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yuradex [85]
2 years ago
13

What is the maximum number of terms a fourth-degree polynomial function in standard form can have?

Mathematics
1 answer:
horrorfan [7]2 years ago
3 0
A polynomial of four terms is sometimes called a quadrinomial, but there's really no need for such words.
Likewise, what is a 4th degree polynomial? Fourth degree polynomials are also known as quartic polynomials. Quartics have these characteristics: Zero to four roots. One, two or three extrema. Zero, one or two inflection points.
The degree of a polynomial is the highest of the degrees of the polynomial's monomials (individual terms) with non-zero coefficients. The degree of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer.
Zero Polynomial. The constant polynomial. whose coefficients are all equal to 0. The corresponding polynomial function is the constant function with value 0, also called the zero map. The zero polynomial is the additive identity of the additive group of polynomials.
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Consider the attached figure. All the triangles shown are similar, so

CB/CA = CD/CB
30/50 = projection/30
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projection = 18

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2 years ago
The center of a hyperbola is located at (0, 0). One focus is located at (0, 5) and its associated directrix is represented by th
Lady bird [3.3K]
For a hyperbola   \dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1
where   a^{2}+b^{2}=c^{2}
the directrix is the line   y=\dfrac{a^{2}}{c}
and the focus is at (0, c).

Here, we have c = 5, a² = 9, so b² = 5² - 9 = 16.
  a = √9 = 3
  b = √16 = 4

Your hyperbola's constants are ...
  a = 3
  b = 4


______
Please note that the equation of a hyperbola has a negative sign for one of the terms. The equation given in your problem statement is that of an ellipse.
8 0
2 years ago
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An image of a parabolic lens is projected onto a graph.The y-intercept of the graph is (0, 90), and the zeros are 5 and 9. Which
Gnesinka [82]

Step-by-step explanation:

If the zeros are 5 and 9, then the equation will have the form:

y = a (x–5) (x–9)

We know the point (0, 90) is on the curve, so we can use this to find the coefficient a:

90 = a (0–5) (0–9)

90 = 45a

a = 2

y = 2 (x – 5) (x – 9)

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