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Tom [10]
3 years ago
7

What is a polynomial function in standard form with zeroes 0, 1, 4, and –1?

Mathematics
1 answer:
mr Goodwill [35]3 years ago
3 0
The polynomial for this case is given by:
 (x) * (x-1) * (x + 1) * (x-4) = 0
 Rescribing we have:
 (x) * (x ^ 2-1) * (x-4) = 0
 (x) * (x ^ 3 - 4x ^ 2 - x + 4) = 0
 (x ^ 4 - 4x ^ 3 - x ^ 2 + 4x) = 0
 Therefore, the polynomial in standard form is:
 f (x) = x ^ 4 - 4x ^ 3 - x ^ 2 + 4x
 Answer:
 
a polynomial function in standard form with zeroes 0, 1, 4, and -1 is:
 
f (x) = x ^ 4 - 4x ^ 3 - x ^ 2 + 4x
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\huge\color{green}\boxed{\colorbox{black}{Answer ☘}}

To find - sum of the given two terms.

(4 + 3x) + (5 - 4x + 2x {}^{2} ) \\ 4 + 3x + 5 - 4x + 2x {}^{2}  \\  =  > 2x {}^{2}  - x + 9

<h2>—————————————————</h2>

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<h3>Here ,</h3>

we basically added the like terms together..

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since there's none other term having "x²" as the coefficient.. so 2x² will be left as it is!

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a family pays 45 dollars each month for cable television.the cost increase 7 percent.how many dollars is the increase?
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8 0
3 years ago
There are 3 islands A,B,C. Island B is east of island A, 8 miles away. Island C is northeast of A, 5 miles away and northwest of
Nostrana [21]

Answer:

The bearing needed to navigate from island B to island C is approximately 38.213º.

Step-by-step explanation:

The geometrical diagram representing the statement is introduced below as attachment, and from Trigonometry we determine that bearing needed to navigate from island B to C by the Cosine Law:

AC^{2} = AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot \cos \theta (1)

Where:

AC - The distance from A to C, measured in miles.

AB - The distance from A to B, measured in miles.

BC - The distance from B to C, measured in miles.

\theta - Bearing from island B to island C, measured in sexagesimal degrees.

Then, we clear the bearing angle within the equation:

AC^{2}-AB^{2}-BC^{2}=-2\cdot AB\cdot BC\cdot \cos \theta

\cos \theta = \frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC}

\theta = \cos^{-1}\left(\frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC} \right) (2)

If we know that BC = 7\,mi, AB = 8\,mi, AC = 5\,mi, then the bearing from island B to island C:

\theta = \cos^{-1}\left[\frac{(7\mi)^{2}+(8\,mi)^{2}-(5\,mi)^{2}}{2\cdot (8\,mi)\cdot (7\,mi)} \right]

\theta \approx 38.213^{\circ}

The bearing needed to navigate from island B to island C is approximately 38.213º.

8 0
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