If Melody used the table and still made a mistake, you can
imagine how badly we'll do if we can't even SEE the table.
Two polynomials are equal if the coefficients are equal.

Answer:
7/12 in exact form
Step-by-step explanation:
I think you just multiply the numerator by the resipical of the denominator.
Answer:
Ur mom
Step-by-step explanation:
Ur mom JK here's the answer 11:45 A.M. because 15 + 15 + 15 = 45 15 is a quarter so yeah Bois is 11:45 A.M Good luck
Answer:
∴Given Δ ABC is not a right-angle triangle
a= AB = √45 = 3√5
b = BC = 12
c = AC = √45 = 3√5
Step-by-step explanation:
Given vertices are A(3,3) and B(6,9)

AB = 
Given vertices are B(6,9) and C( 6,-3)
= 
BC = 12
Given vertices are A(3,3) and C( 6,-3)

AC² = AB²+BC²
45 = 45+144
45 ≠ 189
∴Given Δ ABC is not a right angle triangle