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jolli1 [7]
4 years ago
14

Graph f(x) = \xi. Click on the graph until the graph of f(x) = \xi appears.

Mathematics
1 answer:
marusya05 [52]4 years ago
5 0

Answer:

The graph of IxI is:

y = x for values of x ≥ 0

y = -x for values of x ≤ 0

Then you will see a "V", with the arms pointing up and the vertex in the point (0, 0)

(Something like in the image, but with the arms pointing upside instead of downside)

The actual graph is:

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c square


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The number of cars sold at a dealership over several weeks is given below.
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Answer:

A. 6.9

Step-by-step explanation:

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4 years ago
Slove system of equation.<br><br> 2x - 6y = 8<br> -3x + 9y = 12
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7 0
2 years ago
Given the equation 4x2 − 8x 20 = 0, what are the values of h and k when the equation is written in vertex form a(x − h)2 k = 0?
Westkost [7]

The vertex form of the given quadratic equation is

4(x-1)^{2} + 16 =0.

According to the given question.

We have a quadratic equation

4x^{2} -8x + 20 = 0

Since, for the standard quadratic form is ax^{2} +bx + c = y, the vertex form of a quadratic equation is y = a(x -h)^{2} + k where (h, k) is the vertex.    

And h and k can be calculated as

h = -b/2a and y = k

So, for the given equation 4x^{2} -8x + 20 =0 the vertex (h, k) is given by

h = -(-8)/2(4) = 8/8 = 1   (X coordinate of vertex)

and,

y =4x^{2} -8x+ 20

substitute x = 1 in the above equation for the value of k

Y = 4(1)(1) - 8(1) + 20

⇒ Y = 4 - 8 + 20

⇒ Y = -4 + 20

⇒ Y = 16

so, k = 16          (Y coordinate of vertex)

Now, substitute the value of a, k and h in y = a(x-h)^{2} + k.

⇒  0 =4(x-1)^{2} + 16

Therefore, the vertex form of the given quadratic equation is

4(x-1)^{2} + 16 =0.

Find out more information about vertex form of a quadratic equation here:

brainly.com/question/12223454

#SPJ4

6 0
2 years ago
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