When two shapes are mathematically similar, that means that all their sides are in a proportion.
We know the sieds of EH and AD, and so we can find the proportion of the shapes using these two sides.
EH/AD = 6/9 = 2/3
The ratio of EFGH to ABCD is 2:3
Now we have to find BC.
Becuase they're in proportion, we can form this equation:
= 
Cross multiply:
24 = 2*BC
Divide both sides by 2:
BC = 12 cm
Good luck!
Answer:
2/7x +3=9
x=21
Step-by-step explanation:
Let the unknown be x.
Take 3 to the other side so: 2/7 multiplied by x= 6
Divide 6 by 2/7 to get the answer;
Therefore x=21
Sub back into equation to check answer.
Answer:
1 yard=94.44 cm
Step-by-step explanation:
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.
<h3>
Answer:</h3>
(x, y) = (7, -5)
<h3>
Step-by-step explanation:</h3>
It generally works well to follow directions.
The matrix of coefficients is ...
![\left[\begin{array}{cc}2&4\\-5&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%264%5C%5C-5%263%5Cend%7Barray%7D%5Cright%5D)
Its inverse is the transpose of the cofactor matrix, divided by the determinant. That is ...
![\dfrac{1}{26}\left[\begin{array}{ccc}3&-4\\5&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B26%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-4%5C%5C5%262%5Cend%7Barray%7D%5Cright%5D)
So the solution is the product of this and the vector of constants [-6, -50]. That product is ...
... x = (3·(-6) +(-4)(-50))/26 = 7
... y = (5·(-6) +2·(-50))/26 = -5
The solution using inverse matrices is ...
... (x, y) = (7, -5)
