Answer:
The equations that represent the reflected function are
Step-by-step explanation:
The correct question in the attached figure
we have the function
we know that
A reflection across the y-axis interchanges positive x-values with negative x-values, swapping x and −x.
therefore
The reflection of the given function across the y-axis will be equal to
(Remember interchanges positive x-values with negative x-values)
An equivalent form will be
![f(x)=5(\frac{1}{5})^{(-1)(x)}=5[(\frac{1}{5})^{-1})]^{x}=5(5)^{x}](https://tex.z-dn.net/?f=f%28x%29%3D5%28%5Cfrac%7B1%7D%7B5%7D%29%5E%7B%28-1%29%28x%29%7D%3D5%5B%28%5Cfrac%7B1%7D%7B5%7D%29%5E%7B-1%7D%29%5D%5E%7Bx%7D%3D5%285%29%5E%7Bx%7D)
therefore
The equations that represent the reflected function are
Lizzy's new neckalce cost 0.4 cents less than lucy's neckalce that cost 0.29. cents.How many cents did lizzy's neckalce cost.
Answer:
√(p²-4q)
Step-by-step explanation:
Using the Quadratic Formula, we can say that
x = ( -p ± √(p²-4(1)(q))) / 2(1) with the 1 representing the coefficient of x². Simplifying, we get
x = ( -p ± √(p²-4q)) / 2
The roots of the function are therefore at
x = ( -p + √(p²-4q)) / 2 and x = ( -p - √(p²-4q)) / 2. The difference of the roots is thus
( -p + √(p²-4q)) / 2 - ( ( -p - √(p²-4q)) / 2)
= 0 + 2 √(p²-4q)/2
= √(p²-4q)
