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.
Step-by-step explanation:
Using log properties
![log_{7}(rz {}^{12} )](https://tex.z-dn.net/?f=%20log_%7B7%7D%28rz%20%7B%7D%5E%7B12%7D%20%29%20)
![log_{7}(r) + log_{7}(z {}^{12} )](https://tex.z-dn.net/?f=%20log_%7B7%7D%28r%29%20%20%2B%20%20log_%7B7%7D%28z%20%7B%7D%5E%7B12%7D%20%29%20)
Then
![log_{7}(r) + 12 log_{7}(z)](https://tex.z-dn.net/?f=%20log_%7B7%7D%28r%29%20%20%2B%2012%20log_%7B7%7D%28z%29%20)
Plug in the knowns
![- 7.87 + 12( - 12.59)](https://tex.z-dn.net/?f=%20-%207.87%20%2B%2012%28%20-%2012.59%29)
![- 158.95](https://tex.z-dn.net/?f=%20-%20158.95)
A) is counting down by 3
B) is counting up by 5
A is the answer. Reflection and rotation.
Answer:
x=1, y=7
Step-by-step explanation:
y = -3x + 10 - first equation
y=-3x + 4 - second equation
rearrange the expression to
-3x-y= -10
3x-y= -4
pick an equation and simplify; lets pick the second equation
3x-y= -4
divide 3 by both sides
x=
- third equation
substitute the value of x into an equation; lets pick the first equation
-3x-y= -10
-3
- y = -10
simplify. -3 cancels 3 so we are left with
-1(-4+y)-y = -10
simplify
4-y-y= -10
4-2y= -10
subtract 4 from both sides
-2y= -10-4
-2y= -14
divide -2 by both sides
y=7
substitute the value of y, (y=7) in an equation, we are using the second equation
3x-y= -4
3x-7= -4
3x = -4+7
3x= 3
divide 3 by both sides
x=1
so the answer is x=1, y=7