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3 years ago

Help plzzz....………………….…....................................

Mathematics

1 answer:

Nookie1986 [14]3 years ago

5 0

Answer:

Step-by-step explanation:

You automatically know that it is not A. The dot is filled in on the number line, so it must have a line under the inequality mark.

The number line says that x is greater than or equal to 2.5

Key:

gtoet means greater than or equal to

ltoet means less than or equal to.

3x + 2 gtoet 17

3x gtoet 15

x gtoet 5

This is incorrect. X is not gtoet 5.

12 + 4x gtoet 22

4x gtoet 10

x gtoet 2.5

This is correct. X is gtoet 2.5

-6x - 15 gtoet 0

-6x gtoet 15

(The sign switches in this one because you are dividing by a negative)

x ltoet 2.5

Incorrect. X is not ltoet 2.5

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Find the solutions of the quadratic equation 3x^2-5x+1=0.

Black_prince [1.1K]

Answer:

The solutions of the quadratic equation are $x_{1} = \frac{5 + \sqrt{13}}{6}, x_{2} = \frac{5 - \sqrt{13}}{6}$

Step-by-step explanation:

This is a second order polynomial, and we can find it's roots by the Bhaskara formula.

Explanation of the bhaskara formula:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

For this problem, we have to find $x_{1} \text{and} x_{2}$ .

The polynomial is $3x^{2} - 5x +1$ , so a = 3, b = -5, c = 1.

Solution

$\bigtriangleup = b^{2} - 4ac = (-5)^{2} - 4*3*1 = 25 - 12 = 13$

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a} = \frac{-(-5) + \sqrt{13}}{2*3} = \frac{5 + \sqrt{13}}{6}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a} = \frac{-(-5) - \sqrt{13}}{2*3} = \frac{5 - \sqrt{13}}{6}$

The solutions of the quadratic equation are $x_{1} = \frac{5 + \sqrt{13}}{6}, x_{2} = \frac{5 - \sqrt{13}}{6}$

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------ = ------
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Flura [38]

Answer:

Step-by-step explanation:

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