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
The answer is 36.
Explanation:
Let the unknown number be x
x^3=x*36
or, x^3/x=36
or, x^2=36
or, x=6
.•. x=36
Answer:
The answer should be 420.
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Step-by-step explanation:
Answer:
{- 2, - 4, - 6, - 8, - 10 }
Step-by-step explanation:
Given
f(x) = 2x - 6 with domain { - 2, - 1, 0, 1, 2 }
To obtain the range substitute the values of x from the domain into f(x)
f(- 2) = 2(- 2) - 6 = - 4 - 6 = - 10
f(- 1) = 2(- 1) - 6 = - 2 - 6 = - 8
f(0) = 2(0) - 6 = 0 - 6 = - 6
f(1) = 2(1) - 6 = 2 - 6 = - 4
f(2) = 2(2) - 6 = 4 - 6 = - 2
Range is { - 2, - 4, - 6, - 8, - 10 }
