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.
Answer:
Step-by-step explanation:
find the inverse of the function y=9x+12
To find the inverse function, swap x and y, and solve the resulting equation for x.
If the initial function is not one-to-one, then there will be more than one inverse.
So, swap the variables: y=9x+12 becomes x=9y+12.
Now, solve the equation x=9y+12 for y.
(x) = (x-12)/9
y=(x−12)/9
Answer: Is there some answer choices? Because I can't help if I don't have any choices.
Step-by-step explanation:
Answer:
A.32
Step-by-step explanation:
