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4 years ago

The power line can be described by the equation shown below:

Mathematics

1 answer:

scZoUnD [109]4 years ago

3 0

Answer:

$\begin{array}{cc}x&y\\ \\-2&-1\\0&-2\\2&-3\\6&-5\\9&-6.5\end{array}$

Step-by-step explanation:

Given:

$y=-\dfrac{1}{2}x-2$

First, find y for given x:

When x = 0, then

$y=-\dfrac{1}{2}\cdot 0-2\\ \\y=-2$

When x = 6, then

$y=-\dfrac{1}{2}\cdot 6-2\\ \\y=-3-2\\ \\y=-5$

When x = 9, then

$y=-\dfrac{1}{2}\cdot 9-2\\ \\y=-4.5-2\\ \\y=-6.5$

Now find x for given y:

When y = -1, then

$-1=-\dfrac{1}{2}x-2\\ \\-1+2=-\dfrac{1}{2}x\\ \\1=-\dfrac{1}{2}x\\ \\-2=x\\ \\x=-2$

When y = -3, then

$-3=-\dfrac{1}{2}x-2\\ \\-3+2=-\dfrac{1}{2}x\\ \\-1=-\dfrac{1}{2}x\\ \\2=x\\ \\x=2$

So, the table is

$\begin{array}{cc}x&y\\ \\-2&-1\\0&-2\\2&-3\\6&-5\\9&-6.5\end{array}$

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ludmilkaskok [199]

The answer is A. 25
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3 years ago

Given that the formula for the area of a circle is A=π r2.If the radius of a circular park is 15 feet,what is the area of the pa

Romashka-Z-Leto [24]

Answer:

706.5

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Step-by-step explanation:

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2 years ago

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3 years ago

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4 years ago

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

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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 <em>n </em>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

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

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3 years ago