Answer:
2/11
Step-by-step explanation:
6 = 2 x 3
6/11 x 1/3
= ( 6 x 1 )/( 11 x 3 )
= ( 2 x 3 x 1 )/( 11 x 3 )
Cancel 3 in both numerator and denominator.
= ( 2 x 1 )/( 11 )
= 2/11
Answer:
This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by
2.1
gives the next term. In other words,
a
n
=
a
1
⋅
r
n
−
1
.
Geometric Sequence:
r
=
2.1
This is the form of a geometric sequence.
a
n
=
a
1
r
n
−
1
Substitute in the values of
a
1
=
5
and
r
=
2.1
.
a
n
=
(
5
)
⋅
(
2.1
)
n
−
1
Remove parentheses around
2.1
.
a
n
=
5
⋅
2.1
n
−
1
Step-by-step explanation:
Answer:
d = Nc + x
Step-by-step explanation:
Given
N = 
Multiply both sides by c to clear the fraction
Nc = d - x ( add x to both sides )
Nc + x = d
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.
