Function is p(x)=(x-4)^5(x^2-16)(x^2-5x+4)(x^3-64)
first factor into (x-r1)(x-r2)... form
p(x)=(x-4)^5(x-4)(x+4)(x-4)(x-1)(x-4)(x^2+4x+16)
group the like ones
p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)
multiplicity is how many times the root repeats in the function
for a root r₁, the root r₁ multiplicity 1 would be (x-r₁)^1, multility 2 would be (x-r₁)^2
so
p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)
(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the root 1 and has multiplity 1
(x^2+4x+16) is not on the real plane, but the roots are -2+2i√3 and -2-2i√3, each multiplicity 1 (but don't count them because they aren't real
baseically
(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the root 1 and has multiplity 1
Semi --> twice
semimonthly --> occuring twice a month
Thus:
You are paid twice a month
12 months in a year
12 x 2 = 24 paychecks an year
Answer: -14
Step-by-step explanation:
You should get the right answer by following PEMDAS
We can use the SSS congruence theorem to prove that the two triangles in the attached figure are congruent. The SSS or side-side-side theorem states that each side in the first triangle must have the same measurement or must be congruent on each of the opposite side of another triangle. In this problem, for the first triangle, we have sides AC, CM, AM while in the second triangle we have sides BC, CM, and BM. By SSS congruent theorem, we have the congruent side as below:
AC = BC
CM = CM
AM = BM
The answer is SSS theorem.
