Answer:
Transitive property
Step-by-step explanation:
Given
See attachment
Required
The missing property?
In step 1, we have:
![\angle A \cong \angle B,\ \angle C \cong \angle B](https://tex.z-dn.net/?f=%5Cangle%20A%20%5Ccong%20%5Cangle%20B%2C%5C%20%5Cangle%20C%20%5Ccong%20%5Cangle%20B)
Transitive property states that:
If
and ![b = c](https://tex.z-dn.net/?f=b%20%3D%20c)
Then ![a = c](https://tex.z-dn.net/?f=a%20%3D%20c)
Similarly:
If
and ![\angle C \cong \angle B](https://tex.z-dn.net/?f=%5Cangle%20C%20%5Ccong%20%5Cangle%20B)
Then
--- step 3
<em>Hence, we can conclude that the transitive property is the missing justification</em>
Answer:
(4, 7/2)
Step-by-step explanation:
midpoint = x values/2, y values/2
Answer:
I think it's would be D
Step-by-step explanation:
1/2 is 50% So you add 19 and 50 to get 69% or 69/100
Answer:
![\bigtriangleup= 19.84\ in^2](https://tex.z-dn.net/?f=%5Cbigtriangleup%3D%2019.84%5C%20in%5E2)
Step-by-step explanation:
We calculate the initial area before the dimension modifications.
-The surface area(given the slant height) is calculated as:
![A=Base \ Area +Triangle:s \ Area\\\\=s^2+4(0.5sh)\\\\s=base \ side, h=\ slant \ height\\\\\therefore A=4^2+4(0.5\times 8\times 4)\\\\=80\ cm^2](https://tex.z-dn.net/?f=A%3DBase%20%5C%20Area%20%2BTriangle%3As%20%5C%20Area%5C%5C%5C%5C%3Ds%5E2%2B4%280.5sh%29%5C%5C%5C%5Cs%3Dbase%20%5C%20side%2C%20h%3D%5C%20slant%20%5C%20height%5C%5C%5C%5C%5Ctherefore%20A%3D4%5E2%2B4%280.5%5Ctimes%208%5Ctimes%204%29%5C%5C%5C%5C%3D80%5C%20cm%5E2)
-If the slant height is tripled, the new height will be 3*8=24, and the base lengths remain unchanged:
![A_n=Base \ Area +Triangle:s \ Area\\\\=s^2+4(0.5sh)\\\\s=base \ side, h=\ slant \ height\\\\\therefore A_n=4^2+4(0.5\times 24\times 4)\\\\=208\ cm^2](https://tex.z-dn.net/?f=A_n%3DBase%20%5C%20Area%20%2BTriangle%3As%20%5C%20Area%5C%5C%5C%5C%3Ds%5E2%2B4%280.5sh%29%5C%5C%5C%5Cs%3Dbase%20%5C%20side%2C%20h%3D%5C%20slant%20%5C%20height%5C%5C%5C%5C%5Ctherefore%20A_n%3D4%5E2%2B4%280.5%5Ctimes%2024%5Ctimes%204%29%5C%5C%5C%5C%3D208%5C%20cm%5E2)
-The change in area is calculated as (1 sq cm=0.155 sq in):
![\bigtriangleup=A_n-A\\\\=128\ cm^2\\\\=>1 \ cm^2=0.155\ in^2\\\\\therefore 128\ cm^2=0.155\ in^2\times 128\\\\=19.84\ in^2](https://tex.z-dn.net/?f=%5Cbigtriangleup%3DA_n-A%5C%5C%5C%5C%3D128%5C%20cm%5E2%5C%5C%5C%5C%3D%3E1%20%5C%20cm%5E2%3D0.155%5C%20in%5E2%5C%5C%5C%5C%5Ctherefore%20128%5C%20cm%5E2%3D0.155%5C%20in%5E2%5Ctimes%20128%5C%5C%5C%5C%3D19.84%5C%20in%5E2)