<h2>Answer:</h2>
please see analysis
<h2>Step-by-step explanation:</h2>F
<em>I hope this helps you:)</em>
Subtract 2m – 7q + 5 from 4p + 2q and add your result to 7m² - 3m + 9q – 1
2 answers:
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Answer:
m∠Q ≈ 53°
Step-by-step explanation:
To find the measure of ∠Q, the law of cosines will need to be used. Lowercase letters represent the side lengths, while upper case letters represent angles.
In this situation, 'A' will be ∠Q. Therefore:
17² = 18² + 20² -2(18)(20)cosQ
Simplify:
289 = 324 + 400 -2(360)cosQ
Continue simplifying down:
-435 = -720cosQ
Divide both sides by '-720':
0.604 = cosQ
m∠Q ≈ 52.83 or 53° rounded to the nearest whole degree.
Answer:
Step-by-step explanation:
The question is as following:
The verticies of a triangle on the coordinate plane are
A(0, 0), B(2, 0) and C(0, 2).
What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of 1/3 ?
=============================================
Given: the vertices of a triangle ABC are A(0, 0), B(2, 0) and C(0, 2).
IF the triangle is dilated by a factor of k about the origin, then
(x,y) → (kx , ky)
that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.
It is given that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.
If a figure dilated by a factor of 1/3 about the origin
So,
<u>So, The coordinates of the triangle A'B'C' are:</u>