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.
Step-by-step explanation:
the area is
length × width
increasing the length by 30% means multiplying the original length by 1.3.
increasing the width by 10% means multiplying the original width by 1.1.
so, the new area is (by using the old length and width)
length × 1.3 × width × 1.1 = length × width × 1.3×1.1 =
= old area × 1.3 × 1.1 = old area × 1.43
so, the area increases by 43%.
Answer:
Step-by-step explanation:
Pattern 1 consists of
small squares (here n = 1 and 
).
Pattern 2 consists of
small squares (here n = 2 and 
).
Pattern 3 consists of
small squares (here n = 3 and 
).
Thus, pattern n consists of
small squares.
Answer:
then 2x = 0 and x - 3 = 0
Step-by-step explanation:
Step-by-step explanation:
measure all the sides then multiply you will get the answer
