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:
theta = 9/19 radians or approximately .4737 radians
Step-by-step explanation:
The formula for arc length is
S = r theta
where theta is in radians. Lets put in what we know. The radius would be 1.9 since the ropes would be the radius.
.9 = 1.9 theta
Divide by 1.9 on each side.
.9/1.9 = theta
theta = 9/19 radians
or in decimal form it is approximately .4737 radians
What is the "in" for in your equation? Repost.
Answer:
The outlier on the the scatter plot is point L(6,2).
Step-by-step explanation:
M(3,3)
P(5,5)
N(5,7)
L(6,2)
Other points are : (1,3), (2,3), (2,4), (3,4), (3,5), (4,5), (4,6), (5,6)
Now, To find the outliers on the scatter plot we plot all the given points on the graph.
The resultant graph is attached below :
An outlier is a value in a data set that is very different from the other values. That is, outliers are values which are usually far from the middle.
As we can see the graph the point L(6,2) is the most unusual or the farthest point on the scatter plot as compared with all the other points on the scatter plot.
Hence, The outlier on the the scatter plot is point L(6,2).
Dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd