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4 years ago

Domain and range of y=0.025x^2

1 answer:

6 0

Since it's not a domain restricted function anywhere, the domain of this function is (-∞, ∞). Because the x is squared, you will never get a negative answer, and therefore, the range is either (0, ∞) or {x: x > 0}. Hope this helps you out!

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Identify the slope of the given line. y=x-5

laiz [17]

Answer:

m is the slope and b is the y-intercept. Therefore, for the equation y=x+5 , the slope is 1 , and the y-intercept is 5

Step-by-step explanation:

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3 years ago

The scale factor of the volume of two cubes is 8 to 27. What is the scale factor of the sides of the cubes?

tekilochka [14]

The scale factor of the length of the sides of the cubes is 2:3

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4 years ago

The base of an open rectangular box is of length(2x+5)cm and width xcm.

Tju [1.3M]

Step-by-step explanation:

For solving, use quadratic formula

8 0

3 years ago

Plz help me and answer right

azamat

I think it might be 30 but I don’t know

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3 years ago

Read 2 more answers

A student claims that 2i is the only imaginary root of a polynomial equation that has real coefficients. Explain the student's m

____ [38]

Answer:

The Fundamental Theorem of Algebra assures that any polynomial f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i

Step-by-step explanation:

1) This claim is mistaken.

2) The Fundamental Theorem of Algebra assures that any polynomial f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i with real coefficients.

$a_{0}x^{n}+a_{1}x^{2}+....a_{1}x+a_{0}$

For example:

3) Every time a polynomial equation, like a quadratic equation which is an univariate polynomial one, has its discriminant following this rule:

$\Delta < 0\\b^{2}-4*a*c$

We'll have n different complex roots, not necessarily 2i.

For example:

Taking 3 polynomial equations with real coefficients, with

$\Delta < 0$

$-4x^2-x-2=0 \Rightarrow S=\left \{ x'=-\frac{1}{8}-i\frac{\sqrt{31}}{8},\:x''=-\frac{1}{8}+i\frac{\sqrt{31}}{8} \right \}\\-x^2-x-8=0 \Rightarrow S=\left\{\quad x'=-\frac{1}{2}-i\frac{\sqrt{31}}{2},\:x''=-\frac{1}{2}+i\frac{\sqrt{31}}{2} \right \}\\x^2-x+30=0\Rightarrow S=\left \{ x'=\frac{1}{2}+i\frac{\sqrt{119}}{2},\:x''=\frac{1}{2}-i\frac{\sqrt{119}}{2} \right \}\\(...)$

2.2) For other Polynomial equations with real coefficients we can see other complex roots ≠ 2i. In this one we have also -2i

$x^5\:-\:x^4\:+\:x^3\:-\:x^2\:-\:12x\:+\:12=0 \Rightarrow S=\left \{ x_{1}=1,\:x_{2}=-\sqrt{3},\:x_{3}=\sqrt{3},\:x_{4}=2i,\:x_{5}=-2i \right \}\\$

4 0

3 years ago