For this case what you should see is that for the interval [9, 11] the behavior of the function is almost linear.
Therefore, we can find the average rate of change as follows:
m = (y2-y1) / (x2-x1)
m = (11-6) / (11-9)
m = (5) / (2)
m = 5/2
Answer:
the average rate of speed over the interval [9, 11] is:
D. 5 / 2
Answer:
In February, 423 daytime minutes is used
Step-by-step explanation:
Let the base plan charges be x
And cost per daytime minute be y
In December,
x + 510y = 92.25------------------(1)
In January,
x + 397y = 77.56---------------------(2)
Subtracting eq(2) from eq(1)
x + 510y = 92.25
x + 397y = 77.56
-------------------------------
0 + 113y = 14.69
-------------------------------
y = \frac{14.69}{113}
y = 0.13----------------------------------(3)
Substituting (3) in (1)
x + 510(0.13) = 92.25
x + 66.3 = 92.25
x = 92.25 - 66.3
x = 25.95
So In February
base plan + (daytime minute)(cost per daytime minute) = 80.9
25.95 + (daytime minute)(0.13) = 80.9
(daytime minute)(0.13) = 80.9 - 25.95
(daytime minute)(0.13) = 54.95
(daytime minute) =
daytime minutes = 422.69
daytime minute 
<span>Using the Pythagorean theorum we can solve this v a^2+b^2= c^2. A is the distance from base of house to ladder A= 1.5, c ifls length of ladder, c= 10. (1.5)^2 + ;b^2 = (10)^2. Solve for b. B= 9.88. 12 foot height of house - 9.88 feet to top of angled ladder = 2.11 from top of ladder to edge of roof</span>
This is<span> not the exact, precise </span>definition<span> of a </span>limit. If you would like to see the more precise and mathematical definition<span> of a </span>limit<span> you should check out the The </span>Definition<span> of a </span>Limit<span> section at the end of this chapter. The </span>definition<span> given above </span>is<span> more of a “working” </span>definition<span>.</span>
