Answer:
M and P
Step-by-step explanation:
they do not cross each other
Answer:
part A) The scale factor of the sides (small to large) is 1/2
part B) Te ratio of the areas (small to large) is 1/4
part C) see the explanation
Step-by-step explanation:
Part A) Determine the scale factor of the sides (small to large).
we know that
The dilation is a non rigid transformation that produce similar figures
If two figures are similar, then the ratio of its corresponding sides is proportional
so
Let
z ----> the scale factor

The scale factor is equal to

substitute

simplify

Part B) What is the ratio of the areas (small to large)?
<em>Area of the small triangle</em>

<em>Area of the large triangle</em>

ratio of the areas (small to large)

Part C) Write a generalization about the ratio of the sides and the ratio of the areas of similar figures
In similar figures the ratio of its corresponding sides is proportional and this ratio is called the scale factor
In similar figures the ratio of its areas is equal to the scale factor squared
Answer:
The average rate of change of the function over the interval is 5.
Step-by-step explanation:
Average rate of change of a function:
The average rate of change of a function f(x) over an interval [a,b] is given by:

Interval -3 less-than-or-equal-to x less-than-or-equal-to 3
This means that 

So


Average rate of change

The average rate of change of the function over the interval is 5.
Answer:
See Below.
Step-by-step explanation:
We want to show that the function:

Increases for all values of <em>x</em>.
A function is increasing whenever its derivative is positive.
So, find the derivative of our function:
![\displaystyle f'(x) = \frac{d}{dx}\left[e^x - e^{-x}\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5Be%5Ex%20-%20e%5E%7B-x%7D%5Cright%5D)
Differentiate:

Simplify:

Since eˣ is always greater than zero and e⁻ˣ is also always greater than zero, f'(x) is always positive. Hence, the original function increases for all values of <em>x.</em>