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pishuonlain [190]
3 years ago
6

How to know when the domain and range are all real numbers?

Mathematics
1 answer:
sdas [7]3 years ago
6 0
Linear functions such as  2x - 6 . 1/2 x    and  -6x  will have domain and range  = all real numbers.
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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
Select the two values of x that are roots of this equation.<br> 2x^2+ 1 = 5x
iren [92.7K]

Answer:

\frac{5+\sqrt{17}}{4},      \frac{5-\sqrt{17}}{4}

Step-by-step explanation:

One is asked to find the root of the following equation:

2x^2+1=5x

Manipulate the equation such that it conforms to the standard form of a quadratic equation. The standard quadratic equation in the general format is as follows:

ax^2+bx+c=0

Change the given equation using inverse operations,

2x^2+1=5x

2x^2-5x+1=0

The quadratic formula is a method that can be used to find the roots of a quadratic equation. Graphically speaking, the roots of a quadratic equation are where the graph of the quadratic equation intersects the x-axis. The quadratic formula uses the coefficients of the terms in the quadratic equation to find the values at which the graph of the equation intersects the x-axis. The quadratic formula, in the general format, is as follows:

\frac{-b(+-)\sqrt{b^2-4ac}}{2a}

Please note that the terms used in the general equation of the quadratic formula correspond to the coefficients of the terms in the general format of the quadratic equation. Substitute the coefficients of the terms in the given problem into the quadratic formula,

\frac{-b(+-)\sqrt{b^2-4ac}}{2a}

\frac{-(-5)(+-)\sqrt{(-5)^2-4(2)(1)}}{2(2)}

Simplify,

\frac{-(-5)(+-)\sqrt{(-5)^2-4(2)(1)}}{2(2)}

\frac{5(+-)\sqrt{25-8}}{4}

\frac{5(+-)\sqrt{17}}{4}

Rewrite,

\frac{5(+-)\sqrt{17}}{4}

\frac{5+\sqrt{17}}{4},      \frac{5-\sqrt{17}}{4}

8 0
3 years ago
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Ghella [55]

Answer:

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