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Butoxors [25]
3 years ago
13

Select the correct answer. Which equation could be solved using this application of the quadratic formula?

Mathematics
2 answers:
iVinArrow [24]3 years ago
7 0

Answer:

x² + 2x - 4

Step-by-step explanation:

The quadratic equation is ax² + bx + c =0

To solve this, you plug it into the quadratic formula (-b±√(b²-4ac))/(2a).

As it shows in the picture, -B = 2. There is no negative on the two in the equation because the quadratic formula makes any positive number negative in the first part.

The values of A and C are give in the parenthesis because they are being multiplied by -4.

The value for A = 1 and we can leave that as x² because 1 and x have the same value.

The value for C = -4 and that goes at the back of the equation, turning the plus sign after 2x negative.

When you put the equation together, you get x² + 2x - 4

Elena L [17]3 years ago
4 0

Answer:

p=x8*2

Step-by-step explanation:

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Graph quadrilateral abcd whose vertex matrix is shown below. Then graph the dilation of quadrilateral abcd with a scale factor o
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Answer:

The answer is below

Step-by-step explanation:

Transformation is the movement of a point from its initial location to a new location. Types of transformation is reflection, translation, rotation and dilation.

Dilation is the increase or decrease in size of an object.

Given the vertex matrix of quadrilateral ABCD as:

\left[\begin{array}{cccc}-1&5&5&-4\\2&5&3&-1\end{array}\right]

Therefore the vertex of quadrilateral ABCD is at A(-1, 2), B(5, 5), C(5, 3) and D(-4, -1).

Quadrilateral ABCD is dilated by a scale factor of 3 to produce quadrilateral A'B'C'D'.

Hence, the vertex matrix of quadrilateral A'B'C'D' is:

3\left[\begin{array}{cccc}-1&5&5&-4\\2&5&3&-1\end{array}\right]=\left[\begin{array}{cccc}-3&15&15&-20\\6&15&9&-3\end{array}\right]

Therefore the vertex of quadrilateral A'B'C'D' is at A'(-3, 6), B'(15, 15), C'(15, 9) and D'(-20, -3).

The correct option is figure A

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3 years ago
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Answer:

Step-by-step explanation:

4x = 38

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