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
Answer:
C is the anser on edege
Step-by-step explanation:
Answer:
<em>1) y = (x-3)²-12</em>
<em>2) y = 5(x-2)²-9 </em>
Step-by-step explanation:
Given the quadratic expressions
y=x²-6x-3
We are to transform it in the form y=ax²+bx+c
y=(x²-6x)-3
Complete the square of the expression in parenthesis
y = (x²-6x + (6/2)² - (6/2)²)-3
y = (x²-6x+3²-3²)-3
y = (x²-6x+9-9)-3
y = (x²-6x+9)-12
y = (x²-3x-3x+9)-12
y = x(x-3)-3(x-3)-12
y = (x-3)(x-3)-12
y = (x-3)²-12
Hence the transformation is y = (x-3)²-12 where a = 1 and k = -12
For the quadratic equation
y=(5x²-20x)-5
y = 5(x²-4x)- 5
Complete the square of the expression in parenthesis
y = 5(x²-4x+4-4)- 5
y = 5(x²-4x+4)-9
y = 5(x-2)²-9
<em>Hence the transformation is y = 5(x-2)²-9 where a = 5 and k = -9</em>
4/12 equals 2/6 which is smaller than 4/6 so the correct answer is 4/12 <4/6
10 dollars, since you are most likely to get a 15 dollar bill but pay 5 dollars to draw.
