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belka [17]
1 year ago
13

Triangle JKL is transformed to create triangle J'K'L'. The angles in both triangles are shown.

Mathematics
1 answer:
Viefleur [7K]1 year ago
5 0

The correct statement is AngleL = 25° AngleL' = 25°.

We have given that,

Triangle JKL is transformed to create triangle J'K'L'.

We have to choose the correct option.

<h3>What is the information about the transformed triangle?</h3>

The following information should be considered:

In a rigid transformation, the image & pre-image are congruent.

Reflection, translation, and rotation are rigid transformations.

In a non-rigid transformation, the image and pre-image are similar.

Dilation is a non rigid transformation.

In a rigid or nonrigid transformation, the corresponding angles are the same.

If the corresponding sides are the same, then it is a rigid transformation.

If the corresponding sides are proportional, then it is a nonrigid transformation.

It can be a rigid or a nonrigid transformation based on whether the corresponding side lengths have the same measures.

Therefore option 3 is correct.

To learn more transformation of the triangle visit:

brainly.com/question/17429689

#SPJ1

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To find the factor a a polynomial from its roots, we are going to seat each one of the roots equal tox, and then we are going to factor backwards.

We know for our problem that one of the roots of our polynomial is -3, so lets set -3 equal tox and factor backwards:
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(x+3) is a factor of our polynomial.

We also know that another root of our polynomial is 1+ \sqrt{2}, so lets set 1+ \sqrt{2} equal to x and factor backwards:
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We can conclude that there is no correct answer in your given choices.
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I need a answer to 77.5% of 13.270​
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A road crew took 3 days to pave a road. On the first day they paved 2/5 of the road, and on the second day they 1/3 of the road.
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3 0
3 years ago
Find sinϴ and cosϴ if tanϴ=1/4 and sinϴ&gt;0
eduard
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2787701

_______________


\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\&#10;\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\&#10;\mathsf{4\,sin\,\theta=cos\,\theta\qquad\quad(i)}


Square both sides:

\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\&#10;\mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\&#10;\mathsf{16\,sin^2\,\theta=cos^2\,\theta\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\&#10;\mathsf{16\,sin^2\,\theta=1-sin^2\,\theta}

\mathsf{16\,sin^2\,\theta+sin^2\,\theta=1}\\\\&#10;\mathsf{17\,sin^2\,\theta=1}\\\\&#10;\mathsf{sin^2\,\theta=\dfrac{1}{17}}\\\\\\&#10;\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{1}{17}}}\\\\\\&#10;\mathsf{sin\,\theta=\pm\,\dfrac{1}{\sqrt{17}}}


Since \mathsf{sin\,\theta} is positive, you can discard the negative sign. So,

\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}


Substitute this value back into \mathsf{(i)} to find \mathsf{cos\,\theta:}

\mathsf{4\cdot \dfrac{1}{\sqrt{17}}=cos\,\theta}\\\\\\&#10;\mathsf{cos\,\theta=\dfrac{4}{\sqrt{17}}\qquad\quad\checkmark}


I hope this helps. =)


Tags:   <em>trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus</em>

7 0
2 years ago
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