Answer: Seven hundred thirteen point forty-nine.
Step-by-step explanation: Hope this helps ;)
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:
Quadratic
Step-by-step explanation:
It looks like it’s forming a very wide arch so quadratic formulas are arches
Answer:
r = 2.75
Step-by-step explanation:
1.25 + r = 4
Subtract 1.25 on both sides.
1.25 + r - 1.25 = 4 - 1.25
r = 4 - 1.25
r = 2.75
The solution to the equation is r = 2.75.
Answer:
m=-16
Step-by-step explanation:
-5=m+11
m=-5-11
m=-16