Average rate of change over interval [a,b]: r=[f(b)-f(a)]/(b-a)
In this case the interval is [0,2], then a=0, b=2
r=[f(2)-f(0)]/(2-0)
r=[f(2)-f(0)]/2
1) First function: h(x)
r=[h(2)-h(0)]/2
x=2→h(2)=(2)^2+2(2)-6
h(2)=4+4-6
h(2)=2
x=0→h(0)=(0)^2+2(0)-6
h(0)=0+0-6
h(0)=-6
r=[h(2)-h(0)]/2
r=[2-(-6)]/2
r=(2+6)/2
r=(8)/2
r=4
2) Second function: f(x)
A function, f, has an
x-intercept at (2,0)→x=2, f(2)=0
and a y-intercept at (0,-10)→x=0, f(0)=-10
r=[f(2)-f(0)]/2
r=[0-(-10)]/2
r=(0+10)/2
r=(10)/2
r=5
3) Third function: g(x)
r=[g(2)-g(0)]/2
From the graph:
g(2)=6
g(0)=2
r=(6-2)/2
r=(4)/2
r=2
4) Fourth function: j(x)
r=[j(2)-j(0)]/2
From the table:
x=2→j(2)=-8
x=0→j(0)=4
r=(-8-4)/2
r=(-12)/2
r=-6
Answer:
Pairs
1) h(x) 4
2) f(x) 5
3) g(x) 2
4) j(x) -6
For this case we have the following equations:
y = 60x + 40
y = 50x + 80
Equaling both equations we have:
60x + 40 = 50x + 80
From here, we clear the value of x:
60x - 50x = 80 - 40
10x = 40
x = 40/10
x = 4 weeks
Substituting x = 4 in any of the equations we have:
y = 60 (4) + 40
y = 240 + 40
y = 280 $
Answer:
$ 280 4 weeks
Answer:
How many drinks should be sold to get a maximal profit? 468
Sales of the first one = 345 cups
Sales of the second one = 123 cups
Step-by-step explanation:
maximize 1.2F + 0.7S
where:
F = first type of drink
S = second type of drink
constraints:
sugar ⇒ 3F + 10S ≤ 3000
juice ⇒ 9F + 4S ≤ 3600
coffee ⇒ 4F + 5S ≤ 2000
using solver the maximum profit is $500.10
and the optimal solution is 345F + 123S
Using a savings interest calculator, I found that either way your savings grow by exactly $1,020.20, so they share the same interest.
Answer:
All values are identical.
Step-by-step explanation:
We are given the following in the question:
If the standard deviation of a set of data is zero.
Then, all the values in data are identical.
This can be shown as:
Let all the terms in data be x.
Formula:
where 
are data points, 
is the mean and n is the number of observations.
Sum of squares of differences =
Thus, the correct answer is
All values are identical.
