Answer:
$945.08
Step-by-step explanation:
please give brainliest
Using transformations and congruency concepts, it is found that with these following transformations, the triangles will be congruent.
- A reflection, then a translation.
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A rotation, then a reflection.
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- Two triangles are congruent if they have the <u>same lengths</u> of the sides and the <u>same angles.</u>
- In a reflection, there is a rule that changes the <u>coordinates (x,y)</u>, but does <u>not </u>change <u>the lengths</u> of the sides of the triangles, thus they will still be congruent.
- A <u>reflection is also a special case of rotation</u>, thus, in a rotation, the triangles are also congruent.
- A translation is also similar to a reflection, using rules to shift the triangle up, down, left or right according to it's coordinates, not changing the sides or angles, thus congruent.
- In a dilation, the <u>lengths of the sides are changed</u>, thus, the triangles will not be congruent.
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Thus, from the bullet points above, the correct options are:
- A reflection, then a translation.
- A rotation, then a reflection.
A similar problem is given at brainly.com/question/24267298
Answer:
-65
Step-by-step explanation:
use Khan academy it's very helpful
Perhaps the easiest way to find the midpoint between two given points is to average their coordinates: add them up and divide by 2.
A) The midpoint C' of AB is
.. (A +B)/2 = ((0, 0) +(m, n))/2 = ((0 +m)/2, (0 +n)/2) = (m/2, n/2) = C'
The midpoint B' is
.. (A +C)/2 = ((0, 0) +(p, 0))/2 = (p/2, 0) = B'
The midpoint A' is
.. (B +C)/2 = ((m, n) +(p, 0))/2 = ((m+p)/2, n/2) = A'
B) The slope of the line between (x1, y1) and (x2, y2) is given by
.. slope = (y2 -y1)/(x2 -x1)
Using the values for A and A', we have
.. slope = (n/2 -0)/((m+p)/2 -0) = n/(m+p)
C) We know the line goes through A = (0, 0), so we can write the point-slope form of the equation for AA' as
.. y -0 = (n/(m+p))*(x -0)
.. y = n*x/(m+p)
D) To show the point lies on the line, we can substitute its coordinates for x and y and see if we get something that looks true.
.. (x, y) = ((m+p)/3, n/3)
Putting these into our equation, we have
.. n/3 = n*((m+p)/3)/(m+p)
The expression on the right has factors of (m+p) that cancel*, so we end up with
.. n/3 = n/3 . . . . . . . true for any n
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* The only constraint is that (m+p) ≠ 0. Since m and p are both in the first quadrant, their sum must be non-zero and this constraint is satisfied.
The purpose of the exercise is to show that all three medians of a triangle intersect in a single point.
Answer:
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