Answer:
The looks of the Triangles are different.
Step-by-step explanation:
Answer: see proof below
<u>Step-by-step explanation:</u>

Use the following Identities:
sec Ф = 1/cos Ф
cos² Ф + sin² Ф = 1
<u>Proof LHS → RHS</u>






Answer:
C
Step-by-step explanation:
10×-28=-280
35-8=27
35×(-8)=-280
10x²+27x-28
=10x²+(35-8) x-28
=10x²+35x-8x-28
=5x(2x+7)-4(2x+7)
=(2x+7)(5x-4)
D
the answer is D because it is equilateral triangle
What is the theoretical probability of 5 students from your school being selected as contestants out of 10 possible contestant spots?
17.9%