The slope in question seven is 1. Below are the solutions to the remaining questions
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In what means can Transformation occur ?</h3>
Transformation of an object can occur in different ways. Such as translation, rotation, enlargement of different scale factors, reflection and dilation.
Given that Blue points, blue line segments, red points, and red line segments arranged on a Cartesian coordinate plane. Here are the blue points and their coordinates. Point A: (negative nine, 3). Point B: (negative 11, 3). Point C: (negative 10, 3). Point D: (negative 10, 5). Point F: (negative 10, 4). Point E: (negative 11, 4). Here are the red points and their coordinates. Point A-1: (negative 1, 6). Point A-2: (negative 3, 4). Point B-1: (negative 2, 4). Point B-2: (negative 5, 4). Point C-2: (negative 3, 3). Point D-1: (negative 5, 5). Point D-2: (negative 5, 3). Point E-1: (negative 6, 6). Point F-1: (negative 5, 6). A semicircle that lies below its line of symmetry AB. A semicircle that lies above its line of symmetry B-2 A-2. A triangle DEF. A triangle D-1 E-1 F-1. Line segments are drawn from C to D, from A-1 to B-1, from A-2 to B-2, and from C-2 to D-2. Triangle DEF, segment CD, and the semicircle with line of symmetry BA are arranged so that they look like a boat.
1. The transformations to use on the blue segment CD to get it to match with the red segment C2D2 is ROTATION of 180° Clockwise
2.The transformations to use on the blue triangle to get it to match with the red triangle is ROTATION of 360° Clockwise Explain your movement using the coordinates of the vertices.
3.The line segments on the boat that is parallel is line B2 to A2 because it is parallel to line D2 to C2
4.The line segments on the boat is perpendicular is line BCA because they are perpendicular to line CFD and AB.
5.The line segments on the boat have a slope of 0 is line BCA because change in y axis is 0
6.The line segments on the boat have an undefined slope is line DFC because change in x axis is 0. When denominator is zero, the fraction is undefined.
7.Given that E is at (−11,4) and D is at (−10,5), the slope of ED will be
Slope = Δy ÷ Δx
Slope = (5 - 4) ÷ (-10 + 11)
Slope = 1 ÷ 1
Slope = 1
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, where m is the slope and
is a point lies on the line which represent the function ⇒ slope-point form