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
<h2>
Answer:</h2>
Ques 1)

Ques 2)

<h2>
Step-by-step explanation:</h2>
Ques 1)
We know that if a graph is stretched by a factor of a then the transformation if given by:
f(x) → a f(x)
Also, we know that the translation of a function k units to the right or to the left is given by:
f(x) → f(x+k)
where if k>0 then the shift is k units to the left
and if k<0 then the shift is k units to the right.
Here the graph of f(x) is transformed into the graph of g(x) by a vertical stretch of 4 units and a translation of 4 units right.
This means that the function g(x) is given by:

Ques 2)
We know that the transformation of the type:
f(x) → f(x)+k
is a shift or translation of the function k units up or down depending on k.
If k>0 then the shift is k units up.
and if k<0 then the shift is k units down.
Here, The graph of the function f(x)=|3x| is translated 4 units up.
This means that the transformed function g(x) is given by:

You have two 30-60-90 triangles, ADC and BDC.
The ratio of the lengths of the sides of a 30-60-90 triangle is
short leg : long leg : hypotenuse
1 : sqrt(3) : 2
Using triangle ADC, we can find length AC.
Using triangle BDC, we can find length BC.
Then AB = AC - BC
First, we find length AC.
Look at triangle ACD.
DC is the short leg opposite the 30-deg angle.
DC = 10sqrt(3)
AC = sqrt(3) * 10sqrt(3) = 3 * 10 = 30
Now, we find length BC.
Look at triangle BCD.
For triangle BCD, the long leg is DC and the short leg is BC.
BC = 10sqrt(3)/sqrt(3) = 10
AB = AC - BC = 30 - 10 = 20