Answer:
1.
a = 112
b = 68
c = 68
2.
a = 127
3.
a=35
b=40
c=35
d=70
4.
a= 30
b=70
c = 30
d=70
e = 130
I'll help you with the rest later
Step-by-step explanation:
a = 112 because of allied angles rule
b and c = 68 because of angles at a point
360-112-112 ÷2
2. a = 127 because of angles on a straight line rule.
180-38-15
3. d= 70, vertically opposite angle
using angles on a straight line, 180 - 70 - 40 ÷ 2
we now have the two angles and because they are vertically opposite a and c = 35
b = 40 because of vertically opposite angles
4. a=30 because 90-70
since a=30, take 90 - 30 to get b, 70
d= 70, vertically opposite angles
e = 130 because a+b+c, vertically opposite angles
They all have a Functional.
Answer:
![\large\boxed{_6P_2=30}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B_6P_2%3D30%7D)
Step-by-step explanation:
![_nP_k=\dfrac{n!}{(n-k)!}\\\\n!=1\cdot2\cdot3\cdot...\cdot n\\======================\\\\_6P_2=\dfrac{6!}{(6-2)!}=\dfrac{6!}{4!}=\dfrac{4!\cdot5\cdot6}{4!}\\\\\text{cancel}\ 4!\\\\=5\cdot6=30](https://tex.z-dn.net/?f=_nP_k%3D%5Cdfrac%7Bn%21%7D%7B%28n-k%29%21%7D%5C%5C%5C%5Cn%21%3D1%5Ccdot2%5Ccdot3%5Ccdot...%5Ccdot%20n%5C%5C%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%3D%5C%5C%5C%5C_6P_2%3D%5Cdfrac%7B6%21%7D%7B%286-2%29%21%7D%3D%5Cdfrac%7B6%21%7D%7B4%21%7D%3D%5Cdfrac%7B4%21%5Ccdot5%5Ccdot6%7D%7B4%21%7D%5C%5C%5C%5C%5Ctext%7Bcancel%7D%5C%204%21%5C%5C%5C%5C%3D5%5Ccdot6%3D30)
Step-by-step explanation:
Given: X is midpoint of UV and Y is midpoint of VW.
(By mid-point theorem)
![\therefore \angle XYV \cong \angle UWY\\.. (corresponding\: \angle 's) \\\therefore m\angle XYV = m\angle UWY\\\therefore 98 - 6x = 63- x\\\therefore 98 - 63= 6x- x\\\therefore 35= 5x\\\\\therefore x = \frac{35}{5}\\\\\huge \orange {\boxed {\therefore x = 7}} \\\\m\angle UWY = (63 - x) °\\\\\therefore m\angle UWY = (63 - 7) °\\\\\huge \purple {\boxed {\therefore m\angle UWY = 56 °}}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Cangle%20XYV%20%5Ccong%20%5Cangle%20UWY%5C%5C..%20%28corresponding%5C%3A%20%5Cangle%20%27s%29%20%5C%5C%5Ctherefore%20m%5Cangle%20XYV%20%3D%20m%5Cangle%20UWY%5C%5C%3C%2Fp%3E%3Cp%3E%5Ctherefore%2098%20-%206x%20%3D%2063-%20x%5C%5C%3C%2Fp%3E%3Cp%3E%5Ctherefore%2098%20-%2063%3D%206x-%20x%5C%5C%3C%2Fp%3E%3Cp%3E%5Ctherefore%2035%3D%205x%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3E%5Ctherefore%20x%20%3D%20%5Cfrac%7B35%7D%7B5%7D%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3E%5Chuge%20%5Corange%20%7B%5Cboxed%20%7B%5Ctherefore%20x%20%3D%207%7D%7D%20%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3Em%5Cangle%20UWY%20%3D%20%2863%20-%20x%29%20%C2%B0%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3E%5Ctherefore%20%20m%5Cangle%20UWY%20%3D%20%2863%20-%207%29%20%C2%B0%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3E%5Chuge%20%5Cpurple%20%7B%5Cboxed%20%7B%5Ctherefore%20%20m%5Cangle%20UWY%20%3D%2056%20%C2%B0%7D%7D%20%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E)
A
No you cannot use HL to prove the triangles are congruent because there is no way to know if the hypotenuses of BE and EC are congruent