The answers would be (0,10) , (20,0) , and (100,100)
Mid term :
Q1 = (88 + 85)/2 = 86.5
Q2 = (92 + 95)/2 = 93.5
Q3 = 100
IQR = Q3 - Q1 = 100 - 86.5 = 13.5
final exams :
Q1 = (65 + 78)/2 = 71.5
Q2 = (88 + 82)/2 = 85
Q3 = (95 + 93)/2 = 94
IQR = Q3 - Q1 = 94 - 71.5 = 22.5
so the final exams has the largest IQR
Answer:
See Below.
Step-by-step explanation:
We are given that ΔAPB and ΔAQC are equilateral triangles.
And we want to prove that PC = BQ.
Since ΔAPB and ΔAQC are equilateral triangles, this means that:
![PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ](https://tex.z-dn.net/?f=PA%5Ccong%20AB%5Ccong%20BP%5Ctext%7B%20and%20%7D%20QA%5Ccong%20AC%5Ccong%20CQ)
Likewise:
![\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ](https://tex.z-dn.net/?f=%5Cangle%20P%5Ccong%20%5Cangle%20PAB%5Ccong%20%5Cangle%20ABP%5Ccong%20Q%5Ccong%20%5Cangle%20QAC%5Ccong%5Cangle%20ACQ)
Since they all measure 60°.
Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:
![m\angle PAC=m\angle PAB+m\angle BAC](https://tex.z-dn.net/?f=m%5Cangle%20PAC%3Dm%5Cangle%20PAB%2Bm%5Cangle%20BAC)
Likewise:
![m\angle QAB=m\angle QAC+m\angle BAC](https://tex.z-dn.net/?f=m%5Cangle%20QAB%3Dm%5Cangle%20QAC%2Bm%5Cangle%20BAC)
Since ∠QAC ≅ ∠PAB:
![m\angle PAC=m\angle QAC+m\angle BAC](https://tex.z-dn.net/?f=m%5Cangle%20PAC%3Dm%5Cangle%20QAC%2Bm%5Cangle%20BAC)
And by substitution:
![m\angle PAC=m\angle QAB](https://tex.z-dn.net/?f=m%5Cangle%20PAC%3Dm%5Cangle%20QAB)
Thus:
![\angle PAC\cong \angle QAB](https://tex.z-dn.net/?f=%5Cangle%20PAC%5Ccong%20%5Cangle%20QAB)
Then by SAS Congruence:
![\Delta PAC\cong \Delta BAQ](https://tex.z-dn.net/?f=%5CDelta%20PAC%5Ccong%20%5CDelta%20BAQ)
And by CPCTC:
![PC\cong BQ](https://tex.z-dn.net/?f=PC%5Ccong%20BQ)
Answer:
C
Step-by-step explanation:
The range of a function is going to be the output of the table. 3, 4, 5, and 6 are all outputs in the table. C is the correct answer.
Hope this helps you.
Have a nice day.
Answer:
See explanation below
Step-by-step explanation:
x = 16sin49° = 12.1
y = 16cos49° = 10.5