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Olin [163]
4 years ago
13

Which of these functions below could be defined by a cubic equation?

Mathematics
2 answers:
Taya2010 [7]4 years ago
6 0
Hello,

Answer B

y=x(x-3)(x+3) we know 3 points of the graph

A: we need a third point
erik [133]4 years ago
4 0

Answer:

Graph 2 is the cubic equation.

Step-by-step explanation:

We have been given  the two graphs

So, we will conclude the function by end behavior

The cubic function enters from the left downwards and exists the graph from the right.

Cubic function will have 3 zeroes being of 3rd degree

And since the second function cuts the graph at three points

Hence, second graph is the cubic equation.

Zeroes are: -3 ,0 and 3.

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A spectroscope helps to identify elements by separating(WHAT)into lines of color.
jarptica [38.1K]
This is actually physics but the answer is:
1: light.

Hope this helps :)
8 0
4 years ago
Read 2 more answers
Simplify x^2+3x-4/x+4 Show work and explain why x doesn’t equal -4
marin [14]
Take the numerator of the fraction and factor it out.

x^{2} + 3x - 4 = (x+4)(x-1)

Your fraction should now look like this:

\frac{(x+4)(x-1)}{(x+4)}

Because (x+4) is present in both the numerator and denominator, you can cross it out. You should be left with only (x-1) in the numerator.

x cannot equal -4, as if it is inputted in the denominator, -4 + 4 would equal 0. Dividing by 0 is impossible.

5 0
3 years ago
What are the x-intercepts of the function f(x) = –2x2 – 3x + 20?
Bumek [7]
If you would like to find the x-intercepts of the function f(x) = - 2 * x^2 - 3 * x + 20, you can calculate this using the following steps:

f(x) = - <span>2 * x^2 - 3 * x + 20
</span>f(x) = - (2x - 5) * (x + 4)
1. x = - 4
2. x = 5/2

(x, y) = (-4, 0)

The correct result would be (-4, 0).
6 0
3 years ago
Read 2 more answers
What is the modulus of the complex number 5 + 2i ?
kicyunya [14]

Answer:

modulus of complex number is \sqrt{29}

Step-by-step explanation:

z=x+iy is a complex numbers

the given complex number is 5+2i

the value of x is 5  and y is 2

the modulus of complex number formula is

|z|=\sqrt{x^2+y^2} =\sqrt{5^2+2^2} =\sqrt{25+4} =\sqrt{29}

modulus of complex number is \sqrt{29}

6 0
3 years ago
Lim x-1 x³-2x²+3x-2/(2x^4-3x+1)
UkoKoshka [18]

Since the limit becomes the undetermined form

\displaystyle \lim_{x\to 1} \dfrac{x^3-2x^2+3x-2}{2x^4-3x+1} \to \dfrac{0}{0}

it means that both polynomials have a root at x=1. So, we can fact both numerator and denominator:

x^3-2x^2+3x-2 = (x-1)(x^2-x+2)

2x^4-3x+1 = (x-1)(2x^3+2x^2+2x-1)

So, the fraction becomes

\dfrac{(x-1)(x^2-x+2)}{(x-1)(2x^3+2x^2+2x-1)} = \dfrac{x^2-x+2}{2x^3+2x^2+2x-1}

Now, as x approaches 1, you have no problems anymore:

\displaystyle \lim_{x\to 1} \dfrac{x^2-x+2}{2x^3+2x^2+2x-1} \to \dfrac{2}{5}

4 0
3 years ago
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