The polynomial functions in their expanded form is given as follows. It is right to state that there are no breaks in the domain of h(x).
<h3>
What is a polynomial function?</h3>
In an equation such as the quadratic equation, cubic equation, etc., a polynomial function is a function that only uses non-negative integer powers or only positive integer exponents of a variable.
For instance, the polynomial 3x+4 has an exponent of 1.
Part A: F(x) has zero at 2 and multiplicity of 1; and
1 at the multiplicity of 2
f(x) = x-2) (x-1)²
= (x-2) (x² - 2x + 1)
= x³ - 4x² + 5x -2
Part B: h (x) = ![\left \{ {{x^3 -4x^2 + 5x -2; X < 0} \atop {\sqrt[3]{x-2} ; X\geq 0 }} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%5E3%20-4x%5E2%20%2B%205x%20-2%3B%20X%20%3C%200%7D%20%5Catop%20%7B%5Csqrt%5B3%5D%7Bx-2%7D%20%3B%20X%5Cgeq%200%20%7D%7D%20%5Cright.)
The domain of X is X ∈ R
Hence it is correct to state that there are no breaks in the domain of h(x).
Learn more about polynomial functions:
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Answer: -44.4°F
Step-by-step explanation:
17,750-8500=9,250
9250/1000=9.25
9.25x-4.8=-44.4
Check the picture below.
if we know the endpoints of the circle, then its center must be the midpoint of those two endpoints.
This equation has some nested grouping symbols on the left-hand side. As usual, I'll simplify from the inside out. I'll start by inserting the "understood" 1 in front of that innermost set of parentheses:
3 + 2[4x – (4 + 3x)] = –1
3 + 2[4x – 1(4 + 3x)] = –1
3 + 2[4x – 1(4) – 1(3x)] = –1
3 + 2[4x – 4 – 3x] = –1
3 + 2[1x – 4] = –1
3 + 2[1x] + 2[–4] = –1
3 + 2x – 8 = –1
2x + 3 – 8 = –1
2x – 5 = –1
2x – 5 + 5 = –1 + 5
2x = 4
x = 2
It is not required that you write out this many steps; once you get comfortable with the process, you'll probably do a lot of this in your head. But until you reach that comfort zone, you should write things out at least this clearly and completely.
Always remember, by the way, that you can check your answers in "solving" problems by plugging the numerical answer back in to the original equation. In this case, the variable is only in terms on the left-hand side (LHS) of the equation; my "check" (that is, my evaluation at the solution value) looks like this:
LHS: 3 + 2[4x – (4 + 3x)]:
3 + 2[4(2) – (4 + 3(2))]
3 + 2[8 – (4 + 6)]
3 + 2[8 – (10)]
3 + 2[–2]
3 – 4
–1
Since this is what I was supposed to get for the right-hand side (that is, I've shown that the LHS is equal to the RHS), my solution value was correct.
