Answer:
see below
Step-by-step explanation:
<h3>Proposition:</h3>
Let the diagonals AC and BD of the Parallelogram ABCD intercept at E. It is required to prove AE=CE and DE=BE
<h3>Proof:</h3>
1)The lines AD and BC are parallel and AC their transversal therefore,
![\displaystyle \angle DAC = \angle ACB \\ \ \qquad [\text{ alternate angles theorem}]](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20%5Cangle%20DAC%20%3D%20%20%5Cangle%20ACB%20%5C%5C%20%20%5C%20%5Cqquad%20%5B%5Ctext%7B%20alternate%20angles%20theorem%7D%5D)
2)The lines AB and DC are parallel and BD their transversal therefore,
![\displaystyle \angle BD C= \angle ABD \\ \ \qquad [\text{ alternate angles theorem}]](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20%5Cangle%20BD%20C%3D%20%20%5Cangle%20ABD%20%5C%5C%20%20%5C%20%5Cqquad%20%5B%5Ctext%7B%20alternate%20angles%20theorem%7D%5D)
3)now in triangle ∆AEB and ∆CED
therefore,

hence,
Proven
Answer:
False
Step-by-step explanation:
Parentheses should be evaluated first according to PEMDAS in every situation.
Parentheses
Exponents
Multiplication
Division
Addition
Subtraction
Answer:
(x, y) = (0, 1/2) or (1, 3)
Step-by-step explanation:
The first equation factors as ...
x(3x -y) = 0
This has solutions x=0 and y=3x.
__
<u>x = 0</u>
Using this in the second equation gives ...
2y -0 = 1
y = 1/2
(x, y) = (0, 1/2) is a solution
__
<u>y = 3x</u>
Using the expression for y in the second equation, we get ...
2(3x) -5x = 1
x = 1 . . . . . . . . . simplify
y = 3x = 3 . . . . using x=1 in the first equation
(x, y) = (1, 3) is a solution
_____
Interestingly, the (red line) graph of 3x^2 -xy = 0 produced by this graphing calculator has a "hole" at x=0, It says that point is (0, undefined). In a sense, y is undefined, in that it can be <em>anything</em>. A more appropriate graph would graph that equation as the two lines x=0 and y=3x.