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viva [34]
2 years ago
5

40 POINTS !!! What is the extraneous solution (undefined) to the equation

Mathematics
2 answers:
aleksklad [387]2 years ago
5 0

Answer:

x = 4

x= -6

(I apologize if it is wrong.)

NikAS [45]2 years ago
4 0
X should equal -6 and 4 but 4 doesn’t work.
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The point (-7,4) is reflected over the line x=-3. Then the resulting point is reflected over the line y=x. Where is the point lo
Ne4ueva [31]

Answer:

  (4, 1)

Step-by-step explanation:

Since the first reflection is over the vertical line x=-3, the y-coordinate remains the same. The x-coordinate of A' will make the point (-3, 4) on the line of reflection be the midpoint between A and A':

  (-3, 4) = (A +A')/2

  2(-3, 4) -A = A' = (-6-(-7), 8 -4) = (1, 4)

The reflection over the line y=x simply interchanges the two coordinate values:

  A'' = (4, 1)

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Answer:

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Step-by-step explanation:

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2 years ago
Complete the given diagram by dragging expressions to each leg of the triangle. Then, correctly complete the equation to derive
Nimfa-mama [501]

The equation to derive the distance d is \sqrt{(x2-x1)^2+(y2-y1)^2}. The lengths of the other legs of the given triangle are (y2 - y1) and (x2 - x1).

<h3>What is the formula for calculating the distance between two points?</h3>

Consider the two points (x1, y1) and (x2, y2)

The formula used for calculating the distance between the two points is

distance = \sqrt{(x2-x1)^2+(y2-y1)^2}

<h3>Calculation:</h3>

Given that,

The triangle in the graph has vertices (x1, y1), (x2, y2), and (x2, y1)

Since this triangle makes 90°, it is a right-angled triangle.

Hypotenuse = (x1, y1) to (x2, y2), Adjacent = (x1, y1) to (x2,y1), and Opposite = (x2, y1) to (x2, y2).

Consider the length of the hypotenuse = d

So, using the distance formula, the length of the hypotenuse(d) is,

d = \sqrt{(x2-x1)^2+(y2-y1)^2}

And the lengths of the other two legs of the given triangle are,

Length of the adjacent side: (x1, y1) to (x2,y1)

= \sqrt{(x2-x1)^2+(y1-y1)^2}

= \sqrt{(x2-x1)^2+0}

= (x2-x1)

Length of the opposite side: (x2, y1) to (x2, y2)

= \sqrt{(x2-x2)^2+(y2-y1)^2}

= \sqrt{0+(y2-y1)^2}

= (y2-y1)

Therefore, the derived distances for the given triangle are:

d=\sqrt{(x2-x1)^2+(y2-y1)^2}, (x2 - x1), and (y2 - y1).

Learn more about the distance between two points here:

brainly.com/question/661229

#SPJ1

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1 year ago
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