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Stels [109]
2 years ago
7

Please help!!!!!!!!! in the similarity transformation of abc to def, abc was dilated by a scale factor of ?, reflected across th

e , and moved through the translation. i need help with the last portion of the question.
Mathematics
1 answer:
shusha [124]2 years ago
5 0

The complete similarity transformation is: triangle abc was dilated by a scale factor of [2], reflected across the [x-axis], and moved through the translation [0].​

<h3><u>What is Similarity?</u></h3>
  • When it comes to Euclidean geometry, two things are said to be comparable if they have the same shape or the same shape as each other's mirror image.
  • More specifically, by uniformly scaling (enlarging or decreasing), maybe with additional translation, rotation, and reflection, one can be created from the other.
  • This indicates that one object may be properly aligned with the other object by rescaling, moving, and reflecting it. When two things are comparable to one another, one is consistent with the outcome of a specific uniform scaling of the other.

From the figure, we have:

AB = 2 units

DE = 4 units

So, the scale factor (k) is:

k = DE/AB

k = 4/2

k = 2

Also, both shapes are on either sides of the x-axis, and they are equidistant from the x-axis.

This means that the triangle is reflected across the x-axis with no translation.

Hence, the complete similarity transformation is: abc was dilated by a scale factor of [2], reflected across the [x-axis], and moved through the translation [0].​

So, the scale factor is 2

Know more about Similarity with the help of the given link:
brainly.com/question/12670209

#SPJ4

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A number subtracted from 90 is three times that number
sergeinik [125]
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erik [133]
For the answer to the question above, 
1 + nx + [n(n-1)/(2-factorial)](x)^2 + [n(n-1)(n-2)/3-factorial] (x)^3 

<span>1 + nx + [n(n-1)/(2 x 1)](x)^2 + [n(n-1)(n-2)/3 x 2 x 1] (x)^3 </span>

<span>1 + nx + [n(n-1)/2](x)^2 + [n(n-1)(n-2)/6] (x)^3 </span>

<span>1 + 9x + 36x^2 + 84x^3 </span>

<span>In my experience, up to the x^3 is often adequate to approximate a route. </span>

<span>(1+x) = 0.98 </span>

<span>x = 0.98 - 1 = -0.02 </span>

<span>Substituting: </span>

<span>1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 </span>

<span>approximation = 0.834 </span>

<span>Checking the real value in your calculator: </span>

<span>(0.98)^9 = 0.834 </span>

<span>So you have approximated correctly. </span>

<span>If you want to know how accurate your approximation is, write out the result of each in full: </span>

<span>1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 = 0.833728 </span>

<span> (0.98)^9 = 0.8337477621 </span>

<span>So it is correct to 4</span>
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g100num [7]

Answer:

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In truth, when given the graph you don't really need to use the points, but if needed it you would use the formula:

y-y₁=m(x-x₁)

Plug in your information and solve.

-1 - 5 = m (2 - (-1))

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m = -6 / 3

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up from left to right: it's positive

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Marianna [84]

Answer:

your welcome your answer gonna be the letter C

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