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Julli [10]
2 years ago
13

On a coordinate plane, 2 triangles are shown. triangle 1 has points at a (negative 3, 4), b (negative 2, 1), c (negative 4, 1).

triangle 2 has points at a prime (4, negative 2), b prime (3, negative 5), c prime (5, negative 5). which rule represents the translation from the pre-image, δabc, to the image, δa'b'c'? (x, y) → (x 7, y 6) (x, y) → (x 7, y – 6) (x, y) → (x – 6, y 7) (x, y) → (x 6, y 7)
Mathematics
1 answer:
Vikki [24]2 years ago
6 0

The rule which represents the translation from the pre-image (Δabc), to the image (Δa'b'c') is (x, y) → (y, x).

<h3>What is a triangle?</h3>

A triangle simply refers to a two-dimensional geometric shape that comprises three (3) sides, three (3) vertices and three (3) angles only.

Based on this triangle, we can deduce the following points:

  • The coordinates of a are given by (-3, 4).
  • The coordinates of a' are given by (4, -2).

By critically observing the triangle, we can also deduce that points were swapped as (x, y) → (y, x) and this is a reflection across line y = x.

Read more on triangles here: brainly.com/question/14527861

#SPJ4

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<h2>Answer:  A trapezoid with bases of 6 mm and 14 mm and a height of 8 mm </h2>

The parallelogram in the figure has an area of 80mm^{2}, according to the following formula, which works for all rectangles and parallelograms:

A_{parallelogram}=(b)(h)   (1)

Where b is the base and h is the height

The<u> area of a triangle</u> is given by the following formula:

A_{triangle}=\frac{1}{2}(b)(h)   (2)

So, for option A:

A_{triangle}=\frac{1}{2}(4mm)(20mm)=40mm^{2} \neq 80mm^{2}    

Now, the <u>area of a trapezoid </u>is:

A_{trapezoid}=\frac{1}{2}(b_{1}+ b_{2})(h)   (3)

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A_{trapezoid}=\frac{1}{2}(15mm+25mm)(2 mm)=40mm^{2} \neq 80mm^{2}    

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A_{trapezoid}=\frac{1}{2}(6mm+14mm)(8 mm)=80mm^{2}>>>>This is the correct option!

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