Find the difference 7/x^2-3x - 1/x^2-9

1 answer:

6 0

To solve this problem you must apply the proccedure shown below:

1. You have the following expression given in the problem above:

7/x^2-3x - 1/x^2-9

2. When you factor the denominator, you obtain:

7/x(x-3) - 1/(x+3)(x-3)

3. By simplifying the expression, you obtain:

3(2x+7)/x(x+3)(x-3)
 3(2x+7)/x(x^2-9)

The answer is: 3(2x+7)/x(x^2-9)

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HELP!!!! I think its C but I'm not sure!

MA_775_DIABLO [31]

Answer:

The fundamental theorem of algebra tells you that the equation will have two complex roots since the degree of the polynomial is 2. The roots are $x=1\pm i\sqrt{7}$ .

Step-by-step explanation:

Consider the provided information.

Algebra's fundamental theorem states that: Every polynomial equation of degree n with complex coefficients has n roots in the complex numbers.

Now consider the provided equation.

$2x^2-4x+16=0$

The degree of the polynomial equation is 2, therefore according to Algebra's fundamental theorem the equation have two complex roots.

Now find the root of the equation.

For the quadratic equation of the form $ax^2+bx+c=0$ the solutions are: $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $a=2,\:b=-4,\:\ and \ c=16$ in above formula.

$x_{1,\:2}=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:2\cdot \:16}}{2\cdot \:2}$

$x_{1,\:2}=\frac{4\pm \sqrt{16-128}}{4}$

$x_{1,\:2}=\frac{4\pm \sqrt{-112}}{4}$

$x_{1,\:2}=\frac{4\pm 4i\sqrt{7}}{4}$

$x_{1,\:2}=1\pm i\sqrt{7}$

Hence, the fundamental theorem of algebra tells you that the equation will have two complex roots since the degree of the polynomial is 2. The roots are $x=1\pm i\sqrt{7}$ .

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{ .. , -3 , -2 , -1 , 0 , 1 , 2 } finite or infinite

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The left side should be infinite, based on the ..., which represents that the list goes on. The right side, however, should be finite, because that’s where the pattern stops.

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Simplifie 8m2+3m+m3+4

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Answer:

$15+m^{6}$