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RideAnS [48]
4 years ago
8

Can you please do the mean median mode and range for this chart please.

Mathematics
1 answer:
AlexFokin [52]4 years ago
5 0

Median: 9

Mode: 9&11

Range: 11

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Evaluate the expression below for x = 1, y = –2, and z = 3.
galben [10]
(3x - 2y) / (z + 4)........x = 1, y = -2, z = 3
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3 0
3 years ago
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Which equations are true for x = –2 and x = 2? Select two options x2 – 4 = 0 x2 = –4 3x2 + 12 = 0 4x2 = 16 2(x – 2)2 = 0
Svetllana [295]

The equation x^{2} -4=0~~and~~4x^2=16 are true for  x = -2 and x = 2.

The given equations are given as:

x^2-4=0\\\\x^2=-4\\\\3x^2+12=0\\\\4x^2=16\\\\2(x-2)^2=0

We need to select two equations that are true for x = -2 and x = 2.

<h3>What are the solutions to an equation ?</h3>

The solutions of an equation are the values that satisfy the given equation or make the equation true when substituted for unknowns in the equations.

Example:

x - 2 = 2

Here the solution for x - 2 = 2 is 4 because x = 4 will make the equation true.

4 - 2 = 2

2 = 2

Let's find the solutions for each equation.

1.

x^2 - 4=0

x^{2} = 4

x = \sqrt{4} = ± 2

x = = 2 and x = -2

2.

x^2 =-4

x = \sqrt{-4}

x = \sqrt{-1 \times 4} = \sqrt{-1}\times\sqrt{4}

x = ± 2 \sqrt{-1}

x = 2i and x = -2i         where i = \sqrt{-1}

3.

3x^{2} + 12  = 0

x^{2} = -12 / 3 = -4

x = \sqrt{-1} \sqrt{4}

x = ± 2\sqrt{-1}

x = 2i and x = -2i

4.

4x^{2} = 16

x^{2} = 16 / 4

x^{2} = 4

x = \sqrt{4}

x = ±2

x = 2 and x = -2

5.

2(x-2)^2 = 0

(x-2)^2 = 0

x - 2 = 0

x = 2.

Thus the equation x^{2} -4=0~~and~~4x^2=16 have solutions x = -2 and x = 2.

Learn more about solutions of equations here:

brainly.com/question/14506845

#SPJ1

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