Answer:
2c put me as brainlest
Step-by-step explanation:
Let's solve for c.
b=2c+6
Step 1: Flip the equation.
2c+6=b
Step 2: Add -6 to both sides.
2c+6+−6=b+−6
2c=b−6
Step 3: Divide both sides by 2.
2c
2
=
b−6
2
c=
1
2
b−3
Answer:
c=
1
2
b−3
now if ur just looking for the funtipn the asnwer would be 2c i think
If both triangles are similar the ratio of sides will be same
The equation for a trapezoid is A = a+b all over 2 times h
So, you would simply substitute.
A=15
B=12
H=8
15+12/2 (8)=
27/2 (8)=
13.5 (8)=
108.
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
_____
* 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.
