No, having the greater rate of change does not mean a function will necessarily reach an output with the smallest input. It also depends on the initial values of the functions. If Jeremy's aunt lives closer to school, Jeremy's rate of change is smaller, but he could still get to school before Amy.
Answer: 
Step-by-step explanation:
Given: A mirror with a parabolic cross section is used to collect sunlight on a pipe located at the focus of the mirror.
The pipe is located 8 inches from the vertex of the mirror.
Assume the vertex is at the origin.
If the parabola opens upwards
then the coordinates of focus= (0,8)
We know that equation of parabola with focus (0,a) and open upards is of the form (vertex=(0,0)) is
Substitute the value of a=8 in equation, we get
Therefore, equation of the parabola that models the cross section of the mirror is
The dot plot or histogram will be skewed.
The mean is pulled up or down toward the tail.
The mean is affected more than the median.
Sample Response: When there is an outlier in the data set, the dot plot or histogram will be skewed. In a skewed representation, the mean is pulled up or down toward the tail of the data. Therefore, skewed data affects the mean more than the median.
I hour because, one of the 5 carpenters built one chair in one hour, so one of the 100 carpenters would build one chair in one hour, there are only 95 more carpenters, nothing else different.
Hope I helped!
NOTE THIS IS AN EXAMPLE:
Let t = time, s = ostrich, and g = giraffe.
Here's what we know:
s = g + 5 (an ostrich is 5 mph faster than a giraffe)
st = 7 (in a certain amount of time, an ostrich runs 7 miles)
gt = 6 (in the same time, a giraffe runs 6 miles)
We have a value for s, so plug it into the first equation:
(g + 5)t = 7
gt = 6
Isolate g so that we can plug that variable value into the equation:
g = 6/t
so that gives us:
(6/t + 5)t = 7
Distribute:
6 + 5t = 7
Subtract 6:
5t = 1
Divide by 5:
t = 1/5 of an hour (or 12 minutes)
Now that we have a value for time, we can plug them into our equations:
1/5 g = 6
multiply by 5:
g = 30 mph
s = 30 + 5
s = 35 mph
Check by imputing into the second equation:
st = 7
35 * 1/5 = 7
7 = 7
