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3 years ago

Which image that would portray the rotation of the given preimage and the rotation arrow.

Mathematics

2 answers:

Karo-lina-s [1.5K]3 years ago

5 0

Answer:

First figure as shown below is the correct option.

Step-by-step explanation:

We are given that,

The pre-image is rotated in the direction of the arrow to form an image.

It is required to find the image formed.

Since, the arrow is rotating counter-clockwise, we get that,

The star in the pre-image will rotate upwards towards the left.

So, according to the options, we have,

<em>The first image where the star is at the highest height will be the image after rotating the pre-image.</em>

Hence, the first figure as shown below is the correct option.

Brut [27]3 years ago

4 0

Is the arrow the length of the rotation? 90 degrees? if so, the second picture is, with the star at the highest point of the other pictures.
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Ede4ka [16]

X^2 - 6x + 9 = -1 + 9. We need to solve for x.
First, add one on each side of the equation:
x^2 - 6x + 9 + 1 = -1 + 9 + 1
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Then subtract 9 from each side:
x^2 - 6x + 10 - 9 = 9-9
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Now we got an equation = 0. This an equation from the second degree, it represented by a parabola which turns up since a>0.
This equation is the developed form as ax^2 + bx + c with a=1; b = -6; and c=1.
Now, to find the zeroes of this equation, we need to find delta Δ.
Δ = b^2 - 4ac.
If Δ>0, the equation admits 2 zeroes: x=(-b-√Δ)/2a and x = (-b+√Δ)/2a.
If Δ<0, the equation doesn't admits any zero.
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Δ = (-6)^2 - 4(1)(1)
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Δ = 32
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So the zeroes of the equation are:
x = (-b-√Δ)/2a = (6-4√2)/2
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Hope this Helps! :)

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