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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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Step-by-step explanation:

We are given the following information in the question:

The radius of a cone is decreasing at a constant rate.

\displaystyle\frac{dr}{dt} = -7\text{ inch per second}

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525 = \displaystyle\frac{1}{3}\pi r^2h = \frac{1}{3}\pi (99)^2h\\\\\text{Instant heigth} = h = \frac{525\times 3}{\pi(99)^2}

Differentiating with respect to t,

\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)

Putting all the values, we get,

\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)\\\\-948 = \frac{1}{3}\pi\bigg(2(99)(-7)(\frac{525\times 3}{\pi(99)^2}) + (99)(99)\frac{dh}{dt}\bigg)\\\\\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi} = (99)^2\frac{dh}{dt}\\\\\frac{1}{(99)^2}\bigg(\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi}\bigg) = \frac{dh}{dt}\\\\\frac{dh}{dt} = -0.085131

Thus, the height of cone is decreasing at a rate of 0.085131 inch per second.

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Step-by-step explanation:

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