why polar coordinates seem better suited for circular-type functions, while the rectangular coordinate system seems better suite
1 answer:
Because "circular-type" functions can be more easily described by the periodic functions 
and 
, which are deeply involved in the behavior of "circular-type" functions. Also, using a polar coordinate plane allows you to draw graphs of equations that typically aren't functions.
For example, the unit circle requires two equations to be graphed to fully represent the circle in the rectangular coordinate plane.
On the other hand, in polar coordinates, substituting
and 
reduces this to a constant function,
(since by definition, 
)
For example, the unit circle requires two equations to be graphed to fully represent the circle in the rectangular coordinate plane.
On the other hand, in polar coordinates, substituting
You might be interested in
Answer:
Null hypothesis:
Alternative hypothesis:
So the p value is a very low value and using any significance level for example always
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions NOT differs significantly.
Step-by-step explanation:
Data given and notation
represent the number of homeowners who would buy the security system
represent the number of renters who would buy the security system
sample 1
sample 2
represent the proportion of homeowners who would buy the security system
represent the proportion of renters who would buy the security system
z would represent the statistic (variable of interest)
represent the value for the test (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to check if the two proportions differs , the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
We need to apply a z test to compare proportions, and the statistic is given by:
(1)
Where
Calculate the statistic
Replacing in formula (1) the values obtained we got this:
Statistical decision
For this case we don't have a significance level provided , but we can calculate the p value for this test.
Since is a two sided test the p value would be:
So the p value is a very low value and using any significance level for example always
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions NOT differs significantly.
Options
A. UV = 14 ft and m∠TUV = 45°
B. TU = 26 ft
C. m∠STU = 37° and m∠VTU = 37°
D. ST = 20 ft, UV = 14 ft, and m∠UST = 98°
E. m∠UST = 98° and m ∠TUV = 45°
Answer:
A. UV = 14 ft and m∠TUV = 45°
D. ST = 20 ft, UV = 14 ft, and m∠UST = 98°
Step-by-step explanation:
Given
See attachment for triangle
Required
What proves that: ΔSTU ≅ ΔVTU using SAS
To prove their similarity, we must check the corresponding sides and angles of both triangles
First:
must equal
So:
Next:
UV must equal US.
So:
Also:
ST must equal VT
So:
Lastly
must equal
So:
Hence: Options A and D are correct
