponential%20function%20and%200%20to%20the%20quadratic%20function.%20Let%27s%20call%20the%20exponential%20f%28x%29%2C%20and%20the%20quadratic%20g%28x%29.%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%5Btex%5D%5Cbegin%7Bgathered%7D%20f%28-2%29%3Dg%28-2%29%3D4%20%5C%5C%20f%280%29%3D1%20%5C%5C%20g%280%29%3D0%20%5Cend%7Bgathered%7D)
To compare the decay rates we need to check the variation on the y-axis of both functions.
Now, we calculate their ratio to find how they compare:
This tell us that the exponential function decays at three-fourths the rate of the quadratic function.
And this is the fourth option.
I hope this helps you find the answer you're looking for
