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pshichka [43]
3 years ago
10

What describes the end behavior of the polynomial function?

Mathematics
1 answer:
zimovet [89]3 years ago
3 0

Answer:

x \to +\infty, f(x) \to +\infty and x \to -\infty, f(x) \to -\infty.

Step-by-step explanation:

A polynomial is an algebraic function of the form:

y = \Sigma_{i=0}^{n}c_{i}\cdot x^{i} (1)

Where:

c_{i} - i-th Coefficient.

x^{i} - i-th Power.

n - Grade of the polynomial.

y - Dependent variable.

Mathematically speaking, polynomials are unbounded functions, and from graphic we notice that polynomial is of order 3 due to the fact that function pass through the x axis three times, where each point is a root of the polynomial.

Then, we may conclude that:  

(i) \lim_{n \to +\infty} f(x) = N.E.

(ii) \lim_{n \to - \infty} f(x) = N.E.

Then, the right answer is: x \to +\infty, f(x) \to +\infty and x \to -\infty, f(x) \to -\infty.

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Find the volume. Answer without units. *<br> 2<br> 4- in<br> 5<br> 3 in<br> 3<br> 16 - in<br> 4
Harman [31]

Answer:

the answer is 221.1

Step-by-step explanation:

8 0
3 years ago
Answer the question in the picture
nekit [7.7K]

Recall the angle sum identities:

\sin(x+y)=\sin x\cos y+\cos x\sin y

\cos(x+y)=\cos x\cos y-\sin x\sin y

Now,

\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}

Divide through numerator and denominator by \cos x\cos y to get

\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}

Next, we use the fact that x,y lie in the first quadrant to determine that

\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2

\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}

So we then have

\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}

\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1

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4 0
3 years ago
Which sets of points represent one-to-one functions?
madreJ [45]
<h3>Two Answers: </h3><h3>Function h (third choice)</h3><h3>Function k (sixth choice, or last choice)</h3>

==========================================================

Explanation:

A function is only possible if the x values do not repeat. A relation is considered one-to-one only if the y values do not repeat. We combine the two ideas. If we want a one-to-one function, then neither x nor y can repeat.

  • Function f has y = 8 repeating, so it is not one-to-one
  • Function g is a similar story but y = 4 repeats.
  • Function h has all unique x values, and all unique y values. This function is one-to-one.
  • Function i has y = 3 repeating, so it is not one-to-one.
  • Function j has y = 16 show up more than once, so it is not one-to-one.
  • Function k has all unique x values, and all unique y values. This function is one-to-one.

Note: if we had two points like (1,2) and (3,1), then it is still one-to-one (even though the digit 1 shows up twice). This is because the set of x values {1,3} is unique, and so is the set of y values {2,1}. We look at each set separately and not combine the two sets.

4 0
3 years ago
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