How do you solve -28-6x=-2(1-3x)-2
2 answers:
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The transformations would map EFGH to is a reflection across line k followed by a translation down. Option (a) is correct.
Further explanation:
Given:
The compositions of transformations from EFGH to are as follows,
(a). A reflection across line k followed by a translation down.
(b). A translation down followed by a reflection across line k.
(c). A rotation about point G followed by a translation to the right.
(d). A translation to the right followed by a rotation about point G.
Explanation:
Translation can be defined as to move the function to a certain displacement. If the points of a line or any objects are moved in the same direction it is a translation.
Rotation is defined as a movement around its own axis. A circular movement is a rotation.
The transformations would map EFGH to is a reflection across line k followed by a translation down. Option (a) is correct.
Option (a) is correct.
Option (b) is not correct.
Option (c) is not correct.
Option (d) is not correct.
Learn more:
- Learn more about inverse of the functionhttps://brainly.com/question/1632445.
- Learn more about equation of circle brainly.com/question/1506955.
- Learn more about range and domain of the function brainly.com/question/3412497
Answer details:
Grade: Middle School
Subject: Mathematics
Chapter: Triangles
Keywords: rotation, transformation, map, EFGH, composition, translation, triangle, rotation about point A, mapped, triangle pair, mapping, equal angles, sides, congruent, two triangles, common point.
DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!
<h2>(a)</h2>Case 1: both balls are white.
At the beginning we have balls. We want to pick a white one, so we have a probability of
of picking a white one.
If this happens, we're left with white balls and still
black balls, for a total of
balls. So, now, the probability of picking a white ball is
The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability
Case 2: both balls are black
The exact same logic leads to a probability of
These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is
Case 1: both balls are white.
In this case, nothing changes between the two picks. So, you have a probability of of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability
of picking a black ball with both picks.
This leads to an overall probability of
Of picking two balls of the same colour.
<h2>(c)</h2>We want to prove that
Expading all squares and products, this translates to
As you can see, this inequality comes in the form
With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:
And this is our case, because in our case we have
so, y has an extra piece and it is larger
which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2